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3 years ago

Broooooo help meeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee!!!!!!!!!!!!!!!

Mathematics

1 answer:

tensa zangetsu [6.8K]3 years ago

3 0

9514 1404 393

Answer:

(b) not a function

Step-by-step explanation:

Look at the list of first values in the ordered pairs:

-1, 4, -1, 0

If there are any duplicates, the relation is NOT a function.

Here, -1 is duplicated, so this relation is not a function.

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Ryan the trainer has two solo workout plans that he offers his clients: Plan A and Plan B. Each client does either one or the ot

Ivahew [28]

Answer:

Both plans last for 1.25 hours (1 hour 15 minutes)

Step-by-step explanation:

Let x hours be the time needed for plan A and y hours be the time needed for plan B.

On Wednesday there were 5 clients who did Plan A and 3 who did Plan B. Thus, 5x+3y=10.

On Thursday there were 2 clients who did Plan A and 6 who did Plan B. Thus, 2x+6y=10.

Solve the sytem of two equation. Multiply the first equation by 2, the second by 5 and subtract them:

$10x+6y-10x-30y=20-50,\\ \\-24y=-30,\\ \\y=\dfrac{30}{24}=\dfrac{5}{4}=1.25\ hours.$

Therefore,

$5x+3\cdot 1.25=10,\\ \\5x=10-3.75=6.25,\\ \\x=1.25\ hours.$

8 0

4 years ago

Consider the following hypothesis test:

postnew [5]

Answer:

a. P-value = 0.039.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

b. P-value = 0.013.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

c. P-value = 0.130.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the population mean significantly differs from 100.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean significantly differs from 100.

Then, the null and alternative hypothesis are:

$H_0: \mu=100\\\\H_a:\mu\neq 100$

The significance level is 0.05.

The sample has a size n=65.

The degrees of freedom for this sample size are:

$df=n-1=65-1=64$

a. The sample mean is M=103.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.5.

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

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11.5}{\sqrt{65}}=1.4264$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{103-100}{1.4264}=\dfrac{3}{1.4264}=2.103$

This test is a two-tailed test, with 64 degrees of freedom and t=2.103, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>2.103)=0.039$

As the P-value (0.039) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

b. The sample mean is M=96.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.

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

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11}{\sqrt{65}}=1.3644$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{96.5-100}{1.3644}=\dfrac{-3.5}{1.3644}=-2.565$

This test is a two-tailed test, with 64 degrees of freedom and t=-2.565, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t$

As the P-value (0.013) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

c. The sample mean is M=102.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=10.5.

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

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{10.5}{\sqrt{65}}=1.3024$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{102-100}{1.3024}=\dfrac{2}{1.3024}=1.536$

This test is a two-tailed test, with 64 degrees of freedom and t=1.536, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>1.536)=0.130$

As the P-value (0.13) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the population mean significantly differs from 100.

8 0

3 years ago

How long does it take annabelle to read 45 pages

earnstyle [38]

Answer:

??? I don't quite understand pls explain better

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

For income tax purposes, a man uses a method called straight-line depreciation to show the loss in value of a copy machine he

PtichkaEL [24]

Answer: $500

Step-by-step explanation:

The vertical axis of the graph represents the value of the copy machine and the horizontal axis represents the years.

Using the line, one can see that in the first year, the value of the machine went from $7,000 to $6,500.

The loss in value is therefore;

= 7,000 - 6,500

= $500

<em>The needed graph is attached. </em>

6 0

3 years ago

Help asap!!

disa [49]

4/16(100)

Step-by-step explanation:

5 0

4 years ago