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Sever21 [200]
3 years ago
8

(-8x⁴y³)(2x⁵y²)+7x⁹y⁵

Mathematics
2 answers:
LenaWriter [7]3 years ago
8 0

Answer:

ss

Step-by-step explanation:

STatiana [176]3 years ago
8 0

Answer:

-9x^{9} y^{5}

Step-by-step explanation:

(-8x⁴y³)(2x⁵y²)+7x⁹y⁵  (grouping like terms)

= (-8)(2) (x⁴x⁵) (y³y²) +7x^{9} y^{5}

= -16 x^{4+5} y^{3+2 } + 7x^{9} y^{5}

= -16 x^{9} y^{5} + 7x^{9} y^{5}

= (-16 + 7)x^{9} y^{5}

= -9x^{9} y^{5}

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83 = y - 17 what the value of y
Artemon [7]

The answer is y = 100

To get this answer, you add 17 to both sides to undo the "minus 17" that is happening on the right side.

83 = y-17

83+17 = y-17+17 ... add 17 to both sides

100 = y

y = 100

3 0
3 years ago
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5x^2 +3 = -122 a. no solution b. -5 c. +‐5 d. +- 25​
Finger [1]

Answer:

I put it in the calc and it got

x=5i,−5i

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3 years ago
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Jenna is writing a check to her vehicle registration office for $100.42. How should she pull out this amount in the "written amo
Angelina_Jolie [31]

Answer:

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Step-by-step explanation: you put the change as a fraction

3 0
3 years ago
Suppose that the sample standard deviation was s = 5.1. Compute a 98% confidence interval for μ, the mean time spent volunteerin
NISA [10]

Answer:

The 95% confidence interval would be given by (5.139;5.861)  

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

2) Confidence interval

Assuming that \bar X =5.5 and the ranfom sample n=1086.

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=1086-1=1085

Since the Confidence is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,1085)".And we see that t_{\alpha/2}=2.33

Now we have everything in order to replace into formula (1):

5.5-2.33\frac{5.1}{\sqrt{1086}}=5.139    

5.5+2.333\frac{5.1}{\sqrt{1086}}=5.861

So on this case the 95% confidence interval would be given by (5.139;5.861)    We are 98% confident that the mean time spent volunteering for the population of parents of school-aged children is between these two values.

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3 years ago
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Step-by-step explanation:

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