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4 years ago

Find the product. (8m2 – 1)(8m2 + 1)

Mathematics

2 answers:

Schach [20]4 years ago

8 0

Answer:

the answer is 64m^4 - 1

64m^4 + 8m^2 - 8m^2 - 1

Step-by-step explanation:

drek231 [11]4 years ago

5 0

Answer:

64m^4 – 1

Step-by-step explanation:

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How many times stronger is a magnitude 9.2 earthquake than a magnitude 6.4 earthquake? Round your answer to the nearest whole nu

Hatshy [7]

Answer:

i have tried but have no idea

help me too haha

Step-by-step explanation:

8 0

3 years ago

Which scenario involves the value of greatest magnitude?

Sonbull [250]

Answer:

D. A withdrawal of $30 from a bank account

Step-by-step explanation:

When it says magnitude it doesn't matter if it's a positive or negative number.

So it's just asking which number is the greatest which is 30.

I hope my explanation makes sense and helps you out.

5 0

3 years ago

sara making punch for her party her punch bowl holds 600 ounces for 80 servings of punch write and solve an equation to find how

nlexa [21]

There are 7.5 ounces in each serving

Solution:

Given that, Sara making punch for her party her punch bowl holds 600 ounces for 80 servings of punch

To find: Ounces of punch in each serving

From given,

Number of servings = 80

Ounces in bowl = 600 ounces

Then, Ounces of punch in each serving is found by dividing the ounces in bowl by number of servings

The formula used is:

$\text{ Ounces of punch in each serving } = \frac{\text{Ounces in bowl}}{\text{Number of servings}}$

Substituting the values we get,

$\text{ Ounces of punch in each serving } = \frac{600}{80}\\\\\text{ Ounces of punch in each serving } = 7.5$

Thus there are 7.5 ounces in each serving

5 0

3 years ago

Suppose that textbook weights are normally distributed. You measure 33 textbooks' weights, and find they have a mean weight of 7

AleksAgata [21]

Answer:

$75-2.58\frac{13.3}{\sqrt{33}}=69.027$

$75+2.58\frac{13.3}{\sqrt{33}}=80.793$

And the 95% confidence interval would be between (69.027;80.793)

Step-by-step explanation:

Information given

$\bar X=75$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=13.3$ represent the population standard deviation

n=33 represent the sample size

Confidence interval

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

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

The degrees of freedom are given by:

$df=n-1=33-1=32$

The Confidence level is 0.99 or 99%, the significance would be $\alpha=0.01$ and $\alpha/2 =0.005$ , the critical value for this case would be $z_{\alpha/2}=2.58$

And replacing we got:

$75-2.58\frac{13.3}{\sqrt{33}}=69.027$

$75+2.58\frac{13.3}{\sqrt{33}}=80.793$

And the 95% confidence interval would be between (69.027;80.793)

6 0

3 years ago

Stephen can deliver 14 newspapers in 20 minutes. How many HOURS will it take him to deliver 131 newspaper

LiRa [457]

Answer:

approx. 3 hours

Step-by-step explanation:

14 newspapers in 20 minutes means he can deliver 42 newspapers an hour.

60/42 = 1.4285 newspapers per minute

131 x 1.4285 = 187.1335 minutes = 3.1188 hours

7 0

3 years ago