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maria [59]
3 years ago
11

Which equation could represent the relationship shown in the scatter plot? y=−3x−2

Mathematics
2 answers:
madreJ [45]3 years ago
6 0

Answer:  the answer is y=-34x+10

Step-by-step explanation:

Orlov [11]3 years ago
3 0

Answer:

Y=-3/4x + 10

Step-by-step explanation:


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Use the distributive property to simplify the expression 6(s-9
Natalija [7]
You have to multiply the 6 by both s and -9

So you will get 6s - 54 which is the answer
7 0
2 years ago
) Set up a double integral for calculating the flux of F=3xi+yj+zk through the part of the surface z=−5x−2y+2 above the triangle
Fynjy0 [20]

The surface (call it S) is a triangle with vertices at the points

x=0,y=0\implies z=2\implies(0,0,2)

x=0,y=2\implies z=-2\implies(0,2,-2)

x=2,y=0\implies z=-8\implies(2,0,-8)

Parameterize S by

\vec s(u,v)=(1-v)(2,0,-8)+v\bigg((1-u)(0,2,-2)+u(0,0,2)\bigg)=(2-2v,2v-2uv,-8+6v+4uv)

with 0\le u\le1 and 0\le v\le1. Take the normal vector to S to be

\vec s_v\times\vec s_u=(20v,8v,4v)

Then the flux of \vec F across S is

\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^1\int_0^1(6-6v,2v-2uv,-8+6v+4uv)\cdot(20v,8v,4v)\,\mathrm du\,\mathrm dv

=\displaystyle8\int_0^1\int_0^1(11v-10v^2)\,\mathrm du\,\mathrm dv=\boxed{\frac{52}3}

8 0
3 years ago
Flat family roof on goformative answer
juin [17]
Guatafak is that ...
3 0
2 years ago
A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri
nlexa [21]

Answer:

df=categor-1=6-1=5

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got \chi^2_{critc}= 11.0705

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

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".

Solution to the problem

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday  Tuesday  Wednesday Thursday Friday Saturday    Total

 12             9                 11                 10           9            9              60

The level of significance assumed for this case is \alpha=0.05

The statistic to check the hypothesis is given by:

\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{60}{6}= 10 and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

df=categor-1=6-1=5

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got \chi^2_{critc}= 11.0705

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

4 0
2 years ago
Pat shows that 16 1/2= the square root of 16. What is one possible way pat could have shown this correctly?
kifflom [539]
<h3>Answer: Choice C</h3>

The base 16 is the same as 4*4.

From there, the rule (a*b)^c = a^c*b^c is used to get (4*4)^{1/2} = 4^{1/2}*4^{1/2}

Afterward, the exponents are added getting 1/2+1/2 = 2/2 = 1. The rule is a^b*a^c = a^{b+c} which only works if the bases are both the same.

6 0
3 years ago
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