Which of the following lines is parallel to the graph of 5x+2y=8

A rock with mass of 5.0 kg is carried up a small hill 10 meters high. What is the potential energy of the rock at the hilltop?

Sonja [21]

490J

Step-by-step explanation:

Given parameters:

Mass of rock = 5kg

Height of hill = 10m

Unknown:

Potential energy at the hilltop = ?

Solution:

Potential energy is the energy at rest of a body. it is given as:

P.E = mgh

m is the mass of the body

g is the acceleration due to gravity = 9.8m/s²

h is the height of the hilltop

P.E = 5 x 10 x 9.8 = 490J

Learn more:

Potential energy brainly.com/question/10770261

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5 0

3 years ago

Which one of the shapes isn’t like the others ?

Vikki [24]

Answer:

c

Step-by-step explanation:

its more oval

7 0

3 years ago

Pls help 7th grade math

miskamm [114]

48.24/12=4.02
4.02 inches of rain per month

8 0

4 years ago

Read 2 more answers

What is seven minus one and three fifths

scoray [572]

The answer would be 5 2/5

8 0

3 years ago

A researcher wishes to see if the five ways (drinking decaffeinated beverages, taking a nap, going for a walk, eating a sugary s

e-lub [12.9K]

Answer:

$\chi^2 = \frac{(21-12)^2}{12}+\frac{(16-12)^2}{12}+\frac{(10-12)^2}{12}+\frac{(8-12)^2}{12}+\frac{(5-12)^2}{12}=13.833$

$p_v = P(\chi^2_{4} >13.833)=0.00785$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(13.833,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that the drowsiness are NOT equally distributed among office workers .

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Method Beverage Nap Walk Snack Other

Number 21 16 10 8 5

The total on this case is 60

We need to conduct a chi square test in order to check the following hypothesis:

H0: Drowsiness are equally distributed among office workers

H1: Drowsiness IS NOT equally distributed among office workers

The level of significance assumed for this case is $\alpha=0.1$

The statistic to check the hypothesis is given by:

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

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total}{5}$

And the calculations are given by:

$E_{Beverage} =\frac{60}{5}=12$

$E_{Nap} =\frac{60}{5}=12$

$E_{Walk} =\frac{60}{5}=12$

$E_{Snack} =\frac{60}{5}=12$

$E_{Other} =\frac{60}{5}=12$

And the expected values are given by:

Method Beverage Nap Walk Snack Other

Number 12 12 12 12 12

And now we can calculate the statistic:

$\chi^2 = \frac{(21-12)^2}{12}+\frac{(16-12)^2}{12}+\frac{(10-12)^2}{12}+\frac{(8-12)^2}{12}+\frac{(5-12)^2}{12}=13.833$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(catgories-1)=(5-1)=4$

And we can calculate the p value given by:

$p_v = P(\chi^2_{4} >13.833)=0.00785$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(13.833,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that the drowsiness are NOT equally distributed among office workers .

7 0

4 years ago