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3 years ago

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An astronaut who weighs 95 kilograms on Earth weighs 15.8 kilograms on the Moon. How much would a person weigh on the Moon if th

ey weigh 105 kilograms on Earth? Round your answer to the nearest tenth, or one decimal place.

1 answer:

nadezda [96]3 years ago

4 0

If you weighed 105 kg on Earth, you would weigh 17.36 kg on the Moon (Rounded would be 17.4).

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Answer:

x = 10

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Step-by-step explanation:

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3 years ago

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4 years ago

A cylindrical oil storage tank has a height of 10 meters and a diameter of 24 meters. If the tank is full, how much oil does it

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3 years ago

What is the Desmos Graphing App for Algebra and how do you use it. Is it a computer thing or do you download it on your phone?

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4 years ago

Read 2 more answers

Mr. Bond’s class wanted to estimate the mean mass of Snickers Fun Size bars. They randomly selected 74 bars and recorded the mas

AleksAgata [21]

Step 1: Find the standard error (SE)

The standard error is given by

$SE=\frac{s}{\sqrt[]{n}}$ $\begin{gathered} \text{ Where } \\ SE=\text{ the standard error} \\ s=\text{ the sample standard deviation} \\ n=\text{ the sample size} \end{gathered}$

In this case,

$n=74,s=0.76$

Therefore,

$SE=\frac{0.76}{\sqrt[]{74}}\approx0.0883$

Step 2: Find the alpha level (α)

$\alpha=1-\frac{(\text{Confidence level})}{100}$ $\alpha=1-0.99=0.01$

Step 3: Find the critical probability (P*)

$P^{\prime}=1-\frac{\alpha}{2}$

Therefore,

$P^{\cdot}=1-\frac{0.01}{2}=0.995$

Step 4: Find the critical value (CV)

The critical value the z-score having a cumulative probability equal to the critical probability (P*).

Using the cumulative z-score table we will find the z-score with value of 0.995

Hence,

$CV=2.576$

Step 5: Find the margin of error (ME)

$ME=SE\times CV$

Therefore,

$ME=0.0883\times2.576=0.2275$

Step 6: Find the confidence interval (CI)

$\begin{gathered} CI\text{ is given by} \\ CI=(\bar{x}-ME,\bar{x}+ME) \\ \text{ In this case} \\ \bar{x}=17.1 \end{gathered}$

Therefore,

$CI=(17.1-0.2275,17.1+0.2275)=(16.8725,17.3275)$

Hence there is a 99% probability that the true mean will lie in the confidence interval

(16.8725, 17.3275)

4 0

1 year ago