Question 5 of 10
Select the correct answer.
The graph shows a line of best fit for data collected on the average temperature, in degrees Fahrenheit, during a month and the
number of inches of rainfall during that month.
Average Temp.
90
80
70
60
50
404
30
20
10
1 1 2 3 4 5 6
X
Inches of Rain
The equation for the line of best fit is y = -3.32x + 97.05.
Based on the line of best fit, what would be the prediction for the average temperature during a month with 13.25 inches of rainfall?

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:

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]

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]

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]

......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

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:

im struggling with the same one

Answer:

a=1, b=9, c=-2, d=4

Step-by-step explanation:

Answer:

10 Ghuups I believe. I am sorry if this is wrong

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%!

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.

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

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).

Hi there!  

»»————-　★　————-««

I believe your answer is:  

[tex]w = 5, -5[/tex]

»»————-　★　————-««  

Here’s why:  

⸻⸻⸻⸻

Therefore:

[tex]w =\pm5[/tex]

⸻⸻⸻⸻

»»————-　★　————-««  

Hope this helps you. I apologize if it’s incorrect.  

Answer:

You: Your welcome❤

Answer:

12

Step-by-step explanation:

PEMDAS is the order

P = parenthesis

E = exponent

M = *

D = division

A = +

S = -

so first 8*2 = 16

and then -8/2 = -4

and then -4 + 16

= 12

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.

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]

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)

Answer:

- Surveyors

- developers of the GPS system

Step-by-step explanation:

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

Answer:

a) 615

b) 715

c) 344

Step-by-step explanation:

According to the Question,

Z = (x - mean)/standard deviation

Now,

For x = 4171,  Z = (4171 - 3311)/860 = 1  

Next, multiply that by the sample size of 732.

Z = (1581 - 3311)/860 = -2    

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.

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]

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]

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.

Answer:

i hope you understand easily

mark me brainlist

Step-by-step explanation:

Answer:

just keep writing down outcome on a sheet of paper then count total

Step-by-step explanation:

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                                    

1*2^3+0*2^2+1*2^1+1*2^0

8+0+2+1

=11