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3 years ago

Evaluate (–27)(–512)(–56) .

2 answers:

8 0

Since there are an odd number of negative signs here, the final result must be negative:

(negative)(27*512*56) = negative 774144 = -774144

seropon [69]3 years ago

4 0

(-27)(-512)(-56)

Final Result :

-774144

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Mr. Moretti is making welcome bags to give to the 18 students in his homeroom on the first day of school he puts m mechanical pe

Mashcka [7]

Answer:

The total number of pencils is 6

Step-by-step explanation:

According to the given scenario, the computation of the total number of pencils is as follows:

There are 18 students

And let us assume there is m mechanical pencils that put in each bag

also he put twice of the regular pencil

So if he put the 3 mechanical pencil

So, the total number of pencils is

= 3 × 2

= 6

As it twice of the regular pencil

hence, the total number of pencils is 6

5 0

3 years ago

PLEASE HELP ME!!!!!

Irina18 [472]

Answer:

f(6) = 1

Step-by-step explanation:

Function:

f(x) = -2/3x + 5

f(6) = -2/3(6) + 5

f(6) = -4 + 5

f(6) = 1

4 0

3 years ago

Read 2 more answers

Find the measure of side a

matrenka [14]

Answer:

8.22899353m

= 8m(round)

Step-by-step explanation:

14 * sin(36 degree) = 8.22899353

5 0

3 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,

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