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stich3 [128]
3 years ago
10

plz help whats the answer to this question !!! calculate the mass of a gold bar that is 18.00cm long, 9.21cm wide, and 4.45cm hi

gh.
Mathematics
1 answer:
erik [133]3 years ago
7 0

We have to find the mass of the gold bar.

We have gold bar in the shape of a rectangular prism.

The length, width, and the height of the gold bar is 18.00 centimeters, 9.21 centimeters, and 4.45 centimeters respectively.

First of all we will find the volume of the gold bar which is given by the volume of rectangular prism:

Volume of the gold bar = length \times width\times height

Plugging the values in the equation we get,

Volume of the gold bar =18.00 \times 9.21 \times 4.45= 737.721cm^3

Now that we have the volume we can find the mass by using the formula,

Mass= density \times volume

The density of the gold is 19.32 grams per cubic centimeter. Plugging in the values of density and volume we get:

Mass = 19.32\times 737.721=14252.769 grams

So, the mass of the gold bar is 14252.769 grams

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student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
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Answer:

1.2%

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We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
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