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3 years ago

EASYYYYY PLEASE HELP!! SOS

Mathematics

2 answers:

V125BC [204]3 years ago

7 0

Answer:

2n+6=20

-6 -6

2n=14

÷2 ÷2

n=7

Westkost [7]3 years ago

7 0

Answer:

n = 7 for #1

Step-by-step explanation:

2n + 6= 20

-6 -6

2n = 14

/2 /2

n = 7

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Answer options 2 3 4 5

777dan777 [17]

Answer:

k = 5

Step-by-step explanation:

given f(x) then f(x) + k represents a vertical translation of f(x)

• If k > 0 then a shift of k units up

• If k < 0 then a shift of k units down

f(x) crosses the y- axis at - 2

g(x) crosses the y- axis at 3

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3 years ago

A student carried out an experiment to determine the amount of vitamin C in a tablet sample. He performed 5 trials to produce th

ivolga24 [154]

Answer:

There is not enough evidence to support the claim that the amount of vitamin C in a tablet sample is different from 500 mg.

P-value = 0.166.

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{5}(490+502+505+495+492)\\\\\\M=\dfrac{2484}{5}\\\\\\M=496.8\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{4}((490-496.8)^2+(502-496.8)^2+(505-496.8)^2+(495-496.8)^2+(492-496.8)^2)}\\\\\\s=\sqrt{\dfrac{166.8}{4}}\\\\\\s=\sqrt{41.7}=6.5\\\\\\$

Then, we can perform the hypothesis t-test for the mean.

The claim is that the amount of vitamin C in a tablet sample is different from 500 mg.

Then, the null and alternative hypothesis are:

$H_0: \mu=500\\\\H_a:\mu< 500$

The significance level is 0.05.

The sample has a size n=5.

The sample mean is M=496.8.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=6.5.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{6.5}{\sqrt{5}}=2.907$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{496.8-500}{2.907}=\dfrac{-3.2}{2.907}=-1.1$

The degrees of freedom for this sample size are:

$df=n-1=5-1=4$

This test is a left-tailed test, with 4 degrees of freedom and t=-1.1, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t$

As the P-value (0.166) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the amount of vitamin C in a tablet sample is different from 500 mg.

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