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3 years ago

100 POINTS PLEASE HELP ME PLEASE DON'T GO PASS THIS QUESTION!!!!

Mathematics

2 answers:

valina [46]3 years ago

8 0

The two points are
(4,2) , (6,5)

slavikrds [6]3 years ago

6 0

Step-by-step explanation:

I'm sorry if I'm wrong but I think it's (4,2)

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Which expression is equivalent to (m^5n/pq^2)^4

tatuchka [14]

$\left( \cfrac{m^5n}{pq^2}\right)^4 = \cfrac{(m^5n)^4}{(pq^2)^4}= \cfrac{m^{20}n^4}{p^4q^8}$

8 0

3 years ago

Read 2 more answers

(a) determine the null and alternative hypotheses, (b) explain what it would mean to make a Type I error, and (c) explain what i

iren2701 [21]

Null: the mean amount of peanut butter in a jar is equal to 32 oz.
alternative: the mean amount of peanut butter in a Jar is less than 32oz.

type 1 error is is rejecting the null when it is actually true. this means that we would say that the mean amount of peanut butter is not equal to 32 when it actually is.

type 2 error is failing to reject the null when it is actually false. this means that we would say the mean amount of peanut butter is equal to 32 when in reality it is less.

3 0

3 years ago

Intellectual development (Perry) scores were determined for 21 students in a first-year, project-based design course. (Recall th

Anit [1.1K]

Answer:

The 99% confidence interval is (3.0493, 3.4907).

We are 99% sure that the true mean of the students Perry score is in the above interval.

Step-by-step explanation:

Our sample size is 21.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 21-1 = 20$ .

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.99}{2} = \frac{0.01}{2} = 0.005$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 20 and 0.005 in the two-sided t-distribution table, we have $T = 2.528$

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

$s = \frac{0.40}{\sqrt{21}} = 0.0873$

Now, we multiply T and s

$M = 2.528*0.0873 = 0.2207$

Then

The lower end of the interval is the mean subtracted by M. So:

$L = 3.27 - 0.2207 = 3.0493$

The upper end of the interval is the mean added to M. So:

$LCL = 3.27 + 0.2207 = 3.4907$

The 99% confidence interval is (3.0493, 3.4907).

Interpretation:

We are 99% sure that the true mean of the students Perry score is in the above interval.

7 0

3 years ago

Marcus is saving money to build a new robot. He already has $75 saved up from the last competition he won. He deposits $8 per we

Oksi-84 [34.3K]

Answer:

soo hardd

Step-by-step explanation:

6 0

3 years ago

Y'all I am struggling Use the following functions to find each value below. f(x)=5x; g(x)=−2x+1; h(x)=x2+6x+8

Anni [7]

Answer:

see below the first three problems

Step-by-step explanation:

f(g(-2))

First, find g(-2) using function g(x). Then use that value as input for function f(x).

g(x) = -2x + 1

g(-2) = -2(-2) + 1

g(-2) = 5

f(x) = 5x

f(5) = 5(5)

f(5) = 25

f(g(-2)) = 25

g(h(3))

First, find h(3) using function h(x). Then use that value as input for function g(x).

h(x) = x^2 + 6x + 8

h(3) = 3^2 + 6(3) + 8 = 9 + 18 + 8

h(3) = 35

g(x) = -2x + 1

g(35) = -2(35) + 1 = -70 + 1

g(35) = -69

g(h(3)) = -69

f(g(3a))

First, find g(3a) using function g(x). Then use that value as input for function f(x).

g(x) = -2x + 1

g(3a) = -2(3a) + 1

g(3a) = -6a + 1

f(x) = 5x

f(-6a + 1) = 5(-6a + 1)

f(-6a + 1) = -30a + 5

f(g(3a)) = -30a + 5

5 0

3 years ago