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2 years ago

Write the equation in slope-intercept form y−6=−2/5(x−10)

Mathematics

1 answer:

stellarik [79]2 years ago

5 0

Answer: y = -2/5x + 10

Step-by-step explanation:

y − 6 = − 2/5 (x − 10)

do distributive property

y - 6 = -2/5x + 4

add six to both sides

y = -2/5x + 10

boom the answer :v

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The area of a rectangular painting is 13.02 square feet. The width is 4.2 feet. What is the length?

Kay [80]

Answer:

3.1

Step-by-step explanation:

Since this is area and you are given the width must divide area by width ( A/W = L) 13.02 divided by 4.2 = 3.1. You can check this by doing 3.1 multiplied by 4.2. It should give you 13.02

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3 years ago

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What is the slope of the line that passes the points(1,-6) and (-8,9)

VARVARA [1.3K]

Answer:

sorry this is for points

Step-by-step explanation:

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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.

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3 years ago

Simplify the sum 4 / m+9 + 5 / m^2 -81 Please help! Will mark brainliest if its right!

slamgirl [31]

m2 + 4m + 5 ——————————— m2

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3 years ago

Which numbers are greater than 0.7? 0.37, 0.9, 0.08, 0.69, 0.71

Marat540 [252]

The correct ansswer is option B. i.e. 0.9 . Because 0.37 and 0.08 are clearly lesser. And amongst, 0.69, 0.71 and 0.9, the number 0.91 is the greatest of all. So, 0.91 is greater than 0.7 from all the given options.

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3 years ago