What is true about all the points that have 5 as their ordinate

Solve for the values of x and y.

Tasya [4]

Answer:

We can’t solve for x and y

Step-by-step explanation:

5 0

2 years ago

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Please give real answers with an explaination. I will follow + I will give the brainliest. No Docs/No Files/No Links only answer

Norma-Jean [14]

Answer:

129 = x

Step-by-step explanation:

The exterior angle is equal to the sum of the opposite interior angles

47+82 = x

129 = x

6 0

2 years ago

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Please tell me y what 6(2y−9)=-42

Bad White [126]

Answer:

y =1

Step-by-step explanation:

6(2y−9)=-42

Divide each side by 6

6/6(2y−9)=-42/6

2y -9 = -7

Add 9 to each side

2y -9+9 = -7+9

2y =2

Divide by 2

2y/2 = 2/2

y =1

4 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 $\chi^2_{crit}=7.815$

We can calculate also the p value:

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

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

5 0

2 years ago

The circumference of a circle is 6.28 miles. What is the circle's radius?

Pepsi [2]

Answer:

3.14

Step-by-step explanation:

I dont know if thats correct for sure. I kinda forgot what circumference is.

But how i got this was 6.28/2=3.14

Let me know if this is correct or incorrect.

6 0

2 years ago