What's the product of 21 and (6.3 times 10-4)

1 answer:

3 0

21(6.3 x 10 - 4). Remember PEMDAS. Everything done in parenthesis is done first.

Within the parenthesis, there is multiplication and subtraction. In PEMDAS, multiplication/division comes before addition/subtraction, so 6.3 x 10 = 63.
63 - 4 = 59
21(59) = 1,239.

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Solve for b2 in A = 1/2 h ( b1 + b2 ), if A = 16, h = 4, and b1 = 3.

Katarina [22]

Hello from MrBillDoesMath!

Answer:

b2 = 5

Discussion:

A = 1/2 h (b1 + b2) .

Substituting A = 16, h = 4, and b1=3 in the above formula gives:

16 = (1/2) (4)( 3 + b2) => (as (1/2)4 = 2) )

16 = 2 ( 3 + b2) => (divide both sides by 2)

8 = (3 + b2) => (subtract 3 from both sides)

8-3 = b2 =>

5 = b2

Check Area formula:

Does A = 16 = (1/2)(4)(3+5) ?

Does 16 = (1/2) (4)(8) ?

Does 16 = (1/2)(32) ? Yes it does so our calculation for b2 is correct

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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

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,

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