Which of the following equations are the correct answer? Please help due tomorrow

Answers

Answer 1

Answer: (3,2), (-1,-4)

Slope = (y2-y1)/(x2-x1)= (2+4)/(3+1) =3/2
(y-y1) = 3/2(x-x1)
(y-2)=3/2(x-3)
(y+4)=3/2(x+1)
Correct answer: A,E

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HELP PLEASE

NEED HELP ASAP

Answers

It should be A. Line DA
Since each box is one, then if you count line DA then you get 10

g let X be a normally distributed random variable with mean 3 and variance 4. a) Let Y = 5X+2. What is the distribution of Y? What are its mean and variance? b) Find P(Y<10). Find P(X<10). c) What is the 99th percentile of the distribution of Y? d) What is the 99th percentile of the distribution of X? e) What is the distribution of W = exp(Y)? What are its mean and variance?

Answers

a. Let $F_X(x)$ be the CDF of $X$ . The CDF of $Y$ is

$F_Y(y)=P(Y\le y)=P(5X+2\le y)=P\left(X\le\frac{y-2}5\right)=F_X\left(\frac{y-2}5\right)$

which is to say, $Y$ is also normally distributed, but with different parameters. In particular,

$E[Y]=E[5X+2]=5E[X]+2=17$

$\mathrm{Var}[Y]=\mathrm{Var}[5X+2]=5^2\mathrm{Var}[X]=100$

b. Using the appropriate CDFs, we have

$P(Y<10)=F_Y(10)=F_X\left(\frac{10-2}5\right)=F_X(1.6)\approx0.242$

$P(X<10)=F_X(10)\approx0.9998$

c. The 99th percentile for any distribution $D$ is the value of $d_(0.99)$ such that $P(D\le d_(0.99))=0.99$ , i.e. all values of $d$ below $d_(0.99)$ make up the lower 99% of the distribution.

We have

$P(Y\le y_(0.99))=0.99\implies y_(0.99)\approx40.26$

d. On the other hand, the 99th percentile for $X$ is

$P(X\le x_(0.99))=0.99\implies x_(0.99)\approx7.653$

e. We have

$F_W(w)=P(W\le w)=P\left(e^Y\le w\right)=P(Y\le\ln w)=F_Y(\ln w)$

which suggests that $\ln W$ is normally distributed, or $W$ is log-normally distributed. Recall that the moment-generating function for $Y$ is

$M_Y(t)=\exp\left(17t+\frac{100t^2}2\right)$

But we also have

$M_Y(t)=E[e^(tY)]=E[e^(t\ln W)]=E[W^t]$

Then

$E[W]=M_Y(1)=e^(67)$

and

$E[W^2]=M_Y(2)=e^(234)\implies\mathrm{Var}[W]=E[W^2]-E[W]^2=e^(234)-e^(134)$

6x
−
5
y
=
14

Find
x
when
y
=
2

Answers

Answer:

Step-by-step explanation:

HELP!How do you derive the equation of a circle?
How do you identify the center and radius of a circle?
How do you define the radian measure of an angle?
How are arc length and area of a sector related to proportionality?

Answers

Answer:

How do you derive the equation of a circle?

you can reverse the circle formula

How do you identify the center and radius of a circle?

You can identify the center because it is the middle of the circle and the radius can be draw from the center to the round part.

How do you define the radian measure of an angle?

it is the ratio of the length of the arc the angle forms÷the radius of the circle

How are arc length and area of a sector related to proportionality?

sector x radian

area = (r^2 x)/2

arc length = r x

they are not propotional

because Area / arc length = r/2 , (it's not a constant)

High Hopes

Barrii

How do I write 4.18 as a mixed number

Answers

Answer:

4 18/100

Step-by-step explanation

A certain test preparation course is designed to help students improve their scores on the MCAT exam. A mock exam is given at the beginning and end of the course to determine the effectiveness of the course. The following measurements are the net change in 7 students' scores on the exam after completing the course: 37,12,12,17,13,32,23 Using these data, construct a 80% confidence interval for the average net change in a student's score after completing the course. Assume the population is approximately normal. Step 1 of 4 : Calculate the sample mean for the given sample data. Round your answer to one decimal place.

Answers

Answer:

The 80% confidence interval for the average net change in a student's score after completing the course is (15.4, 26.3).

Step-by-step explanation:

The net change in 7 students' scores on the exam after completing the course are:

S = {37 ,12 ,12 ,17 ,13 ,32 ,23}

Compute the sample mean and sample standard deviation as follows:

$\bar x=(1)/(n)\sum x=(1)/(7)* 146=20.857\n\ns=\sqrt{(1)/(n-1)\sum (x-\bar x)^(2)}}=\sqrt{(1)/(7)* 622.8571}=10.189$

As the population standard deviation is not known, a t-interval will be formed.

Compute the critical value of t for 80% confidence interval and 6 degrees of freedom as follows:

$t_(\alpha/2, (n-1))=t_(0.20/2, (7-1))=t_(0.10,6)=1.415$

*Use a t-table.

Compute the 80% confidence interval for the average net change in a student's score after completing the course as follows:

$CI=\bar x\pm t_(\alpha/2, (n-1))*(s)/(√(n))$

$=20.857\pm 1.415*(10.189)/(√(7))\n\n =20.857\pm 5.4493\n\n=(15.4077, 26.3063)\n\n\approx (15.4,26.3)$

Thus, the 80% confidence interval for the average net change in a student's score after completing the course is (15.4, 26.3).

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