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crimeas [40]
3 years ago
8

Which expression represents the correct form for the quotient and remainder, written as partial fractions, of StartFraction 6 x

cubed + 15 x squared + 20 x + 36 Over 3 x squared + 7 EndFraction? 2 x + StartFraction A Over 3 x squared + 7 EndFraction 2 x + StartFraction A x + B Over 3 x squared + 7 EndFraction 2 x + 5 + StartFraction A Over 3 x squared + 7 EndFraction 2 x + 5 + StartFraction A x + B Over 3 x squared + 7 EndFraction
Mathematics
2 answers:
diamong [38]3 years ago
8 0

Answer:

The answer is "quotient =2x+5 and remainder =6x+1".

Step-by-step explanation:

Some of the data is missing that's why defined this question as follows:

Given:

\to \frac{6x^3+15x^2+20x+36} {3x^2+7}

when we divide the above value it will give the quotient =2x+5 and remainder =6x+1.

N76 [4]3 years ago
4 0

Answer:

Its D

Step-by-step explanation:

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3(x + 1) - 2x = -6 what is X i need halp
Mashutka [201]

Answer:x=−9

Step-by-step explanation:

Apply the distributive property.

3 x + 3 ⋅ 1 − 2 x = − 6

Multiply  3  by  1

.

3 x + 3 − 2 x = − 6

Subtract 2x from 3x.

x + 3 = − 6

Move all terms not containing  x  to the right side of the equation.

Subtract 3 from both sides of the equation.

x = − 6 − 3

Subtract 3 from − 6 .

x = − 9

4 0
2 years ago
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The Rocky Mountain News (January 24, 1994) indicated that the 20-year mean snowfall in the Denver/Boulder region is 28.76 inches
ycow [4]

Answer:

The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.

Step-by-step explanation:

20-year mean snowfall in the Denver/Boulder region is 28.76 inches. Test if the snowfall for the 1993-1994 winters has higher than the previous 20-year average.

At the null hypothesis, we test if the average was the same, that is, of 28.76 inches. So

H_0: \mu = 28.76

At the alternate hypothesis, we test if the average incresaed, that is, it was higher than 28.76 inches. So

H_1: \mu > 28.76

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

28.76 is tested at the null hypothesis:

This means that \mu = 28.76

Standard deviation of 7.5 inches. However, for the winter of 1993-1994, the average snowfall for a sample of 32 different locations was 33 inches.

This means that \sigma = 7.5, X = 33, n = 32.

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{33 - 28.76}{\frac{7.5}{\sqrt{32}}}

z = 3.2

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 33, which is 1 subtracted by the p-value of z = 3.2. In this question, we consider the standard level \alpha = 0.05.

Looking at the z-table, z = 3.2 has a p-value of 0.9993.

1 - 0.9993 = 0.0007

The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.

5 0
2 years ago
Which transformation occurred?
nikdorinn [45]
The x coordinate remains the same.
 The Y coordinate changes from a positive number to a negative number.

This means that the figure is reflected across the X axis because positive Y values are above the X axis and negative Y values are below the X axis.

The answer is D.

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3 years ago
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Gemiola [76]
Your answer is A. 15 
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1500÷8
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