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3 years ago

Simplify the expression where possible (x^3 y^2) ^4

Mathematics

1 answer:

Alika [10]3 years ago

5 0

Answer:

$x^{12} y^{8}$

Step-by-step explanation:

$(x^{3} y^{2})^{4}$ You must multiply the outside exponent with the inside exponents.

$x^{12} y^{8}$

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Use the formula F=1.8C+32 to convert -20° into fahrenheit.

ZanzabumX [31]

F=1.8C+32

We are given C=-20 degrees

plugging this value in the given equation

F=1.8(-20)+32

F=-36+32=-4

F=-4 degree fahrenheit

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3 years ago

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PLEASE HELP ASAP!! WILL GIVE BRAINLIEST

AleksAgata [21]

I believe that it is D, so sorry if it's wrong, I haven't done this in a while.

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3 years ago

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What is the width of the rectangle?<br> 30y<br><br> 60y4

Bas_tet [7]

Answer:

2 y^3

Step-by-step explanation:

Area of rectangule is length * width

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2 years ago

HELPPP. DO TODAY SO PLEASE HELP!!!!!!!!!!

oksian1 [2.3K]

Answer:

47/6cm or $7\frac{5}{6}$ cm

Step-by-step explanation:

FE= $2\frac{2}{3}$

FA= 5/12

AB= 2/3

CD=2

ED= $1\frac{1}{4}$

BC= 5/6

All=47/6 or $7\frac{5}{6}$

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2 years ago

A quality analyst of a tennis racquet manufacturing plant investigates if the length of a junior's tennis racquet conforms to th

Burka [1]

Answer:

Confidence interval : $21.506$ to $24.493$

Step-by-step explanation:

A quality analyst selects twenty racquets and obtains the following lengths:

21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

So, sample size = n =20

Now we are supposed to find Construct a 99.9% confidence interval for the mean length of all the junior's tennis racquets manufactured at this plant.

Since n < 30

So we will use t-distribution

Confidence level = 99.9%

Significance level = α = 0.001

Now calculate the sample mean

X=21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

Sample mean = $\bar{x}=\frac{\sum x}{n}$

Sample mean = $\bar{x}=\frac{21+25+23+22+24+21+25+21+23+ 26+ 21+24+22+ 24+23+21+ 21+ 26+23+ 24}{20}$

Sample mean = $\bar{x}=23$

Sample standard deviation = $\sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}$

Sample standard deviation = $\sqrt{\frac{(21-23)^2+(25-23)^2+(23-23)^2+(22-23)^2+(24-23)^2+(21-23)^2+(25-23)^2+(21-23)^2+(23-23)^2+(26-23)^2+(21-23)^2+(24-23)^2+(22-23)^2+(24-23)^2+(23-23)^2+(21-23)^2+(21-23)^2+(26-23)^2+(23-23)^2+(24-23)^2}{20-1}}$

Sample standard deviation= s = $1.72$

Degree of freedom = n-1 = 20-1 -19

Critical value of t using the t-distribution table $t_{\frac{\alpha}{2}$ = 3.883

Formula of confidence interval : $\bar{x} \pm t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}}$

Substitute the values in the formula

Confidence interval : $23 \pm 1.73 \times \frac{1.72}{\sqrt{20}}$

Confidence interval : $23 -3.883 \times \frac{1.72}{\sqrt{20}}$ to $23 + 3.883 \times \frac{1.72}{\sqrt{20}}$

Confidence interval : $21.506$ to $24.493$

Hence Confidence interval : $21.506$ to $24.493$

3 0

3 years ago

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