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3 years ago

PLZ HELP AGAIN DUE TODAY ANSWER CHOICES ARE A.140 B.188 C.132 D.104

Mathematics

1 answer:

spin [16.1K]3 years ago

6 0

A) I don’t know for sure but please let me know :)

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Can you help me do this Q please :)

goldenfox [79]

Answer:

i didnt get it

Step-by-step explanation:

can you explain

3 0

3 years ago

Y is between points p and b. pt= 15 and tb = 10. what is pb

Kruka [31]

PT = 15, and TB = 10

Take the 15 from PT, and add the 10 from TB. Which equals 25.

PB = 25

So, 25 is your answer.

Hope this helps! ☺

4 0

3 years ago

The sum of Pete's and Sam's ages is 30. Five years ago, Pete was 3 times as old as Sam. How old is Sam?

RUDIKE [14]

P-5=3s-15 because Pete is 20 and Sam is 10. 20-5= 15 and 3(10)-15 also equals 15. The answer is C.

4 0

3 years ago

Read 2 more answers

Eric's class consists of 12 males and 16 females. If 3 students are selected at random, find the probability that they

Reptile [31]

Answer:

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

Step-by-step explanation:

Let 'M' be the event of selecting males n(M) = 12

Number of ways of choosing 3 students From all males and females

$n(M) = 28C_{3} = \frac{28!}{(28-3)!3!} =\frac{28 X 27 X 26}{3 X 2 X 1 } = 3,276$

Number of ways of choosing 3 students From all males

$n(M) = 12C_{3} = \frac{12!}{(12-3)!3!} =\frac{12 X 11 X 10}{3 X 2 X 1 } =220$

The probability that all are male of choosing '3' students

$P(E) = \frac{n(M)}{n(S)} = \frac{12 C_{3} }{28 C_{3} }$

$P(E) = \frac{12 C_{3} }{28 C_{3} } = \frac{220}{3276}$

P(E) = 0.067 = 6.71%

<u><em>Final answer</em></u>:-

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

3 0

3 years ago

Rectangle poster is 3 times as long as wide perimeter is 24 in

Alexxandr [17]

w= width

l = 3w

P = 2(l+w)

substitute

24 = 2(3w+w)

24 = 2(4w)

24 =8w

3 =w

The width is 3 inches

the length is 9 inches

3 0

3 years ago