Apply the Distributive property to the expression 4(3x +5y - 4) to produce an equivalent expression.

72/60 in simplest form

Bogdan [553]

The common number is 6.
72/6=12
60/6=10
12/10
simplify that, it is 6/5

7 0

3 years ago

What is the range of the relation?

Alika [10]

Answer:

{- 3, - 2, 0, 2 }

Step-by-step explanation:

The range is the y- coordinates of the points

(- 4, - 3 ) , (- 1, - 2 ) , (- 2, 0 ) , (0, 2 )

The y- coordinates are - 3, - 2, 0, 2

Then the range is { - 3, - 2, 0, 2 }

8 0

2 years ago

If every student got one vote, but there were 25 students absent on the day of the vote, how many

kherson [118]

Answer:

Not enough information.

Step-by-step explanation:

The 25 students absent is irrelevant. We still don't know the total number of students at Riverside Elementary School.

3 0

3 years ago

The Fish and Game Department stocked a lake with fish in the following proportions: 30% catfish, 15% bass, 40% bluegill, and 15%

vagabundo [1.1K]

Answer:

1) $\chi^2 =\frac{(112-150)^2}{150}+\frac{(95-75)^2}{75}+\frac{(210-200)^2}{200}+\frac{(83-75)^2}{75}=16.313$

2) $p_v =P(\chi^2_{3}>16.313)=0.000978$

And we got the same decision reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts

The Chi-Square test of independence is used "to determine if there is a significant relationship between two nominal (categorical) variables". And is defined with the following statistic:

$\chi^2 =\sum_{i=1}^n \frac{(O-E)^2}{E}$

Where O rpresent the observed values and E the expected values.

State the null and alternative hypothesis

Null hypothesis: The distribution is 30% catfish, 15% bass, 40% bluegill, and 15% pike

Alternative hypothesis: The distribution is NOT 30% catfish, 15% bass, 40% bluegill, and 15% pike

The observed values are given by the table given:

Catfish =112, BAss = 95, Bluegill=210, Pike=83

Calculate the expected values

In order to calculate the expected values we can use the following formula for each cell of the table

$E = \% Grand total$

$E_{Catfish}=500*0.3=150$

$E_{Bass}=500*0.15=75$

$E_{Bluegill}=500*0.4=200$

$E_{Pike}=500*0.15=75$

Part 1: Calculate the statistic

$\chi^2 =\frac{(112-150)^2}{150}+\frac{(95-75)^2}{75}+\frac{(210-200)^2}{200}+\frac{(83-75)^2}{75}=16.313$

$\chi^2 =16.313$

Calculate the critical value

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

$df= (categories-1)=(4-1)= 3$

Since the confidence provided is 95% the significance would be $\alpha=1-0.95=0.05$ and we can find the critical value with the following excel code: "=CHISQ.INV(0.95,3)", and our critical value would be $\chi^2_{crit}=7.815$