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3 years ago

6

A store donated cases of crayons to a day care center. Each case holds 24 boxes of crayons; each box holds 8 crayons. How many c

rayons did the center receive?

2 answers:

posledela3 years ago

4 0

Answer:

They recieved 192 crayons?

siniylev [52]3 years ago

4 0

Answer:

The Answer Is 192

Step-by-step explanation:

You might be interested in

What should we multiply the top equation by in order for both equations to have like coefficients?

ASHA 777 [7]

Answer:

Multiply by 2

Step-by-step explanation:

If we multiply 2x-6y=26 by 2, we get 4x-12y=52 and the x's cancel out, making it easier to solve for y, using the elimination process.

Hope this makes sense!

5 0

3 years ago

Mark tossed six balls while playing number game. Three balls landed in one section and three balls landed in another section, hi

Vlad1618 [11]

The answer is 3 for sure 100%sure

7 0

3 years ago

DECK is a parallelogram with m∠1 = 20°, m∠(<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^

maw [93]

9514 1404 393

Answer:

m∠2 = 16° or 25°

Step-by-step explanation:

The sum of angles 1 and 2 is congruent to angle CED.

20° +(x^2)° = (9x)°

x^2 -9x +20 = 0 . . . . . divide by °, put in standard form

(x -5)(x -4) = 0 . . . . . . . factor

x = 5 or 4

m∠2 = (x^2)° = (5^2)° or (4^2)°

m∠2 = 16° or 25°

7 0

2 years ago

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

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

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

2 years ago

Erin found 9 cents<br> each day for 7 days.<br> How much money did<br> she find in all?

kherson [118]

After 7 days she would have eventually found 63 cents.

8 0

2 years ago

Read 2 more answers