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Verizon [17]
3 years ago
5

Please help me with this

Mathematics
1 answer:
harkovskaia [24]3 years ago
5 0

Answer:

<u>The answer is 1. 64 and 2. 60</u>

Step-by-step explanation:

Question 7. 75% of what number is 48?

Number = x

75%x = 48

3x/4 = 48 (75% = 3/4)

3x = 192 (Multiplying by 4 at both sides of the equation)

<u>x = 64</u> (Dividing by 3 at both sides of the equation)

<u>Part 75 Whole 48 Part 100 Whole 64</u>

Question 8. 120% of what number is 72?

Number = x

120%x = 72

12x/10 = 72 (120% = 12/10)

12x = 720 (Multiplying by 10 at both sides of the equation)

<u>x = 60</u> (Dividing by 12 at both sides of the equation)

<u>Part 120 Whole 72 Part 100 Whole 60</u>

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If you roll two fair dice repeatedly, what is the probability that you will get a sum of 4 before you get a sum of 5 ? (a) (b) (
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Answer:

The probability that you will get a sum of 4 before you get a sum of 5 is \frac{1}{12}

Step-by-step explanation:

Given : If you roll two fair dice repeatedly.

To find : What is the probability that you will get a sum of 4 before you get a sum of 5 ?

Solution :

When two dice are rolled the outcomes are

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Total number of outcomes = 36

Favorable outcome get a sum of 4 before you get a sum of 5 is (1,3) ,(2,2) and (3,1) = 3

The probability that you will get a sum of 4 before you get a sum of 5 is

P=\frac{3}{36}

P=\frac{1}{12}

Therefore, The probability that you will get a sum of 4 before you get a sum of 5 is \frac{1}{12}

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Step-by-step explanation:

\dfrac{(4p^-4 q)^-2}{10pq^-3} =

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= \dfrac{p^{7} q}{160}

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Step-by-step explanation:

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