Answers
Answer:
[tex]\approx 15.9[/tex]
Step-by-step explanation:
The length of an arc with measure [tex]\theta[/tex] and radius [tex]r[/tex] is given by [tex]\ell_{arc}=2r\pi\cdot \frac{\theta}{360}[/tex]. From the figure, we know that the radius of arc ADC is 4, but we don't know the measure of the arc. Since there are 360 degrees in a circle, the measure of arc ADC is equal to the measure of the arc formed by [tex]\angle AOC[/tex] subtracted from 360. The measure of the arc formed by [tex]\angle AOC[/tex] consists of two congruent angles, [tex]\angle AOB[/tex] and [tex]\angle COB[/tex]. To find them, we can use basic trigonometry for a right triangle, since by definition, tangents intersect a circle at a right angle.
In any right triangle, the cosine of an angle is equal to its adjacent side divided by the hypotenuse, or longest side, of the triangle.
We have:
[tex]\cos \angle AOB=\cos \angle COB=\frac{4}{10},\\\angle AOB=\arccos(\frac{4}{10})=66.42182152^{\circ}[/tex]
Therefore, [tex]\angle AOC=2\cdot 66.42182152=132.84364304^{\circ}[/tex]
The measure of the central angle of [tex]\widehat{ADC}[/tex] must then be [tex]360-132.84364304=227.15635696^{\circ}[/tex]
Thus, the length of [tex]\widehat{ADC}[/tex] is equal to:
[tex]\ell_{\widehat{ADC}}=2\cdot 4\cdot \pi \cdot \frac{227.15635696}{360},\\\ell_{\widehat{ADC}}=15.8585053832\approx \boxed{15.9}[/tex] (three significant figures as requested by question).
Related Questions
Angle x is conterminal with angle y. If the measure of angle x is greater than the measure of angle y, which statement is true regarding the values of x and y?
x = y - 180n, for any positive integer n
x = y - 360n, for any integer n
x = y + 360n, for any positive integer n
x = y + 180n, for any integer n
Answers
9514 1404 393
Answer:
x = y + 360n, for any positive integer n
Step-by-step explanation:
Since x is greater than y, something must be added to y to get x. Angles have the same co-terminal ray at multiples of 360°. Then the amount added to y must be some multiple of 360°:
x = y + 360n . . . . . for positive integer n
find the area of the following figures:
Answers
Answer:
Area = 156 square cm
Step-by-step explanation:
Would these be similar?
Answers
Hey buddy I am here to help!
Yes these r similar!
Hope this helps!
Plz mark me brainliest!
Geometric Probability
Find the probability that a point chosen randomly inside the larger rectangle is in each given smaller shape. Round to the nearest percent. PLEASE HELP!
1) The circle
2) The smaller rectangle
3) Not the circle or smaller rectangle
Answers
Answer:
Step-by-step explanation:
1) P= Area of Circle/ Area of large rectangle
Area of the circle = pi·r² = pi·2²=4 pi ft.²
Area of large rectangle= l·w -12·10 =120 ft.²
P = 4pi/120 rewrite 120 as 4·30
P= 4 pi/4*30 = pi/30 = 3.14/40 ≈ .1047 ≈10% (because .1047·100 =10.47≅10)
2) P = Area of smaller rectangle/ Area of large rectangle
Area of smaller rectangle = l·w = 2·4 =8 ft.²
Area of large rectangle=l·w = 12·10=120 ft²
P= 8/120 ≅ .0666≅ 7% (because .0666·100 =6.66≅7)
3) P= Not the circle or smaller rectangle/ Area of large rectangle
Not the circle or smaller rectangle area
= Area of large rectangle - Area of circle -Area of smaller rectangle
= 120 -4·pi -8 = 120 - (4· 3.14) -8 = 99.4362939 ft²
Area of large rectangle = l·w = 12·10 =120 ft²
P = 99.4362939 /120 ≅ .8286 ≅83% (because .8286·100 =82.86≅83)
Solve this for me guys
Answers
Answer:
x = 7.348
Step-by-step explanation:
I don't really know how to explain this over a computer, but it is correct. Answer credit of Omnicalculator.
Hope that this helps!
Dr. Lum teaches part-time at two community colleges, Hilltop College and Serra College. Dr. Lum can teach up to 5 classes per semester. For every class he teaches at Hilltop College, he needs to spend 3 hours per week preparing lessons and grading papers. For each class at Serra College, he must do 4 hours of work per week. He has determined that he cannot spend more than 18 hours per week preparing lessons and grading papers. If he earns $6,000 per class at Hilltop College and $7,500 per class at Serra College, how many classes should he teach at each college to maximize his income, and what will be his income?
To maximize his income, Dr. Lum should teach_______classes for Hilltop College and __________classes for Serra College. His maximum income would be________.
Answers
Answer:
z (max) = 34500 $
x₁ = 2
x₂ = 3
Step-by-step explanation:
Hilltop College
3 hours per week preparing lessons and grading papers
Serra College
4 hours per week preparing lessons and grading papers
Total hours to spend per week preparing lessons 18
Let´s call x₁ numbers of class at Hilltop College
and x₂ numbers of class at Serra College then:
Objective function
z = 6000*x₁ + 7500*x₂
Constraints:
1.- x₁ + x₂ ≤ 5 the total number of class
2.- 3*x₁ + 4*x₂ ≤ 18
3. General constraints x₁ ≥ 0 x₂ ≥ 0 integers
After 6 iteration optimal solution is: From on-line solver
z (max) = 34500 $
x₁ = 2
x₂ = 3
In a study, researchers wanted to measure the effect of alcohol on the hippocampal region, the portion of the brain responsible for long-term memory storage, in adolescents. The researchers randomly selected 23 adolescents with alcohol use disorders to determine whether the hippocampal volumes in the alcoholic adolescents were less than the normal volume of 9.02 cm3. An analysis of the sample data revealed that the hippocampal volume is approximately normal with no outliers and x=8.16 cm3 and s=0.7 cm3. Conduct the appropriate test at the α=0.01 level of significance.
Answers
Answer:
We do not reject the Null Hypothesis
Step-by-step explanation:
From the question we are told that:
Sample size [tex]n=23[/tex]
Population mean [tex]\mu=9.02cm^3[/tex]
Sample mean [tex]\=x=8.16[/tex]
Standard deviation [tex]\sigma=0.7cm^3[/tex]
Significance level [tex]\alpha =0.01[/tex]
Generally the Null and and alternative Hypothesis are as follows
[tex]H_0:\mu=9.02cm^3[/tex]
[tex]H_a:\mu<9.02cm^3[/tex]
Therefore t critical Value is
[tex]t\ critical\ Value=(\alpha,df)[/tex]
[tex]t\ critical\ Value=(0.01,22)[/tex]
Where
[tex]df=n-1\\\\df=23-1=>22[/tex]
Therefore
From t Table
[tex]t value=-2.8[/tex]
Generally the equation for Z Critical is mathematically given by
[tex]t=\frac{\=x-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]
[tex]t=\frac{8.16-9.02}{\frac{0.7}{\sqrt{23} } }[/tex]
[tex]t=-5.89[/tex]
Therefore
Since the t test statistics is greater than the Critical value
Hence,we do not reject the Null Hypothesis
Dannette and Alphonso work for a computer repair company. They must include the time it takes to complete each repair in their repair log book. The dot plots show the number of hours each of their last 12 repairs took. Part a. Calculate the median, mean, IQR, and standard deviation of each data set. Part b. Which measure of central tendency and spread should you use to compare the two data sets? Explain your reasoning. Part c. Determine whether there are any outliers in either data set. Dannette's Repair Times х х X X X X Х Х + 9 + 1 0 Relations 2 3 4 8 10 12 5 6 7 Repair Time (hours) Geometry Alphonso's Repair Times Groups X Trigonometry X Х X X X х X х Statistics 7 X + 3 10 9 0 4 12 Series 8 1 2 5 7 Repair Time (hours) Greek
PLZ HELP
Answers
Answer:
(a):
Dannette Alphonso
[tex]\bar x_D = 4.33[/tex] [tex]\bar x_A = 5.17[/tex]
[tex]M_D = 2.5[/tex] [tex]M_A = 5[/tex]
[tex]\sigma_D = 3.350[/tex] [tex]\sigma_A = 1.951[/tex]
[tex]IQR_D = 7[/tex] [tex]IQR_A = 1.5[/tex]
(b):
Measure of center: Median
Measure of spread: Interquartile range
(c):
There are no outliers in Dannette's dataset
There are outliers in Alphonso's dataset
Step-by-step explanation:
Given
See attachment for the appropriate data presentation
Solving (a): Mean, Median, Standard deviation and IQR of each
From the attached plots, we have:
IQR_A = 1.5 ---- Dannette
[tex]A = \{3,4,4,4,4,5,5,5,5,6,6,11\}[/tex] ---- Alphonso
n = 12 --- number of dataset
Mean
The mean is calculated
[tex]\bar x = \frac{\sum x}{n}[/tex]
So, we have:
[tex]\bar x_D = \frac{1+1+1+1+2+2+3+7+8+8+9+9}{12}[/tex]
[tex]\bar x_D = \frac{52}{12}[/tex]
[tex]\bar x_D = 4.33[/tex] --- Dannette
[tex]\bar x_A = \frac{3+4+4+4+4+5+5+5+5+6+6+11}{12}[/tex]
[tex]\bar x_A = \frac{62}{12}[/tex]
[tex]\bar x_A = 5.17[/tex] --- Alphonso
Median
The median is calculated as:
[tex]M = \frac{n + 1}{2}th[/tex]
[tex]M = \frac{12 + 1}{2}th[/tex]
[tex]M = \frac{13}{2}th[/tex]
[tex]M = 6.5th[/tex]
This implies that the median is the mean of the 6th and the 7th item.
So, we have:
[tex]M_D = \frac{2+3}{2}[/tex]
[tex]M_D = \frac{5}{2}[/tex]
[tex]M_D = 2.5[/tex] ---- Dannette
[tex]M_A = \frac{5+5}{2}[/tex]
[tex]M_A = \frac{10}{2}[/tex]
[tex]M_A = 5[/tex] ---- Alphonso
Standard Deviation
This is calculated as:
[tex]\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n}}[/tex]
So, we have:
[tex]\sigma_D = \sqrt{\frac{(1 - 4.33)^2 +.............+(9- 4.33)^2}{12}}[/tex]
[tex]\sigma_D = \sqrt{\frac{134.6668}{12}}[/tex]
[tex]\sigma_D = 3.350[/tex] ---- Dannette
[tex]\sigma_A = \sqrt{\frac{(3-5.17)^2+............+(11-5.17)^2}{12}}[/tex]
[tex]\sigma_A = \sqrt{\frac{45.6668}{12}}[/tex]
[tex]\sigma_A = 1.951[/tex] --- Alphonso
The Interquartile Range (IQR)
This is calculated as:
[tex]IQR =Q_3 - Q_1[/tex]
Where
[tex]Q_3 \to[/tex] Upper Quartile and [tex]Q_1 \to[/tex] Lower Quartile
[tex]Q_3[/tex] is calculated as:
[tex]Q_3 = \frac{3}{4}*({n + 1})th[/tex]
[tex]Q_3 = \frac{3}{4}*(12 + 1})th[/tex]
[tex]Q_3 = \frac{3}{4}*13th[/tex]
[tex]Q_3 = 9.75th[/tex]
This means that [tex]Q_3[/tex] is the mean of the 9th and 7th item. So, we have:
[tex]Q_3 = \frac{1}{2} * (8+8) = \frac{1}{2} * 16[/tex] [tex]Q_3 = \frac{1}{2} * (5+6) = \frac{1}{2} * 11[/tex]
[tex]Q_3 = 8[/tex] ---- Dannette [tex]Q_3 = 5.5[/tex] --- Alphonso
[tex]Q_1[/tex] is calculated as:
[tex]Q_1 = \frac{1}{4}*({n + 1})th[/tex]
[tex]Q_1 = \frac{1}{4}*({12 + 1})th[/tex]
[tex]Q_1 = \frac{1}{4}*13th[/tex]
[tex]Q_1 = 3.25th[/tex]
This means that [tex]Q_1[/tex] is the mean of the 3rd and 4th item. So, we have:
[tex]Q_1 = \frac{1}{2}(1+1) = \frac{1}{2} * 2[/tex] [tex]Q_1 = \frac{1}{2}(4+4) = \frac{1}{2} * 8[/tex]
[tex]Q_1 = 1[/tex] --- Dannette [tex]Q_1 = 4[/tex] ---- Alphonso
So, the IQR is:
[tex]IQR = Q_3 - Q_1[/tex]
[tex]IQR_D = 8 - 1[/tex] [tex]IQR_A = 5.5 - 4[/tex]
[tex]IQR_D = 7[/tex] --- Dannette [tex]IQR_A = 1.5[/tex] --- Alphonso
Solving (b): The measures to compare
Measure of center
By observation, we can see that there are outliers is the plot of Alphonso (because 11 is far from the other dataset) while there are no outliers in Dannette plot (as all data are close).
Since, the above is the case; we simply compare the median of both because it is not affected by outliers
Measure of spread
Compare the interquartile range of both, as it is arguably the best measure of spread, because it is also not affected by outliers.
Solving (c): Check for outlier
To check for outlier, we make use of the following formulas:
[tex]Lower =Q_1 - 1.5 * IQR[/tex]
[tex]Upper =Q_3 + 1.5 * IQR[/tex]
For Dannette:
[tex]Lower = 1 - 1.5 * 7 = -9.5[/tex]
[tex]Upper = 8 + 1.5 * 7 = 18.5[/tex]
Since, the dataset are all positive, we change the lower outlier to 0.
So, the valid data range are:
[tex]Valid = 0 \to 18.5[/tex]
From the question, the range of Dannette's dataset is: 1 to 9. Hence, there are no outliers in Dannette's dataset
For Alphonso:
[tex]Lower = 4 - 1.5 * 1.5 =1.75[/tex]
[tex]Upper = 5.5 + 1.5 * 1.5 =7.75[/tex]
So, the valid data range are:
[tex]Valid = 1.75\to 7.75[/tex]
From the question, the range of Alphonso's dataset is: 3 to 11. Hence, there are outliers in Alphonso's dataset
17. A loan of $8000 was paid back in 2
years in monthly payments of $400.
The interest on the loan as a
percentage, was
A. 5%
B. 8-%
C. 162 %
16
D. 20%
Answers
Answer:
D. 20%
Step-by-step explanation:
2 years = 24 months
400 × 24 = 9600
9600 - 8000 = 1600
1600/8000 = 1/5
1/5 = 20%
What does it mean to the rise over run when the slope is an integer? a. the rise number is one c. the run part of the slope is going to be one b. the run number is always negative d. there will be no slope Please select the best answer from the choices provided A B C D
Answers
Answer: integer is a whole number
Eg
10/2 = 5/1 = 5 = integer
But 12/8 = 3/2 = 1.5 = not an integer
So really, slope = integer
Means rise greater than run
And run is a factor of rise
Or run = 1 will satisfy the above too :)
Answer:
C. The run part of the slope is going to be one
Step-by-step explanation:
.
In the following distribution, P(X<2) = 0.35, and expected value is 1.9
X 0 1 2 3 4
P(X) 0.10 A 0.35 B C
Required:
a. Use the fact that P(X< 2) = 0.35 to find the value of A.
b. Determine the value of B.
c. Determine the value of C.
Answers
Solution :
We have :
X 0 1 2 3 4
P(X) 0.10 A 0.35 B C
a). P(X < 2) = 0.35
P(X < 2) = P(X = 0) + P(X = 1) = 0.35
⇒ 0.10 + A = 0.35
⇒ A = 0.25
So the value of A is 0.25
b). The total probability = 1
So ,
0.10 + A + 0.35 + B + C = 1
0.10 + 0.25 + 0.35 + B + C = 1
B + C = 1 - 0.70
B + C = 0.30 ......(i)
We have the expected value = 1.9
So, [tex]$\sum X P(X) = 19$[/tex] for x = 0, 1, 2, 3, 4
⇒ (0 x 0.10) + (1 x 0.25) + (2 x 0.35) + (3 x B) + (4 x C) = 1.9
⇒ 0 + 0.25 + 0.70 + 3B + 4C = 1.9
⇒ 3B + 4C = 1.9 - 0.95
⇒ 3B + 4C = 0.95 ...................(ii)
From (i), we take the value of B = 0.30 - C and substitute it in the equation (i), we get,
⇒ 3( 0.30 - C) + 4C = 0.95
⇒ 0.90 - 3C + 4C = 0.95
⇒ C = 0.95 - 0.90
= 0.05
Now substituting the value of C = 0.05 in (ii), we get,
⇒ B = 0.05 = 0.30
⇒ B = 0.25
c). The value of C is 0.05
what are T-ratio ? explain
answer my question
plz
Answers
Answer:
The t-ratio is the estimate divided by the standard error. With a large enough sample, t-ratios greater than 1.96 (in absolute value) suggest that your coefficient is statistically significantly different from 0 at the 95% confidence level. A threshold of 1.645 is used for 90% confidence.
Step-by-step explanation:
Hope it help you.
Last month Rudy’s Tacos sold 22 dinner specials. The next month they released a new commercial and sold 250% of last month’s dinners. How many dinner specials did they sell this month?
Answers
Answer:
the answer is 2
Step-by-step explanation: because 250 -22 is i dont even know
Answer:
55
Step-by-step explanation:
Use the graph of the function to find its domain and range. Write the domain and range in interval notation.
Answers
Answer:
Domain = -6 < x < 3
range = -6 < x < -4
Step-by-step explanation:
The domain is the input values along the x-axis. According to the graph, the x values are within the interval;
Domain= -6 < x < 3
The range is the output values along the y-axis. According to the graph, the y values are within the interval;
range = -6 < x < -4
hi how are you? will you kindly help me with this only one?
Answers
Answer:
The answer is J.
Step-by-step explanation:
A bicycle shop owner offers five styles of mountain bikes for $450, $275, $675, $490, and $300. He wants to increase the mean price but keep the median price and range of prices the same. Suggest a new set of prices for the five styles
Answers
Answer:
275, 350, 450, 550, 675
Step-by-step explanation:
Arrange in order
275, 300, 450, 490, 675
range 275 to 675
median 450
mean 438
---------------------------
Raise 300 to 350
Raise 490 to 550
New set of prices
275, 350, 450, 550, 675
range 275 to 675 same
median 450 same
mean 460 increased
Answer:
One set of prices could be: {275,300,450,600,675}
Another set could be: {275,300,450,600,675}
There are many other solutions possible.
====================================================
Explanation:
A = {450, 275, 675, 490, 300}
B = {275, 300, 450, 490, 675}
Set A is the original set of values in the order they were given to you. Set B is the sorted version of set A from smallest to largest.
The mean is found by adding up the values and dividing by 5 (because there are five items in the set).
The mean is (275+300+450+490+675)/5 = 2190/5 = 438. The shopkeeper wants to increase the mean to something larger, but keep the median and range the same.
The median is the middle most number. In set B, we can see that is 450. So the median is 450. We want to keep the median the same at 450.
The range is the difference in min and max
range = max - min = 675-275 = 400
We want to keep the range at 400
---------------------------
There are a number of ways to increase the mean, while keeping the median and range the same.
Let's say we keep the min and max the same. In order to increase the mean, we need to increase the 490 (second largest value) to something larger. Let's bump that up to 600 for instance.
Recomputing the mean gets us
(275+300+450+600+675)/5 = 2300/5 = 460
The old mean was 438 and the new mean is now 460. The mean has increased. This is due to the larger price pulling on the mean to get the mean to increase.
The median is still 450 because it's still in the direct middle of set C
C = {275,300,450,600,675}
The range is still the same as well because we haven't changed the min and max.
---------------------------
So one possible set could be
C = {275,300,450,600,675}
We could also have
D = {275,400,450,500,675}
The difference is that the 300 bumped to 400, and the 600 dropped to 500. You should find that the median and range are the same, while the mean is 460.
There are many possible solutions here.
If the exponential model f(x)=3(2)x is written with the base e, it will take the form A0ekx. What is A0 and what is k?
Answers
Answer:
Answer in pic
Step-by-step explanation:
The required value of A₀ = 3 and k = ln 2, as of the given exponential model.
What is an exponential function?The function which is in format f(x) =a^x where a is constant and x is variable, the domain of this exponential function lies (-∞, ∞).
here,
Since.
[tex]a^x = e^{xlna}[/tex]
According to the question,
[tex]3(2)^x = A_oe^{kx}\\3e^{xln2} = A_oe^{kx}[/tex]
Now
Compare the values,
A₀ = 3 and k = ln 2,
Thus, According to the stated exponential model, A0 = 3 and k = ln 2.
Learn more about exponential function here:
brainly.com/question/15352175
#SPJ2
Solve the inequality c+12<16
Answers
Answer:
c+12<16
c=16-12
c=4
answer is this
Find the equation (in terms of x ) of the line through the points (-2,5) and (3,4)
y=
Answers
Hi there!
»»————- ★ ————-««
I believe your answer is:
[tex]y=-\frac{1}{5}x+\frac{23}{5}[/tex]
»»————- ★ ————-««
Here’s why:
We first need to find the slope of the equation.⸻⸻⸻⸻
[tex]\boxed{\text{\underline{Slope Is...}}}\\\\\rightarrow\frac{y_2-y_1}{x_2-x_1}\\\\\boxed{\text{Key:}}\\\\\rightarrow (x_1,y_1)\text{ and }(x_2,y_2) - \text{Two Points Given}[/tex]
⸻⸻⸻⸻
[tex]\boxed{\text{Calculating the Slope...}}\\\\\rightarrow m=\frac{4-5}{3-(-2)}\\\\\rightarrow m=\frac{-1}{5}\\\\\rightarrow \boxed{m=-\frac{1}{5}}\\\\\\\text{The slope is } -\frac{1}{5}.[/tex]
⸻⸻⸻⸻
We then need to find the y-intercept of the equation.⸻⸻⸻⸻
[tex]\boxed{\text{Calculating the intercept...}}\\\\y=-\frac{1}{5}x+b\\-------------\\\rightarrow 4 = -\frac{1}{5}(3) +b\\\\\rightarrow4= -\frac{3}{5}+b\\\\\rightarrow4+\frac{3}{5}= -\frac{3}{5}+\frac{3}{5} +b\\\\\rightarrow \boxed{\frac{23}{5}=b}[/tex]
⸻⸻⸻⸻
[tex]\text{The equation should be: }\boxed{ y=-\frac{1}{5}x+\frac{23}{5} }.[/tex]
»»————- ★ ————-««
Hope this helps you. I apologize if it’s incorrect.
2hr 57min+3hrs42min
Answers
Answer:
6 hrs 33 minutes
Step-by-step explanation:
2hr 57min
+3hrs42min
----------------------
5 hrs 99 minutes
But 1 hr = 60 minutes so subtract 60 minutes and add 1 hour
6 hrs 33 minutes
Answer:
6hr 39 min
Step-by-step explanation:
Add both
2hr 57 min
+ 3hr 42 min
5hr 99 min
we know that 1 hr = 60 min
then , 99 min = 1hr 39 min
so, 5hr + 1hr 39min
= 6hr 39 min
A matinee ticket costs $6 per adult and $1 per child. On a certain day, the total number of adults (a) and children (c) who saw a movie was 35, and the total money collected was $70. Which of the following options represents the number of children and the number of adults who saw a movie that day, and the pair of equations that can be solved to find the numbers?
7 children and 28 adults
Equation 1: a + c = 35
Equation 2: 6a − c = 70
7 children and 28 adults
Equation 1: a + c = 35
Equation 2: 6a + c = 70
28 children and 7 adults
Equation 1: a + c = 35
Equation 2: 6a + c = 70
28 children and 7 adults
Equation 1: a + c = 35
Equation 2: 6a − c = 70
Answers
Answer:
28 children and 7 adults
Equation 1: a + c = 35
Equation 2: 6a + c = 70
Step-by-step explanation:
If the total number of people at the movie was 35 people, one of the equations will be a + c = 35.
If $70 was collected in total, the other equation will be 6a + c = 70.
Now, solve this system of equations:
a + c = 35
6a + c = 70
Solve by elimination by multiplying the top equation by -1, then adding the equations together:
-a - c = -35
6a + c = 70
Add these together, and solve for a:
5a = 35
a = 7
Since there were 35 people in total, find how many children attended by subtracting 7 from 35:
35 - 7
= 28
So, there were 28 children and 7 adults.
The equations used were: a + c = 35 and 6a + c = 70
So, the correct answer is:
28 children and 7 adults
Equation 1: a + c = 35
Equation 2: 6a + c = 70
What is tan 30? A b c d e f
Answers
Answer:
try all the square roots and wichever gets to 0.58(rounded) is your answer
Step-by-step explanation:
(x² +x-y)dy+ x dy=0
Differential equations
Answers
Answer:
where is dx
I cannot find dx
HELP PLEASE MATH PROBLEM
Answers
Answer:
x=41
Step-by-step explanation:
LM =JM
154=4x-10
154+10=4x
164=4x
164/4=4x/4
41=x
hope this is helpful
if $1995 .00 is Shared equally among 7 men, how much would each get?
Answers
Anwer:$285
Explaination: Division method
$1995.00÷7=$285
A school principal wants to know more about the number of students absent each day. He counts the number of students absent each day for one week: {24, 18, 31,
Answers
Answer:
6.27
Step-by-step explanation:
We are to obtain the standard deviation of the given values :
{24, 18, 31,25, 34}
The standard deviation = √(Σ(x - mean)²/ n)
The mean = (ΣX) /n
Using calculator to save computation time :
The standard deviation, s = 6.27 (2 decimal places)
what of the following functions is graphed below
being timed help quickly will mark brainliest !!!
Answers
There are some Rs2 and Rs5 coins in a box. The ratio of the number of Rs2 coins to the number of Rs5 coins is 1:3. The value of all the Rs5 coins is Rs45. What is the value of all the Rs2 coins in the box?
Answers
Given:
The ratio of the number of Rs2 coins to the number of Rs5 coins is 1:3.
The value of all the Rs5 coins is Rs45.
To find:
The value of all the Rs2 coins in the box.
Solution:
Let x be the number of Rs2 coins and y be the number of Rs5 coins.
The value of all the Rs5 coins is Rs45.
[tex]5y=45[/tex]
[tex]y=\dfrac{45}{5}[/tex]
[tex]y=9[/tex]
The ratio of the number of Rs2 coins to the number of Rs5 coins is 1:3.
[tex]\dfrac{x}{y}=\dfrac{1}{3}[/tex]
[tex]\dfrac{x}{9}=\dfrac{1}{3}[/tex]
Multiply both sides by 9.
[tex]\dfrac{x}{9}\times 9=\dfrac{1}{3}\times 9[/tex]
[tex]x=3[/tex]
The value of all the Rs2 coins in the box is:
[tex]\text{Required value}=2x[/tex]
[tex]\text{Required value}=2(3)[/tex]
[tex]\text{Required value}=6[/tex]
Therefore, the value of all the Rs. 2 coins in the box is Rs. 6.
55
3. Patrick paid $20 for 5 peaches. How much did he pay per peach? Show
your work!
Answers
Answer:
Step-by-step explanation:
LA EDUCACION ES IMPIRTANTE YA QUE PROMUEVE UN MEJOR DESARROLLO DE LOS NIÑOS,NIÑAS Y ADOLESCENTES QUE LOS HACE FOMENTAR UN VINCULO MUY ESPECIAL CON SUS MAESTROS,COMPAÑEROS QUE LOS HACE SENTIR QUE SON PARTE DE SU FAMILIA ADEMAS ELLOS PASAN LA MAYORIA DE TIEMPO EN LA ESCUELA QUE LOS HACE SENTIRSE MÁS CÓMODOS COMO SI FUERA SU PROPIO HOGAR
Answer:
4 peaches
Step-by-step explanation:
$20/5 = 4 peaches
Estimate 9272 - 28 by first rounding each number so that it has only 1 nonzero digit.
Answers
Answer:
8970
Step-by-step explanation:
In order to round 9272 so that it has only 1 nonzero digit, look at the hundred digit, If the number is greater or equal to 5, add 1 to the thousand figure. If this is not the case, add zero
The hundred digit is 2 which is less than 5, so 0 is added to 9. the number becomes 9000
In order to round 28 so that it has only 1 nonzero digit, look at the units digit, If the number is greater or equal to 5, add 1 to the tens figure. If this is not the case, add zero
The units digit is greater than 5, so 1 would be added to tens digit. the number becomes 30
9000 - 30 = 8970