Demonstrate how you can use the slope formula to write the point-slope form of an equation of a line.

The commuting time for a student to travel from home to a college campus is normally distributed with a mean of 30 minutes and a

Rainbow [258]

Answer:

0.1587

Step-by-step explanation:

Let X be the commuting time for the student. We know that $X\sim N(30, (5)^2)$ . Then, the normal probability density function for the random variable X is given by

$f(x) = \frac{1}{\sqrt{2\pi(5)^{2}}}\exp[-\frac{(x-\mu)^{2}}{2(5)^{2}}]$ . We are seeking the probability P(X>35) because the student leaves home at 8:25 A.M., we want to know the probability that the student will arrive at the college campus later than 9 A.M. and between 8:25 A.M. and 9 A.M. there are 35 minutes of difference. So,

$P(X>35) = \int\limits_{35}^{\infty}f(x)dx$ = 0.1587

To find this probability you can use either a table from a book or a programming language. We have used the R statistical programming language an the instruction pnorm(35, mean = 30, sd = 5, lower.tail = F)

6 0

3 years ago

How can you solve this problem <br> 2250=5000 x R x 5

vesna_86 [32]

R is equal to 450

first you subtract 5000 from each side

-2250=Rx5
then divid by 5

R=450

5 0

2 years ago

What is the completely factored form of 4x+20

allsm [11]

Hi there!

Since both 4x and 20 can be divisible by 4 and is greatest able to be divisible by 4, we can divide the 4 and put it outside of the parenthesis in order to factor:

4x / 4 = x
20 / 4 = 5

4 (x + 5)
would be your answer.

Hope this helps!

3 0

2 years ago

Suppose that we are testing a coin to see if it is fair, so our hypotheses are: H0: p = 0.5 vs Ha: p ≠ 0.5. In each of (a) and (

aleksley [76]

Answer:

$H_0: p = 0.5\\H_a: p \neq 0.5$

a. We get 56 heads out of 100 tosses.

We will use one sample proportion test

x = 56

n = 100

$\widehat{p}=\frac{x}{n}$

$\widehat{p}=\frac{56}{100}$

$\widehat{p}=0.56$

Formula of test statistic = $\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

= $\frac{0.56-0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}$

= $1.2$

refer the z table for p value

p value = 0.8849

a. We get 560 heads out of 1000 tosses.

We will use one sample proportion test

x = 560

n = 1000

$\widehat{p}=\frac{x}{n}$

$\widehat{p}=\frac{560}{1000}$

$\widehat{p}=0.56$

Formula of test statistic = $\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

= $\frac{0.56-0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}$

= $3.794$

refer the z table for p value

p value = .000148

p value of part B is less than Part A because part B have 10 times the number the tosses.

6 0

2 years ago

I dont get how to solve this?

Alex

Answer:

Step-by-step explanation:

In standard form this could be rewritten as | b - 0 | > 6, which in words, says,

"b is greater than 6 units away from 0 on a number line". If b is greater than 6 units away, both positive 6 and negative 6 are 6 units away from 0. Because the values we need are greater than 6 units away from 0, we are starting at both 6 and -6 and moving away from 0.

The "greater than" symbol indicates that this is a disjunction so the word "or" is used as opposed to in a conjunction where the word "and" is used.

Therefore, our solutions are

b < -6 or b > 6, the 3rd choice down

5 0

3 years ago