All categories

3 years ago

A.

Mathematics

1 answer:

matrenka [14]3 years ago

8 0

Answer:

net income = (cost of goods sold) - (net sales) - (operating expenses)

Step-by-step explanation:

because income is based on sales, thus the cost of how much you got minus all the expenses must be the net income.

You might be interested in

PLEASE HELP. In the figure below, is there a dilation D that would map figure PQRS to figure P’Q’R’S’?

dybincka [34]

Hihihihihihihihihihi

3 0

3 years ago

Read 2 more answers

When did my mom chop up my glizzy A:2015 B:2017 C:2018 D:2020

Artist 52 [7]

Answer:

imma go wit B. I mean thats my guess

5 0

3 years ago

The average weekly salary of two employees is $5350. One makes $250 more than the other. Find their salaries.

nataly862011 [7]

Answer:

Step-by-step explanation:

7 0

3 years ago

Pls help with other question. PLZ ILL MARK YOU AS BRAINLEST!!!

Reil [10]

Count 12 spaces down and 1 space to right i think

3 0

3 years ago

Read 2 more answers

Joan's Nursery specializes in custom-designed landscaping for residential areas. The estimated labor cost associated with a part

Gnoma [55]

Answer:

A) Null and alternative hypothesis

$H_0: \mu=2\\\\H_a:\mu\neq 2$

B) M = 2.2 hours

C) s = 0.52 hours

D) P-value = 0.255

E) At a significance level of 0.05, there is not enough evidence to support the claim that the mean tree-planting time significantly differs from two hours.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the mean tree-planting time significantly differs from two hours.

Then, the null and alternative hypothesis are:

$H_0: \mu=2\\\\H_a:\mu\neq 2$

The significance level is 0.05.

The sample has a size n=10.

The sample mean is M=2.2.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.52.

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

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.52}{\sqrt{10}}=0.1644$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{2.2-2}{0.1644}=\dfrac{0.2}{0.1644}=1.216$

The degrees of freedom for this sample size are:

$df=n-1=10-1=9$

This test is a two-tailed test, with 9 degrees of freedom and t=1.216, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>1.216)=0.255$

As the P-value (0.255) 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 mean tree-planting time significantly differs from two hours.

Sample mean and standard deviation:

 $M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{10}(1.7+1.5+2.6+. . .+2.3)\\\\\\M=\dfrac{22}{10}\\\\\\M=2.2\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{9}((1.7-2.2)^2+(1.5-2.2)^2+(2.6-2.2)^2+. . . +(2.3-2.2)^2)}\\\\\\s=\sqrt{\dfrac{2.4}{9}}\\\\\\s=\sqrt{0.27}=0.52\\\\\\$

3 0

3 years ago