The assembly room has 6 rows with 30 chair in each row. If third grader fill 3 rows, how many third grader are in the room?

Juan is rewriting the expression 12d-26c as a product. Which statements about the expression are accurate and relevant to his re

Alex777 [14]

2(6d-13c) is an equivalent way to write this expression
One statement that is true when re-writing expression is that the terms cannot be combined because the variables are different.

Are there statements you can chose from, like multiple choice answers?

6 0

4 years ago

Read 2 more answers

Bobbie wants to make $2.00 using dollars half dollars and quarters.how many different ways can she make $2.00?

Solnce55 [7]

Answer:

8 * 1/4

6 * 1/4 + 1/2

4 * 1/4 + 2 * 1/2

2 * 1/4 + 3 * 1/2

4 * 1/2

4 * 1/4 + 1 * 1

2 * 1/4 + 1/2 + 1 * 1

2 * 1/2 + 1 * 1

2 * 1

There are 9 different ways.

5 0

4 years ago

I really need the help lads.

Tpy6a [65]

Answer:

why is the picture all white on my screen

7 0

3 years ago

An accounting firm is trying to decide between IT training conducted in-house and the use of the third party consultants. To get

Jlenok [28]

Answer:

$t=\frac{500-490}{\sqrt{\frac{48^2}{180}+\frac{32^2}{210}}}}=2.379$

$p_v =2*P(t_{388}>2.379)=0.0178$

Comparing the p value with the significance level provided $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and a would be a significant difference in the average of the two groups.

Step-by-step explanation:

Assuming the following info

IT Training N mean Stdev

In-house 210 $490 $32

Consultants 180 $500 $48

1) Data given and notation

$\bar X_{I}=490$ represent the mean for the sample of In-house

$\bar X_{C}=500$ represent the mean for the sample Consutants

$s_{I}=32$ represent the population standard deviation for the sample In-house

$s_{C}=48$ represent the population standard deviation for the sample Consultants

$n_{I}=210$ sample size for the group In-house

$n_{C}=180$ sample size for the group Consultants

t would represent the statistic (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:

Null hypothesis: $\mu_{I}=\mu_{C}$

Alternative hypothesis: $\mu_{I} \neq \mu_{C}$

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{C}-\bar X_{I}}{\sqrt{\frac{s^2_{C}}{n_{C}}+\frac{s^2_{I}}{n_{I}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

3) Calculate the statistic

With the info given we can replace in formula (1) like this:

$t=\frac{500-490}{\sqrt{\frac{48^2}{180}+\frac{32^2}{210}}}}=2.379$

4) Statistical decision

First we need to calculate the degrees of freedom given by:

$df=n_{C}+n_{I}-2=210+180-2=388[/tex[The sample is large enough to assume that the distribution is also normal.Since is a bilateral test the p value would be:
[tex]p_v =2*P(t_{388}>2.379)=0.0178$

Comparing the p value with the significance level provided $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and a would be a significant difference in the average of the two groups.

6 0

4 years ago

23.) What is one benefit or drawback to an hourly wage?

xxMikexx [17]

An hourly wage may result in lower pay

6 0

3 years ago