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IrinaVladis [17]
4 years ago
15

I’m in the middle of a test please answer fast!!! Will give brainliest!!!!

Mathematics
1 answer:
dmitriy555 [2]4 years ago
6 0

Answer:

the third one is the one having the highest y - intercept.

reason ; the equation is of the form y = x +c

substituting the given values of y and x to get the value of c,

the first one will give c = -8

the second will give c = -1

and the third will give c = 6

Step-by-step explanation:

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Answer: ± 1.96

Step-by-step explanation:

Given the following :

Mean (m) time spent = 11.85

Standard deviation = 4 minutes

Since the population of site visits is Normally distributed, Hence, the test statistic will follow a normal distribution :

The test the hypothesis that mean time has changed at 0.05 confidence interval

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Using the critical value calculator, which in general under normal distributions :

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± 1.96

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PLZ HELP ME!!!!!!!!!!!
Nana76 [90]

1. 12/32

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Find the quotient using polynomial long division:
Snowcat [4.5K]

(1) 50y^3=5y\cdot 10y^2, and

10y^2\cdot(5y-4)=50y^3-40y^2

Subtract this from 50y^3+10y^2-35y-7 to get a remainder of

(50y^3+10y^2-35y-7)-(50y^3-40y^2)=50y^2-35y-7

What we've done here is show that

\dfrac{50y^3+10y^2-35y-7}{5y-4}=10y^2+\dfrac{50y^2-35y-7}{5y-4}

We can keep going as long as the degree of the remainder's numerator is at least the same as the degree of its denominator.

Next, 50y^2=5y\cdot 10y, and

10y\cdot(5y-4)=50y^2-40y

Subtract this from the previous remainder to get a new remainder of

(50y^2-35y-7)-(50y^2-40y)=5y-7

which means

\dfrac{50y^3+10y^2-35y-7}{5y-4}=10y^2+10y+\dfrac{5y-7}{5y-4}

Finally, 5y=5y\cdot1, and

1\cdot(5y-4)=5y-4

Subtract this from the previous remainder to get

(5y-7)-(5y-4)=-3

So we end up with

\dfrac{50y^3+10y^2-35y-7}{5y-4}=\boxed{10y^2+10y+1}-\dfrac3{5y-4}

(the quotient is the expression in the box)

(2) Using the same process as before:

8m^4=2m\cdot4m^3

4m^3\cdot(2m+1)=8m^4+4m^3

(8m^4-4m^2+m+4)-(8m^4+4m^3)=-4m^3-4m^2+m+4

\implies\dfrac{8m^4-4m^2+m+4}{2m+1}=4m^3-\dfrac{4m^3+4m^2+m+4}{2m+1}

-4m^3=2m\cdot(-2m^2)

-2m^2\cdot(2m+1)=-4m^3-2m^2

(-4m^3-4m^2+m+4)-(-4m^3-2m^2)=-2m^2+m+4

\implies\dfrac{8m^4-4m^2+m+4}{2m+1}=4m^3-2m^2-\dfrac{2m^2-m-4}{2m+1}

-2m^2=2m\cdot(-m)

-m\cdot(2m+1)=-2m^2-m

(-2m^2+m+4)-(-2m^2-m)=2m+4

\implies\dfrac{8m^4-4m^2+m+4}{2m+1}=4m^3-2m^2-m+\dfrac{2m+4}{2m+1}

2m=2m\cdot1

1\cdot(2m+1)=2m+1

(2m+4)-(2m+1)=3

\implies\dfrac{8m^4-4m^2+m+4}{2m+1}=\boxed{4m^3-2m^2-m+1}+\dfrac3{2m+1}

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