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4 years ago

What is the simplified form of the expression?

Mathematics

1 answer:

AleksAgata [21]4 years ago

6 0

The answer is: 10 x¹³ y¹⁰ .
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1x^8 * 2y^(10) * 5x^5 =

1* 2* 5 * x^8 * x^5 * y^(10) =

10 * x^(8+5) * y^(10) =

10 * x^(13) * y^(10) = 10 x^(13) y^10 ; write as:
_______________________________________________
10 x¹³ y¹⁰ .
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Tara's car accelerated from 10 miles per hour to 70 miles per hour in 12 seconds.

SSSSS [86.1K]

Acceleration=change in velocity/time
a=(70-10)/12
a=60/12
a=5
So it is A

Hope this helps :)

8 0

3 years ago

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Hurry!!!Need help with this question!!! If p(x) = 2x2 – 4x and q(x) = x – 3, what is (p*h)(x)?

nignag [31]

Hey there,

If p(x) = 2x2 – 4x and q(x) = x – 3, what is (p*h)(x)?

Based on my understanding, I believe that your correct answer would
be 2x2 – 16x + 30 because 2x^2-16x+30. Just by plugging in both 2x^2
and also 16x+30, so this would mean that 2x²-16x+30 would be your correct answer.

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4 0

3 years ago

Absolute value please help me

finlep [7]

-7
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3 years ago

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Write as an equation: 1/3 of a shipment of books weights 28 pounds.

FromTheMoon [43]

Answer:

Option D

Step-by-step explanation:

A third of a shipment of books is 28 pounds. Assuming that 'f' will be the total weight of the entire shipment, we would divide 'f' by 3 to get 28.

The answer should be D.

3 0

3 years ago

1. Let the test statistics Z have a standard normal distribution when H0 is true. Find the p-value for each of the following sit

Nonamiya [84]

Answer:

1 a $p -value = 0.030054$

1b $p -value = 0.0029798$

1c $p -value = 0.0039768$

2a $p-value = 0.00099966$

2b $p-value = 0.00999706$

2c $p-value = 0.0654412$

Step-by-step explanation:

Considering question a

The alternative hypothesis is H1:μ>μ0

The test statistics is z =1.88

Generally from the z-table the probability of z =1.88 for a right tailed test is

$p -value = P(Z > 1.88) = 0.030054$

Considering question b

The alternative hypothesis is H1:μ<μ0

The test statistics is z=−2.75

Generally from the z-table the probability of z=−2.75 for a left tailed test is

$p -value = P(Z < -2.75) = 0.0029798$

Considering question c

The alternative hypothesis is H1:μ≠μ0

The test statistics is z=2.88

Generally from the z-table the probability of z=2.88 for a right tailed test is

$p -value = P(Z >2.88) = 0.0019884$

Generally the p-value for the two-tailed test is

$p -value = 2 * P(Z >2.88) = 2 * 0.0019884$

=> $p -value = 0.0039768$

Considering question 2a

The alternative hypothesis is H1:μ>μ0

The sample size is n=16

The test statistic is t = 3.733

Generally the degree of freedom is mathematically represented as

$df = n - 1$

=> $df = 16 - 1$

=> $df = 15$

Generally from the t distribution table the probability of t = 3.733 at a degree of freedom of $df = 15$ for a right tailed test is

$p-value = t_{3.733 , 15} = 0.00099966$

Considering question 2b

The alternative hypothesis is H1:μ<μ0

The degree of freedom is df=23

The test statistic is ,t= −2.500

Generally from the t distribution table the probability of t= −2.500 at a degree of freedom of df=23 for a left tailed test is

$p-value = t_{-2.500 , 23} = 0.00999706$

Considering question 2c

The alternative hypothesis is H1:μ≠μ0

The sample size is n= 7

The test statistic is ,t= −2.2500

Generally the degree of freedom is mathematically represented as

$df = n - 1$

=> $df = 7 - 1$

=> $df = 6$

Generally from the t distribution table the probability of t= −2.2500 at a degree of freedom of $df = 6$ for a left tailed test is

$t_{-2.2500 , 6} = 0.03272060$

Generally the p-value for t= −2.2500 for a two tailed test is

$p-value = 2 * 0.03272060 = 0.0654412$

4 0

3 years ago