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2 years ago

Suppose I rolled a fair six-sided die. Let R=number rolled. What is the sample space,

Mathematics

1 answer:

Mariana [72]2 years ago

6 0

Answer:

No Side Is Being Favored Since The Die Is A Fair Six-Sided Die.

So Each Number Has The Same Chance Of Being Rolled.

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A club had some money to purchase new chairs. After buying 355 chairs at $199 each, there was $1,068 remaining. How much money d

Bumek [7]

Answer:

Explanation:

Find The Total Cost Of The 355 Chairs
355 Chairs - $199 Per Chair

355 · 199 = $70,645

Club Spent $70,645 And Had $1,068 Remaining

$70,645 + $1,068 = $71,713

<h2><em><u>The Club Had $71,713 At First</u></em></h2>

- PNW

7 0

3 years ago

Which of the following are true statements.

Mariulka [41]

Answer:

Second statement is true.

The lengths 7, 40 and 41 can not be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.

Step-by-step explanation:

for first part of statement

The lengths 7, 40 and 41 can not be sides of a right triangle.

If the square of long side is equal to the sum of square of other two sides

then the given length can be sides of a right triangle.

Check the given length by Pythagoras Theorem.

$c^{2} =a^{2} +b^{2}$ ----------(1)

Let $c=41$ and $a = 7$ and $b=40$

Put all the value in equation 1.

$41^{2} =7^{2} +40^{2}$

$1681=49+1600$

$1681=1649$

Therefore, the square of long side is not equal to the sum of square of other two sides, So given lengths 7, 40 and 41 can not be sides of a right triangle.

for second part of statement.

The lengths 12, 16, and 20 can be sides of a right triangle.

Check the given length by Pythagoras Theorem.

Let $c=20$ and $a = 12$ and $b=16$

$20^{2} =12^{2} +16^{2}$

$400=144+256$

$400=400$

Therefore, the square of long side is equal to the sum of square of other two sides, So given the lengths 12, 16, and 20 can be sides of a right triangle.

Therefore, The lengths 7, 40 and 41 can not be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.

8 0

2 years ago

Mel has to put the greatest number of

Dmitry_Shevchenko [17]

Mel should use the least common multiple to solve the problem

<u>Solution:</u>

Given, Mel has to put the greatest number of bolts and nuts in each box so each box has the same number of bolts and the same number of nuts.

We have to find that should Mel use the greatest common factor or the least common multiple to solve the problem?

He should use least common multiple.

Let us see an example, suppose 12 bolts and nuts are to be fit in 6 boxes.

Then, if we took H.C.F of 12 and 6, it is 6, which means 6 bolts and nuts in each box, but, after filling 2 boxes with 6 bolts and nuts, there will be nothing left, which is wrong as remaining boxes are empty.

So the remaining method to choose is L.C.M.

Hence, he should use L.C.M method.

6 0

3 years ago

Equations

sergij07 [2.7K]

Answer:

2x+12=24

first you subtract 12 from both sides

2x+12=24

-12. -12

The 12-12 should cancel itself, the rest of the equation you bring down to get

2x=12 (because 24-12=12)

Now you have 2x=12.
you then divide 2x by both sides.

2x=12

/2x=/2x

The 2x/2x cancels itself out so you then solve for 12/2x.

For this you just divide 12/2 which is 6!

x= 6 is your final answer.
to check this equation you can plug your number back into x to see if it is true! 2(6)+12=24.
6 times 2 is 12 and 12+12 is 24 so your answer (6) is true!

hope this helps! :D

4 0

2 years ago

Read 2 more answers

Your bill from Acme Supply Company is marked "3/15, net 30." The amount of the bill is $289.95. If you pay the bill within 15 da

Sonbull [250]

Answer:

$8,698.50

Step-by-step explanation:

289.95x30= $8,695.50

6 0

2 years ago