What is the difference between a number line and a coordinate plane?

What are the rules to solving an equation with variables on one side

max2010maxim [7]

Answer:

Whatever you do to one side you must do it to the other side of the equation.

Step-by-step explanation:

5 0

4 years ago

25 graduating seniors form Stellar High School are randomly selected. A mean GPA of 3.42 is calculated with a margin of error ±0

Fantom [35]

Answer: it is b

Step-by-step explanation:

3 0

3 years ago

The number y of raccoons in an area after x years can be modeled by the function y= 0.4x^2+2x+2. When were there about 50 raccoo

marysya [2.9K]

Answer:

9 years.

Step-by-step explanation:

For 50 raccoons the equation is

0.4x^2+2x+2 = 50

0.4x^2+2x - 48 = 0

x = [-2 +/- sqrt(2^2- 4* 0.4*(-48)] / (2*0.4)

= (-2 +/- sqrt 9) / 0.8

= 8.74 , - 13.75 (ignore the negative).

So it's 9 years.

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

Hii can someone help with this pls :)

BlackZzzverrR [31]

Answer:

$R(x,y) = (0.8,-1.2)$