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3 years ago

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A small bar of gold measures 20 mm by 300 mm by 2 mm. One cubic millimeter of gold weighs about 0.0005 ounces. Find the volume i

n cubic millimeters and the weight in ounces of this small bar of gold.

1 answer:

Keith_Richards [23]3 years ago

4 0

Answer:

12000 mm^2

6 ounces

Step-by-step explanation:

20*300*2=12000

12000*0.0005=6

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Find the value of x. Round your answer to the nearest tenth.

Fantom [35]

Hello again! So we will be using the pythagorean theorem ( $leg^2+leg^2=hypotenuse^2$ ) to solve for the bottom leg of the triangles. After we do that, we'll subtract the value of the legs from 31, and that'll be x.

Plug in 17 and z (or any variable that isn't x) into the leg variable and 19 into the hypotenuse variable, and from there we can solve.

$17^2+z^2=19^2$

Solve the exponents to get $289+z^2=361$

Subtract 289 on each side to get $z^2=72$

Lastly, square root each side of the equation and your answer should be $z= \sqrt{72}$ (To get more accurate answers, you shouldn't round until the end of the problem)

Nextly, we will need to subtract $\sqrt{72}$ from 31. And since there are 2 triangles, you'll need to multiply $\sqrt{72}$ by 2.

$x=31- \sqrt{72} *2 \\ x=14.0$

In short, x = 14.0

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3 years ago

Determine the length of JK

wlad13 [49]

The length of JK rounded to the nearest hundredth is 4.72

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3 years ago

Please help, due tomorrow, please do all!

zlopas [31]

Answer:

are we supposed to multiply

Step-by-step explanation:

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3 years ago

Simplify 4 + 10 ÷ (–2).

Gnesinka [82]

Answer:

-1 Your answer is A

Step-by-step explanation:

4+ 10/-2=-1

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2 years ago

Read 2 more answers

Ignoring leap days, the days of the year can be numbered 1 to 365. Assume that birthdays are equally likely to fall on any day o

faust18 [17]

Answer:

Follows are the solution to the given question:

Step-by-step explanation:

They can count the days of the year 1 to 365. The random project consists of drawing a sample of n objects from D where elements are n people's birth in a group but instead, D = {1,....365}. And then there's the issue.

$S=365^n$

This because the list of future birthdays of n people was its test point; therefore m points will be in the sequence so each point contains 365 distinct outcomes. The probability function P for \Omega is that any event is likely to happen in 365 days.

$P(x)=\frac{1}{365^{n}}$

if x is between 1 and 365 as well as the occurrence is just all similarly possible

In point i:

That somebody mentions their birthday throughout the party

Guess I was born on day b. Therefore the consequence of "x is in A" is "b is now in the series of x," which is to say, b = bk for some amount k approximately 1 and n.

In point ii:

Any 2 persons share the same birthday at this party". A result x is in B" means which "two of entries in x are same." This means that perhaps the outcome x is in B if or only if bj = bk is in B of two numbers j, and k of 1, of two. , no, n.

In point iii:

Many three students share the same birthday with both the party. The consequence is x at the level of C only when bj = bk = bl at three (different) indices, j, k, l, 1. , no, n.

6 0

2 years ago