Plz help will give brainliest

write the ordered pair for each point. PLS! I will give 20 points did correct same goes with brainliest. DO NOT TYPE THAT YOU DO

patriot [66]

Answer:

M=(0.75,1.25)

H=(-1.5,-0.5)

D=(1,-1.75)

Step-by-step explanation:

Hope this helps.

4 0

3 years ago

Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, 33% of all homeowners are ins

Lina20 [59]

Answer:

(a) The probability mass function of X is:

$P(X=x)={4\choose x}\ (0.33)^{x}\ (1-0.33)^{4-x};\ x=0,1,2,3...$

(b) The most likely value for X is 1.32.

(c) The probability that at least two of the four selected have earthquake insurance is 0.4015.

Step-by-step explanation:

The random variable X is defined as the number among the four homeowners who have earthquake insurance.

The probability that a homeowner has earthquake insurance is, p = 0.33.

The random sample of homeowners selected is, n = 4.

The event of a homeowner having an earthquake insurance is independent of the other three homeowners.

(a)

All the statements above clearly indicate that the random variable X follows a Binomial distribution with parameters n = 4 and p = 0.33.

The probability mass function of X is:

$P(X=x)={4\choose x}\ (0.33)^{x}\ (1-0.33)^{4-x};\ x=0,1,2,3...$

(b)

The most likely value of a random variable is the expected value.

The expected value of a Binomial random variable is:

$E(X)=np$

Compute the expected value of X as follows:

$E(X)=np$

$=4\times 0.33\\=1.32$

Thus, the most likely value for X is 1.32.

(c)

Compute the probability that at least two of the four selected have earthquake insurance as follows:

P (X ≥ 2) = 1 - P (X < 2)

= 1 - P (X = 0) - P (X = 1)

$=1-{4\choose 0}\ (0.33)^{0}\ (1-0.33)^{4-0}-{4\choose 1}\ (0.33)^{1}\ (1-0.33)^{4-1}\\\\=1-0.20151121-0.39700716\\\\=0.40148163\\\\\approx 0.4015$

Thus, the probability that at least two of the four selected have earthquake insurance is 0.4015.

3 0

3 years ago

Amy is 11 ⅚ years old. Camden is 1 ⅓ years younger than amy and Jane is 1 ¼ years younger than Camden How old is Jane

erastova [34]

Answer:

Jane is 9 1/4 years old

Step-by-step explanation:

Camden is 10 1/2 years old and Jane is (10 2/4 - 1 1/4) which equals 9 1/4 years

5 0

3 years ago

Solve 3(x - 2) < 18.

Musya8 [376]

This answer is 290 because when you skip the math you would find the answer

5 0

3 years ago

The amounts a soft drink machine is designed to dispense for each drink are normally distributed, with a mean of 12.1 fluid ounc

Ludmilka [50]

Answer:

$\mu = 12.1\\\sigma = 0.3$

a) Find the probability that the drink is less than 11.9 fluid ounces

We are supposed to find P(x<11.9)

x= 11.9

We will use z score formula :

$z=\frac{x-\mu}{\sigma}$

$z=\frac{11.9-12.1}{0.3}$

$z=-0.67$

Refer the z table

So, P(z<-0.67) = 0.2514

Hence the probability that the drink is less than 11.9 fluid ounces is 0.2514

b)Find the probability that the drink is between 11.6 and 11.9 fluid ounces

x= 11.6

We will use z score formula :

$z=\frac{x-\mu}{\sigma}$

$z=\frac{11.6-12.1}{0.3}$

$z=-1.67$

x= 11.9

We will use z score formula :

$z=\frac{x-\mu}{\sigma}$

$z=\frac{11.9-12.1}{0.3}$

$z=-0.67$

We are supposed to find P(11.6<x<11.9)

So, P(11.6<x<11.9)= P(x<11.9)-P(x<11.6)

P(-1.67<z<-0.67)= P(z<-0.67)-P(z<-1.67)

Refer the z table

P(-1.67<z<-0.67)= P(z<-0.67)-P(z<-1.67)

P(-1.67<z<-0.67)= 0.2514-0.0475

P(-1.67<z<-0.67)= 0.2039

So, the probability that the drink is between 11.6 and 11.9 fluid ounces is 0.2039.

c)Find the probability that the drink is more than 12.6 fluid ounces

We are supposed to find P(x>12.6)

x= 12.6

We will use z score formula :

$z=\frac{x-\mu}{\sigma}$

$z=\frac{12.6-12.1}{0.3}$

$z=1.67$

Refer the z table

P(x>12.6)=1-P(x<12.6)

P(z>1.67)=1-P(z<1.67)

P(z>1.67)=1-0.9525

P(z>1.67)=0.0475

Hence the probability that the drink is more than 12.6 fluid ounces is 0.0475

d) Is the drink containing more than 12.6 fluid ounces an unusual event?

Since p value is less than alpha (0.05)

Answer = No, because the probability that a drink containing more than 12.6 fluid ounces is less than 0.05, this event is not unusual.

3 0

3 years ago