Answer:
0.7031 = 70.31% probability of a bulb lasting for at most 528 hours.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
The standard deviation of the lifetime is 15 hours and the mean lifetime of a bulb is 520 hours.
This means that [tex]\sigma = 15, \mu = 520[/tex]
Find the probability of a bulb lasting for at most 528 hours.
This is the p-value of Z when X = 528. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{528 - 520}{15}[/tex]
[tex]Z = 0.533[/tex]
[tex]Z = 0.533[/tex] has a p-value of 0.7031
0.7031 = 70.31% probability of a bulb lasting for at most 528 hours.
Combine as indicated by the signs. Write answer In descending powers of x.
X+6/x^28x+15+3x/x+5-x-3/x+3
= ?
Answer:
Step-by-step explanation:
Convert 1101, to base 10.
1*2^3+0*2^2+1*2^1+1*2^0
8+0+2+1
=11
Triangle A'B'C' is formed using the translation (x + 2 y + 0) and the dilation by a scale factor of 1/2 from the origin. Which equation explains the relationship between
AB and A"B"?
Answer:
[tex]A"B" = \frac{AB}{2}[/tex]
Step-by-step explanation:
Given
[tex]k = \frac{1}{2}[/tex] --- scale factor
Required
Relationship between AB and A"B"
[tex]k = \frac{1}{2}[/tex] implies that the sides of A"B"C" are smaller than ABC
i.e.
[tex]A"B" = k * AB[/tex]
[tex]A"B" = \frac{1}{2} * AB[/tex]
This gives:
[tex]A"B" = \frac{AB}{2}[/tex]
the shorter side of a rectangle is 60% of the longer side and the perimeter of the rectangle is 96 inches. find the side lengths
Answer:
Length of the rectangle:
[tex]x = \frac{4800}{106} = \frac{2400}{53} [/tex]
Breadth of the rectangle:
[tex]60\%(x) = \frac{60}{100} \times \frac{4800}{106} \: \: \: \: \: \: \: \: \: \: \: \\ =60 \times \frac{48}{106} \\ = \frac{2880}{106} [/tex]
Step-by-step explanation:
Longer side of the rectangle(length) = x
Shorter side of the rectangle(breadth) = (60%)x
Perimeter of the rectangle = 2(l+b) = 96 inches
Hence,
[tex]96 = 2(x + 60\%(x))[/tex]
[tex]96 = 2(x + \frac{6}{100 } x)[/tex]
[tex]96 = 2( \frac{100}{1 00} x + \frac{6}{100} x)[/tex]
[tex]96 = 2( \frac{106}{100} x)[/tex]
[tex]96 = \frac{106}{50} x[/tex]
[tex]96 \div \frac{106}{50} = x[/tex]
[tex]96 \times \frac{50}{106} = x[/tex]
[tex] \frac{4800}{106} = x[/tex]
every student from different schools planted as many plants of their number if they planted 4225 plants how many students were there
Answer:
65 students.
Step-by-step explanation:
Given that :
Every student planted as many plant as their number ;
Then let the number of student = x
Then the number of plant planted by each student will also = x
The total number of plants planted by all the students = 4225
The Number of students can be obtained thus ;
Total number of plants = Number of plants * number of plants per student
4225 = x * x
4225 = x²
√4225 = x
65 = x
Hence, there are 65 students
100 POINTS!!
Directions: Respond to this question to demonstrate your understanding of the topic/content. Be sure to provide adequate and relevant details learned in the module to support your response. Pay close attention to organizing your response so it makes sense and uses correct grammar. Your response should be at least 5-7 sentences at a minimum.
Question: Your friend claims that the only solution to the equation sin(x)=1 is x=90 degrees. Is your friend correct? If there are more solutions, explain how to determine additional solutions.
......hope it helps......
Answer:
yes,.to obtain sinx as 1 the angle must be 90degrees
so the answer is correct
but there are more solutions like when the cosine angle is 45 the answer is 1
and when x is 450 still sinx = 1..that is to say sin450= 1
This graph shows the solution to which inequality?
(32)
(-3.-6);
A ys 1/x - 2
B. y> fx-2
C. yzfx-2
***-2
The figure shows trapezoid ABCD on a coordinate plane.
Which of the following represents the area of this figure, rounded to the nearest square unit?
99
121
198
231
Answer:
121 unit^2.
Step-by-step explanation:
The area = height/2 * ( sum of the opposite parallel lines)
= h/2(BC + AD
h = BF = 14 - 3 = 11 units.
BC = 13 - 5 = 8 units.
AD = 16 - 2 = 14 units.
Area = (11/2)(8 + 14)
= 5.5 * 22
= 121 unit^2.
Answer:
121
Step-by-step explanation:
a. Consider the situation where you have three game chips, each labeled with one of the the numbers 3, 5, and 10 in a hat a. If you draw out 2 chips without replacement between each chip draw, list the entire sample space of po ssible results that can occur in the draw Use the three events are defined as follows, to answer parts b through n below:
Event A: the sum of the 2 drawn numbers is even.
Event B: the sum of the 2 drawn numbers is odd.
Event C: the sum of the 2 drawn numbers is a prime number
Now, using your answer to part a find the following probability values
b. P (A)=
c. P (B)=
d. P (C)=
e. P (A and C)-=
f. P(A or B)=
g. P (B andC)=
h. P(A or C)- =
i. P (C given B)=
j. P(C given A)=
k. P (not B)=
l. P (not C)=
Are events A and B mutually exclusive?Why or why not?
Are events B and C mutually exclusive? Why or why not?
Answer:
a) {3,5}{3,10}{5,10}
b) [tex]P(A)=\frac{1}{3}[/tex]
c) [tex]P(B)=\frac{2}{3}[/tex]
d) [tex]P(C)=\frac{1}{3}[/tex]
e) [tex]P(A and C)=0[/tex]
f) [tex]P(A or B)=1[/tex]
g) [tex]P(B and C)=\frac{1}{3}[/tex]
h) [tex]P(A or C)=\frac{2}{3}[/tex]
i) [tex]P(C given B)=\frac{1}{2}[/tex]
j) [tex]P(C given A)=0[/tex]
k) [tex]P(not B)=\frac{1}{3}[/tex]
l) [tex]P(not C)=\frac{2}{3}[/tex]
Yes, events A and B are mutually exclusive. Because the results can either be even or odd, not both. No, events B and C are not mutually exclusive because the result can be both, odd and prime.
Step-by-step explanation:
a)
In order to solve part a of the problem, we need to find the possible outcomes, in this case, the possible outcomes are:
{3,5}{3,10} and {5,10}
We could think of the oppsite order, for example {5,3}{10,3}{10,5} but these are basically the same as the previous outcomes, so we will just take three outcomes in our sample space. We can think of it as drawing the two chips at the same time.
b)
Now the probability of the sum of the chips to be even. There is only one outcome where the sum of the chips is even, {3,5} since 3+5=8 the other outcomes will give us an odd number, so:
[tex]P=\frac{#desired}{#possible}[/tex]
[tex]P(A)=\frac{1}{3}[/tex]
c) For the probability of the sum of the chips to be odd, there are two outcomes where the sum of the chips is odd, {3,10} since 3+10=13 and {5,10} since 5+10=15 the other outcomes will give us an even number, so:
[tex]P(B)=\frac{2}{3}[/tex]
d) The probability of the sum of the chips is prime. There is only one outcome where the sum of the chips is prime, {3,10} since 3+10=13 the other outcomes will give us non prime results, so:
[tex]P(C)=\frac{1}{3}[/tex]
e) The probability of the sum of the chips to be even and prime. There are no results where we can get an even and prime number, since the only even and prime number there is is number 2 and no outcome will give us that number, so:
P(A and C)=0
f) The probability of the sum of the chips is even or odd. We can either get even or odd results, so no matter what outcome we get, we will get an odd or even result so:
[tex]P(A or B)=1[/tex]
g) The probability of the sum of the chips is odd and prime. There is only one outcome where the sum of the chips is odd and prime, {3,10} since 3+10=13 the other outcomes will give us non prime results, so:
[tex]P(B and C)=\frac{1}{3}[/tex]
h) The probability of the sum of the chips is even or prime. There are two outcomes where the sum of the chips is even or prime, {3,10} since 3+10=13 and {3,5} since 3+5=8 so:
[tex]P(A or C)=\frac{2}{3}[/tex]
i) The probability of the sum of the chips is prime given that the sum of the chips is odd. There are two possible results where the sum of the chips is odd {3,10} and {5,10} and only one of those results is even, {3,10}, so
[tex]P(C given B)=\frac{1}{2}[/tex]
j) The probability of the sum of the chips is prime given that the sum of the chips is even. There is only one possible even result: {3,5} but that result isn't prime, so
[tex]P(C given A)=0[/tex]
k) The probability of the sum of the chips is not odd. There is only one outcome where the sum of the chips is not odd (even), {3,5} so:
[tex]P(not B)=\frac{1}{3}[/tex]
l) The probability of the sum of the chips is not prime. There are two outcomes where the sum of the chips is not prime, {3,5} and {5,10} so:
[tex]P(not C)=\frac{2}{3}[/tex]
Are events A and B mutually exclusive?
Yes, events A and B are mutually exclusive.
Why or why not?
Because the results can either be even or odd, not both.
Are events B and C mutually exclusive?
No, events B and C are not mutually exclusive.
Why or Why not?
Because the result can be both, odd and prime.
Leroy borrowed $1500 at an annual simple interest rate of 12%. He paid $270 in interest. For what time period did Leroy borrow the money?
Answer:
i hope you understand easily
mark me brainlist
Step-by-step explanation:
Use variation of parameters to find a general solution to the differential equation given that the functions y1 and y2 are linearly independent solutions to the corresponding homogeneous equation for t > 0.
ty'' + (2t - 1)y' - 2y = 7t2 e-2t y1 = 2t - 1, y2 = e-2t
Recall that variation of parameters is used to solve second-order ODEs of the form
y''(t) + p(t) y'(t) + q(t) y(t) = f(t)
so the first thing you need to do is divide both sides of your equation by t :
y'' + (2t - 1)/t y' - 2/t y = 7t
You're looking for a solution of the form
[tex]y=y_1u_1+y_2u_2[/tex]
where
[tex]u_1(t)=\displaystyle-\int\frac{y_2(t)f(t)}{W(y_1,y_2)}\,\mathrm dt[/tex]
[tex]u_2(t)=\displaystyle\int\frac{y_1(t)f(t)}{W(y_1,y_2)}\,\mathrm dt[/tex]
and W denotes the Wronskian determinant.
Compute the Wronskian:
[tex]W(y_1,y_2) = W\left(2t-1,e^{-2t}\right) = \begin{vmatrix}2t-1&e^{-2t}\\2&-2e^{-2t}\end{vmatrix} = -4te^{-2t}[/tex]
Then
[tex]u_1=\displaystyle-\int\frac{7te^{-2t}}{-4te^{-2t}}\,\mathrm dt=\frac74\int\mathrm dt = \frac74t[/tex]
[tex]u_2=\displaystyle\int\frac{7t(2t-1)}{-4te^{-2t}}\,\mathrm dt=-\frac74\int(2t-1)e^{2t}\,\mathrm dt=-\frac74(t-1)e^{2t}[/tex]
The general solution to the ODE is
[tex]y = C_1(2t-1) + C_2e^{-2t} + \dfrac74t(2t-1) - \dfrac74(t-1)e^{2t}e^{-2t}[/tex]
which simplifies somewhat to
[tex]\boxed{y = C_1(2t-1) + C_2e^{-2t} + \dfrac74(2t^2-2t+1)}[/tex]
Which of these professionals most directly uses geometry? Choose the best answer
Answer:
- Surveyors
- developers of the GPS system
Step-by-step explanation:
Suppose 49% of the doctors in America are dentists. If a random sample of size 689 is selected, what is the probability that the proportion of doctors who are dentists will be less than 47%
Answer:
[tex]P(<47\%)=0.1468[/tex]
Step-by-step explanation:
From the question we are told that:
Percentage of Dentist Doctors P(D)=49\%
Sample size n=689
Generally the equation for probability that the proportion of doctors who are dentists will be less than [tex]P(<47\%)[/tex] is mathematically given by
[tex]P(<47\%)=Z>(\frac{\=x-P(D)}{\sqrt{\frac{P(D)*1-P(D)}{n}}})[/tex]
[tex]P(<47\%)=Z>(\frac{0.47-0.49}{\sqrt{\frac{0.49*0.51}{689}}})[/tex]
[tex]P(<47\%)=Z>(1.05)[/tex]
Therefore from table
[tex]P(<47\%)=0.1468[/tex]
Trapezoid A B C D is shown. A diagonal is drawn from point B to point D. Sides B C and A D are parallel. Sides B A and C D are congruent. Angle C B D is 24 degrees and angle B A D is 116 degrees.
What is the measure of angle ABD in trapezoid ABCD?
24°
40°
64°
92°
Answer:
40 degrees un edge
Step-by-step explanation:
Answer:
The person above me got this correct, so the answer to this is 40! I just did the Unit Test and got a 100%!
3 coins are flipped.
Answer:
just keep writing down outcome on a sheet of paper then count total
Step-by-step explanation:
Determine whether the given functions are linearly dependent or linearly independent on the specified interval. Justify your decision.
The question is incomplete. The complete question is :
Determine whether the given functions are linearly dependent or linearly independent on the specified interval. Justify your decision.
[tex]$\{ x^5, x^5-1,3 \} \text{ on } (- \infty, \infty)$[/tex]
Solution :
Given :
Function : [tex]$\{ x^5, x^5-1,3 \} \text{ on } (- \infty, \infty)$[/tex]
We have to determine whether the given function is linear dependent or linearly independent for the interval [tex]$(-\infty, \infty)$[/tex].
The given function are linearly dependent because for the constants, [tex]c_1[/tex] and [tex]c_2[/tex], the equation is :
[tex]$c_1x^5 + c_23 = x^5-1$[/tex] has the solution [tex]$c_1 = 1$[/tex] and [tex]$c_2 = -\frac{1}{3}$[/tex]
Therefore,
[tex]$1x^5 + \left(-\frac{1}{3}\right)3 = x^5-1$[/tex]
Type the correct answer in each box. Use numerals instead of words.
Multiply the expressions.
If a = 1, find the values of b, c, and d that make the given expression equivalent to the expression below.
Answer:
a=1, b=9, c=-2, d=4
Step-by-step explanation:
Assume the population is bell-shaped. Between what two values will approximately 95% of the population be
Answer:The 95% Rule states that approximately 95% of observations fall within two ... about 95% will be within two standard deviations of the mean, and about 99.7% will be ... Suppose the pulse rates of 200 college men are bell-shaped with a mean of 72 ... 1.2 - Samples & Populations ... 3.5 - Relations between Multiple Variables.
Step-by-step explanation:
At a snack food manufacturing facility, the quality control engineer must ensure that all products feature the appropriate expiration date. Suppose that a box of 60 candy bars includes 12 which do not have the proper printed expiration date. The quality control engineer, in inspecting the box, grabs a handful of seven candy bars. What is the probability that there are exactly 3 faulty candy bars among the seven
Answer:
0.1108 = 11.08% probability that there are exactly 3 faulty candy bars among the seven.
Step-by-step explanation:
The bars are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.
Hypergeometric distribution:
The probability of x successes is given by the following formula:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]
In which:
x is the number of successes.
N is the size of the population.
n is the size of the sample.
k is the total number of desired outcomes.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
In this question:
60 total candies means that [tex]N = 70[/tex]
12 are faulty, which means that [tex]k = 12[/tex]
Seven are chosen, so [tex]n = 7[/tex]
What is the probability that there are exactly 3 faulty candy bars among the seven?
This is [tex]P(X = 3)[/tex]. So
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 3) = h(3,70,7,12) = \frac{C_{12,3}*C_{48,4}}{C_{60,7}} = 0.1108[/tex]
0.1108 = 11.08% probability that there are exactly 3 faulty candy bars among the seven.
integrate G(x,y,z)=yz over the surface of x+y+z=1 in the first octant.
Parameterize the surface (I'll call it S) by
r(u, v) = (1 - u) (1 - v) i + u (1 - v) j + v k
with 0 ≤ u ≤ 1 and 0 ≤ v ≤ 1.
Take the normal vector to this surface to be
n = ∂r/∂u × ∂r/∂v = ((v - 1) i + (1 - v) j) × ((u - 1) i - u j + k) = (1 - v) (i + j + k)
with magnitude
||n|| = √3 (1 - v)
Then in the integral, we have
[tex]\displaystyle\iint_SG(x,y,z)\,\mathrm ds = \int_0^1\int_0^1 G((1-u)(1-v),u(1-v),v) \|\mathbf n\| \,\mathrm du\,\mathrm dv \\\\= \sqrt3 \int_0^1\int_0^1uv(1-v)^2\,\mathrm du\,\mathrm dv \\\\= \boxed{\frac1{8\sqrt3}}[/tex]
Alternatively, if you're not familiar with parameterizing surfaces, you can use the "projection" formula:
[tex]\displaystyle\iint_S G(x,y,z)\,\mathrm ds = \int_{S_{xy}}G(x,y,z)\sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}\,\mathrm dx\,\mathrm dy[/tex]
where I write [tex]S_{xy}[/tex] to mean the projection of the surface onto the (x, y)-plane, and z = f(x, y). We would then use
x + y + z = 1 ==> z = f(x, y) = 1 - x - y
and [tex]S_{xy}[/tex] is the triangle,
{(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 - x}
Then the integral becomes
[tex]\displaystyle\int_0^1\int_0^{1-x}y(1-x-y)\sqrt{1+(-1)^2+(-1)^2}\,\mathrm dy\,\mathrm dx \\\\= \sqrt3\int_0^1\int_0^{1-x} y(1-x-y)\,\mathrm dy\,\mathrm dx \\\\= \frac{\sqrt3}{24} \\\\= \boxed{\frac1{8\sqrt3}}[/tex]
If asphalt pavement costs $0.70 per square foot, find the cost to pave the circular How much does it cost to pave this road?
road in the figure shown
nents
(Round to the nearest dollar as nooded)
Please help :)
Answer:
Cost to pave the road = $4257
Step-by-step explanation:
Area of the pavement = Area of the outer circle - Area of the internal circle
Area of the outer circle = πr²
= π(55)²
= 3025π square feet
Area of the inner circle = π(33)²
= 1089π square feet
Area of the pavement = 3025π - 1089π
= 1936π
= 6082.12 square feet
Cost of pavement = $0.70 per square feet
Therefore, cost of 6082.12 square feet = 6082.12 × 0.70
= 4257.49
≈ $4257
Cost to pave the road = $4257
What is the true solution to the equation below?
l n e Superscript l n x Baseline + l n e Superscript l n x squared Baseline = 2 l n 8
x = 2
Given:
The equation is:
[tex]\ln e^{\ln x}+\ln e^{\ln x^2}=2\ln 8[/tex]
To find:
The solution for the given equation.
Solution:
We have,
[tex]\ln e^{\ln x}+\ln e^{\ln x^2}=2\ln 8[/tex]
It can be written as:
[tex]\ln x+\ln x^2=2\ln 8[/tex] [tex][\because \ln e^x=x][/tex]
[tex]\ln (x\cdot x^2)=2\ln 8[/tex] [tex][\because \ln a+\ln b=\ln (ab)][/tex]
[tex]\ln (x^3)=\ln 8^2[/tex] [tex][\because \ln x^n=n\ln x ][/tex]
On comparing both sides, we get
[tex]x^3=8^2[/tex]
[tex]x^3=64[/tex]
Taking cube root, we get
[tex]x=\sqrt[3]{64}[/tex]
[tex]x=4[/tex]
Therefore, the required solution is [tex]x=4[/tex].
Answer:
x=4
Step-by-step explanation:
What is the true solution to the equation below?
ln e Superscript ln x Baseline + ln e Superscript ln x squared Baseline = 2 ln 8
x = 2
x = 4
x = 8
Suppose a research company takes a random sample of 45 business travelers in the financial industry and determines that the sample average cost of a domestic trip is $1,192, with a sample standard deviation of $279. Construct a 98% confidence interval for the population mean (for domestic trip) from these sample data. Round your answers to 3 decimal places.
Answer:
98% confidence interval for the population mean =(1095.260,1288.740)
Step-by-step explanation:
We are given that
n=45
[tex]\mu=1192[/tex]
Standard deviation,[tex]\sigma=279[/tex]
We have to construct a 98% confidence interval for the population mean.
Critical value of z at 98% confidence, Z =2.326
Confidence interval is given by
[tex](\mu\pm Z\frac{\sigma}{\sqrt{n}})[/tex]
Using the formula
98% confidence interval is given by
[tex]=(1192\pm 2.326\times \frac{279}{\sqrt{45}})[/tex]
[tex]=(1192\pm 96.740)[/tex]
=[tex](1192-96.740,1192+96.740)[/tex]
=[tex](1095.260,1288.740)[/tex]
Hence, 98% confidence interval for the population mean (1095.260,1288.740)
Newborn babies: A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 732 babies born in New York. The mean weight was 3311 grams with a standard deviation of 860 grams. Assume that birth weight data are approximately bell-shaped. Part 1 of 3 (a) Estimate the number of newborns whose weight was less than 4171 grams. Approximately of the 732 newborns weighed less than 4171 grams. Part 2 of 3 (b) Estimate the number of newborns whose weight was greater than 1591 grams. Approximately of the 732 newborns weighed more than 1591 grams. Part 3 of 3 (c) Estimate the number of newborns whose weight was between 3311 and 5031 grams. Approximately of the 732 newborns weighed between 3311 and 5031 grams.
Answer:
a) 615
b) 715
c) 344
Step-by-step explanation:
According to the Question,
Given that, A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 732 babies born in New York. The mean weight was 3311 grams with a standard deviation of 860 gramsSince the distribution is approximately bell-shaped, we can use the normal distribution and calculate the Z scores for each scenario.Z = (x - mean)/standard deviation
Now,
For x = 4171, Z = (4171 - 3311)/860 = 1
P(Z < 1) using Z table for areas for the standard normal distribution, you will get 0.8413.Next, multiply that by the sample size of 732.
Therefore 732(0.8413) = 615.8316, so approximately 615 will weigh less than 4171For part b, use the same method except x is now 1591.
Z = (1581 - 3311)/860 = -2
P(Z > -2) , using the Z table is 1 - 0.0228 = 0.9772 . Now 732(0.9772) = 715.3104, so approximately 715 will weigh more than 1591.For part c, we now need to get two Z scores, one for 3311 and another for 5031.
Z1 = (3311 - 3311)/860 = 0
Z2 = (5031 - 3311)/860= 2
P(0 ≤ Z ≤ 2) = 0.9772 - 0.5000 = 0.4772
approximately 47% fall between 0 and 1 standard deviation, so take 0.47 times 732 ⇒ 732×0.47 = 344.
PLSHELPASAPDFFFFFFFFFFFFFFFFFFFFFFFFFF
im struggling with the same one
What are all the values of w such that|-W | = 5?
Hi there!
»»————- ★ ————-««
I believe your answer is:
[tex]w = 5, -5[/tex]
»»————- ★ ————-««
Here’s why:
⸻⸻⸻⸻
Absolute value is simply how far a digit is from zero.The digits '-5' and '5' are 5 away from zero.Therefore:
[tex]w =\pm5[/tex]
⸻⸻⸻⸻
»»————- ★ ————-««
Hope this helps you. I apologize if it’s incorrect.
Three more than twice a number is 35.
Answer:
x = 16, or if you didn't want the value for x,
2x + 3 = 35
Step-by-step explanation:
Three more: +3
Twice a number: 2x
Combined:
2x + 3 = 35.
Get rid of the 3 by subtracting it from both sides:
2x = 32
Get rid of the 2 by dividing it from both sides:
x = 16
Answer:
The number is 16.
Step-by-step explanation:
Let the unknown number be x.
Now we translate the sentence into an equation piece by piece.
Three more than twice a number is 35.
2x
Three more than twice a number is 35.
2x + 3
Three more than twice a number is 35.
2x + 3 = 35
Now we solve the equation.
Subtract 3 from both sides.
2x = 32
Divide both sides by 2.
x = 16
Answer: The number is 16.
P.S. Notice that x was a variable that was introduced solely to solve the problem. The original problem is a word problem, not an equation, and has no x in it. The correct answer makes no reference to x since x was used to solve the equation but is not part of the given problem. The person asking the question has no idea what x is. He just wants a number as an answer.
On planet Ghaap, two Gheeps are worth three Ghiips, two Ghiips are worth five Ghoops, and three Ghoops are worth two Ghuups. How many Ghuups are seven Gheeps worth?
Answer:
10 Ghuups I believe. I am sorry if this is wrong
El arquitecto Gómez, dirige el proyecto de remodelación del parque municipal del distrito La Esperanza. La forma del parque está representada por la ecuación polar r(5-3sensθ)=16. El arquitecto planea construir un camino que une los extremos de la parte más ancha del terreno y necesita saber la distancia que existe entre los extremos (considerar que las medidas están en cientos de metros), además en el centro del camino colocará una pileta. Por ello, se requiere obtener las coordenadas de los extremos y del centro en coordenadas rectangulares. Para ayudar al arquitecto Gómez a lograr su objetivo, se deberá seguir la siguiente estrategia:
Pasar la ecuación polar a rectangular (en su forma ordinaria) (2 Puntos)
Hallar el centro, los vértices de la parte más ancha del terreno en la forma rectangular y determinar la distancia entre los vértices (considerar que las medidas están en cientos de metros), utilizando la ecuación cartesiana, hallada en a). (2 Puntos)
Graficar la cónica en el plano cartesiano ubicando las coordenadas de los vértices y del centro. (1 Punto)
Answer:
thank you for the point too mucheeeYou: Your welcome❤
Solve by using matrices. 2x – y +2 + w = -3 x + 2y – 3z + w = 12 3x - y - + 2w = 3 -2x + 3y + 2 – 3w = -3
Some symbols and numbers are missing. I assume the system is supposed to read
2x - y + 2z + w = -3
x + 2y - 3z + w = 12
3x - y - z + 2w = 3
-2x + 3y + 2z - 3w = -3
In matrix form, this is
[tex]\begin{bmatrix}2&-1&2&1\\1&2&-3&1\\3&-1&-1&2\\-2&3&2&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\\w\end{bmatrix}=\begin{bmatrix}-3\\12\\3\-3\end{bmatrix}[/tex]
which we can strip down to the augmented matrix,
[tex]\left[\begin{array}{cccc|c}2&-1&2&1&-3\\1&2&-3&1&12\\3&-1&-1&2&3\\-2&3&2&-3&-3\end{array}\right][/tex]
Now for the row operations:
• swap rows 1 and 2
[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\2&-1&2&1&-3\\3&-1&-1&2&3\\-2&3&2&-3&-3\end{array}\right][/tex]
• add -2 (row 1) to row 2, -3 (row 1) to row 3, and 2 (row 1) to row 4
[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&-7&8&-1&-33\\0&7&-4&-1&21\end{array}\right][/tex]
• add 7 (row 2) to -5 (row 3), and row 3 to row 4
[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&16&-2&-24\\0&0&4&-2&-12\end{array}\right][/tex]
• multiply through rows 3 and 4 by 1/2
[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&8&-1&-12\\0&0&2&-1&-6\end{array}\right][/tex]
• add -4 (row 4) to row 3
[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&0&3&12\\0&0&2&-1&-6\end{array}\right][/tex]
• swap rows 3 and 4
[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&2&-1&-6\\0&0&0&3&12\end{array}\right][/tex]
• multiply through row 4 by 1/3
[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&2&-1&-6\\0&0&0&1&4\end{array}\right][/tex]
• add row 4 to row 3
[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&2&0&-2\\0&0&0&1&4\end{array}\right][/tex]
• multiply through row 3 by 1/2
[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&1&0&-1\\0&0&0&1&4\end{array}\right][/tex]
• add -8 (row 3) and row 4 to row 2
[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&0&0&-15\\0&0&1&0&-1\\0&0&0&1&4\end{array}\right][/tex]
• multiply through row 2 by -1/5
[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&1&0&0&3\\0&0&1&0&-1\\0&0&0&1&4\end{array}\right][/tex]
• add -2 (row 2) and 3 (row 3) and -1 (row 4) to row 1
[tex]\left[\begin{array}{cccc|c}1&0&0&0&-1\\0&1&0&0&3\\0&0&1&0&-1\\0&0&0&1&4\end{array}\right][/tex]
Then the solution to the system is (x, y, z, w) = (-1, 3, -1, 4).