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sergey [27]
3 years ago
15

What is 9 of 25% in the form of a whole number.

Mathematics
1 answer:
Yuliya22 [10]3 years ago
8 0
I believe you are trying to say 25% of a number is 9, what is the number? in that case the answer is 36. just divide 9 by .25
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2/6 +36 Whats the answer ?​
777dan777 [17]

36 2/6

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4 0
2 years ago
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A birthday cake was cut into equal pieces, and four pieces were eaten. The fraction is 3/7 how many are left
GaryK [48]
3 are left.
3/7 means there were 7 slices, but it's missing 4 slices. (Eaten)

i hope this helps :)

7 0
3 years ago
24*43x22*8x67x890x10*3x6*1x8*1
lukranit [14]

Answer:

1032x176x67x890x30x6x8

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4 0
2 years ago
In a study of 100 new cars, 29 are white. Find and g, where
sleet_krkn [62]

Question

<em>In a study of 100 new cars, 29 are white. Find p and q , where p is the proportion of new cars that are white. </em>

Answer:

p = 0.29  and q = 0.71

Step-by-step explanation:

Given

Total new cars =  100

White new cars = 29

Required

Determine p and q

From the question;

<em>p represents white new cars</em>

Hence;

p = 29

Note that;

p + q = 100

Substitute 29 for p

29 + q = 100

29 - 29 + q = 100 - 29

q = 100 - 29

q = 71

The proportion of p is calculate by dividing p by the total number of new cars (Same process is done for q)

For proportion of p

Proportion,\ p = \frac{p}{new\ cars}

Proportion,\ p = \frac{29}{100}

Proportion,\ p = 0.29

For proportion of q

Proportion,\ q = \frac{q}{new\ cars}

Proportion,\ q = \frac{71}{100}

Proportion,\ q = 0.71

6 0
3 years ago
Suppose you decided to keep flipping a coin until tails came up, at which
MAVERICK [17]

Answer:

B. 6.3%

Step-by-step explanation:

For each time that the coin is tosse, there are only two possible outcomes. Either it comes up tails, or it does not. The probability of coming up tails on a toss is independent of any other toss. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Fair coin:

Equally as likely to come up heads or tails, so p = 0.5

Probability that the first tails comes up on the 4th flip of the coin?

0 tails during the first three, which is P(X = 0) when n = 3.

Tails in the fourth, with probability 0.5. So

p = 0.5P(X = 0)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

p = 0.5P(X = 0) = 0.5*(C_{3,0}.(0.5)^{0}.(0.5)^{3}) = 0.0625

0.0625 * 100 = 6.25%

Rounding to the nearest tenth of a percent, the correct answer is:

B. 6.3%

4 0
2 years ago
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