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3 years ago

The weight of an object on the moon is

Mathematics

1 answer:

Tanzania [10]3 years ago

7 0

Answer:

40 2/5Ibs or $40\frac{2}{5}$ lbs

Step-by-step explanation:

Let weight of object on the moon be represented = Wm

weight of object on the Earth = We

In the above question, we are told:

The weight of an object on the moon is

5 of its weight on Earth.

Hence, mathematically, this is represented as:

5Wm = We

If a moon rock weighs 202 lbs on Earth

5Wm = 202lbs

Wm = 202lbs/5

Wm = $40\frac{2}{5}$ Ibs

Therefore, the moon rock weighs $40\frac{2}{5}$ Ibs on the moon.

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A small box of oranges costs $9 and weighs 12 pounds. A large box of oranges costs $18 and weighs 30 pounds. Right now one of th

jeka94

Answer:

See Below

Step-by-step explanation:

Small Box:

9 dollars for 12 pounds

Per pound cost is:

9/12 = $0.75 per pound

Large Box:

18 dollars for 30 pounds

Per pound cost is:

18/30 = $0.60 per pound

Better deal is large box.

To make the deals same, we can change the number of pounds in the large box. We can lessen it so it becomes the same ratio as dollars to pound.

Small box was

9 to 12

Large box is 18 dollars to WHAT?

18 is double of 9 so, "WHAT" should be double of 12, which is:

12 * 2 = 24

So, we can change large box "30" pounds to "24" pounds.

So, the ratio becomes (cost per pound):

18/24 = $0.75 per pound (same)

4 0

3 years ago

10.

Kay [80]

Answer:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.535 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.2 \leq \sigma \leq 2.8$

And the best option would be:

A. 2.2 < σ < 2.8

Step-by-step explanation:

Information provided

$\bar X=32.1$ represent the sample mean

$\mu$ population mean

s=2.4 represent the sample standard deviation

n=83 represent the sample size

Confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom are given by:

$df=n-1=83-1=82$

The Confidence is given by 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , the critical values for this case are:

$\chi^2_{\alpha/2}=62.132$

$\chi^2_{1- \alpha/2}=104.139$

And replacing into the formula for the interval we got:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.535 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.2 \leq \sigma \leq 2.8$

And the best option would be:

A. 2.2 < σ < 2.8

3 0

3 years ago

SHOW WORK ILL MARK BRAINLIST !!!!

Greeley [361]

Answer:

I'm thinking it's Associative or Identity property(preferably Associative) but I'm not sure

Step-by-step explanation:

Education is important, don't cheat <3

search up the different properties and Associative makes sense more with Step 2!

3 0

2 years ago

Answer quickly thank you so much!!!!!!!!!!!!!!!!!!!!!

marin [14]

The answer is C I think

3 0

2 years ago

What is the mean score of the vaulter?

dsp73

~Hello There!~

6.856 + 7.923 + 8.135 = 22.914
22.914/3 = 7.638 <=== Mean score

Hope This Helps You!
Good Luck :)
Have A Great Day ^_^

- Hannah ❤

6 0

2 years ago

Read 2 more answers