Simone purchased a circular exercise pen with a radius of 2.5 feet to keep her new pet safe. Find the area inside the pen

The surface area is the amount a solid can hold. True False

sveticcg [70]

This statement is false.

What you described is the volume of a solid. The surface area is the outermost layer of something or the outside part of it.

7 0

2 years ago

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What is the solution to the system of equations 3x+2y=39 5x-y=13

Vadim26 [7]

Hiya. I'm going to rewrite the second equation.

By subtracting 5x by both sides, I'll be able to have y by itself:

-y=-5x+13

I'm going to then divide both sides by -1 to get:

y=5x-13.

Then, I'm going to plug that equation into the first equation

3x+2(5x-13)=39

Factor:

3x+10x-26=39

Combine like terms:

13x=65

Divide both sides by 13 to get x by itself to get x=5

Plug this back into the equation of y=5x-13

y=5(5)-13 to get y=12

4 0

2 years ago

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2. The Welcher Adult Intelligence Test Scale is composed of a number of subtests. On one subtest, the raw scores have a mean of

IgorC [24]

Answer:

a) 37.31 b) 42.70 c) 0.57 d) 0.09

Step-by-step explaanation:

We are regarding a normal distribution with a mean of 35 and a standard deviation of 6, i.e., $\mu$ = 35 and $\sigma$ = 6. We know that the probability density function for a normal distribution with a mean of $\mu$ and a standard deviation of $\sigma$ is given by

$f(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp[-\frac{(x-\mu)^{2}}{2\sigma^{2}}]$

in this case we have

$f(x) = \frac{1}{\sqrt{2\pi}6}\exp[-\frac{(x-35)^{2}}{2(6^{2})}]$

Let $X$ be the random variable that represents a row score, we find the values we are seeking in the following way

a) we are looking for a number $x_{0}$ such that

$P(X\leq x_{0})$ = $\int\limits^{x_{0}}_{-\infty} {f(x)} \, dx = 0.65$ , this number is $x_{0}$ =37.31

you can find this answer using the R statistical programming languange and the instruction qnorm(0.65, mean = 35, sd = 6)

b) we are looking for a number $x_{1}$ such that

$P(X\leq x_{1})$ = $\int\limits^{x_{1}}_{-\infty} {f(x)} \, dx = 0.9$ , this number is $x_{1}$ =42.70

you can find this answer using the R statistical programming languange and the instruction qnorm(0.9, mean = 35, sd = 6)

c) we find this probability as

$P(28\leq X\leq 38)$ = $\int\limits^{38}_{28} {f(x)} \, dx = 0.57$

you can find this answer using the R statistical programming languange and the instruction pnorm(38, mean = 35, sd = 6) -pnorm(28, mean = 35, sd = 6)

d) we find this probability as

$P(41\leq X\leq 44)$ = $\int\limits^{44}_{41} {f(x)} \, dx = 0.09$

you can find this answer using the R statistical programming languange and the instruction pnorm(44, mean = 35, sd = 6) -pnorm(41, mean = 35, sd = 6)

6 0

2 years ago

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Two different simple random samples are drawn from two different populations. The first sample consists of 30 people with 16 hav

Furkat [3]

Answer:

There is no significant evidence that p1 is different than p2 at 0.01 significance level.

99% confidence interval for p1-p2 is -0.171 ±0.237 that is (−0.408, 0.066)

Step-by-step explanation:

Let p1 be the proportion of the common attribute in population1

And p2 be the proportion of the same common attribute in population2

$H_{0}$ : p1-p2=0

$H_{a}$ : p1-p2≠0

Test statistic can be found using the equation:

$z=\frac{p2-p1}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}$ where

p1 is the sample proportion of the common attribute in population1 ( $\frac{16}{30} =0.533$ )

p2 is the sample proportion of the common attribute in population2 ( $\frac{1337}{1900} =0.704$ )

p is the pool proportion of p1 and p2 ( $\frac{16+1337}{30+1900}=0.701$ )

n1 is the sample size of the people from population1 (30)

n2 is the sample size of the people from population2 (1900)

Then $z=\frac{0.704-0.533}{\sqrt{{0.701*0.299*(\frac{1}{30} +\frac{1}{1900}) }}}$ ≈ 2.03

p-value of the test statistic is 0.042>0.01, therefore we fail to reject the null hypothesis. There is no significant evidence that p1 is different than p2.

99% confidence interval estimate for p1-p2 can be calculated using the equation

p1-p2± $z*\sqrt{\frac{p1*(1-p1)}{n1}+\frac{p2*(1-p2)}{n2}}$ where

z is the z-statistic for the 99% confidence (2.58)

Thus 99% confidence interval is

0.533-0.704± $2.58*\sqrt{\frac{0.533*0.467}{30}+\frac{0.704*0.296}{1900}}$ ≈ -0.171 ±0.237 that is (−0.408, 0.066)

7 0

2 years ago

So a man goes to sleep at 8 am and wakes up on the following day and wakes up at 1 am. can you tell me what happened?

Nata [24]

Answer:

He was on an airplane.

Step-by-step explanation:

Hope I helped!

Please mark Brainliest!!!

5 0

2 years ago