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3 years ago

Simplify the given expression. (See attachment)

Mathematics

1 answer:

stepladder [879]3 years ago

4 0

Working on x and y first, we get: x6 / x3= x3 and y6/ y12 = 1/ y6 so, we have: 2 x3 / y6
Entering values in original equation, we get:
(2 x3/ y6)^2=4 x^6 / y^12

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The figure below was made with a scale of 1 unit =9 m. You can place your figure anywhere on the grid on the right.

Zielflug [23.3K]

Answer & Step-by-step explanation:

I can't draw the image, but I can explain it to you.

Since 1 unit was 3 units and now it's 9, you would have to divide each side by 3. So, the width would be 1 unit and the length would be 2 units.

I hope this helped!!!

5 0

3 years ago

Evan was broke didn't he got his weekly allowance but because he has a hard time saving money he spent half his weekly allowance

grin007 [14]

Answer:

$16

Step-by-step explanation:

Let his weekly allowance = x

Amount spent playing golf = x / 2

Amount earned for cleaning car = $4

Final amount at hand = $12

Mathematically ;

x/2 + 4 = 12

x/2 = 12 - 4

x/2 = 8

x = 2 * 8

x = $16

Weekly allowance = $16

8 0

2 years ago

8. Graph the function

goldfiish [28.3K]

Answer:

Last option is the correct choice.

Step-by-step explanation:

See the attachment below.

Best Regards!

3 0

2 years ago

A class sorted 45 books in 90 minutes. How long would it take to sort 120 books?

pav-90 [236]

$45 \ books\ in\ 90\ minutes\\\\
120\ bookes\ in\ x\ minutes\\\\
making\ proportion\\
45-90\\120-x\\\\cross\ multiplication\\\\
45x=90*120\\\\
45x=10800\ \ \ | divide\ by\ 45\\\\x=240\\\\
It\ will\ take\ 240\ minutes.$

8 0

3 years ago

Read 2 more answers

Construct a 95% confidence interval for the population mean, μ. Assume the population has a normal distribution. A random sample

GuDViN [60]

Answer: $(628.48,\ 661.52)$

Step-by-step explanation:

Given : Sample size : $n=16$ , which is a small sample (, 30) so we use t-test.

Sample mean : $\overline{x}=645 \text{ hours}$

Standard deviation : $\sigma = 31 \text{ hours}$

Significance level : $\alpha=1-0.95=0.05$

Critical value : $t_{n-1,\alpha/2}=2.131$

The confidence interval for population mean is given by :-

$\overline{x}\pm t_{n-1,\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\=645\pm(2.131)\dfrac{31}{\sqrt{16}}\\\\\approx645\pm16.52\\\\=(645-16.52,\ 645+16.52)\\\\=(628.48,\ 661.52)$

Hence, a 95% confidence interval for the population mean $\mu$ = $(628.48,\ 661.52)$

6 0

3 years ago