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goldenfox [79]
3 years ago
10

32 is 25% of x. Hurry please

Mathematics
2 answers:
Keith_Richards [23]3 years ago
5 0
32 = 0.25x
32/0.25 = x
x = 128
kondor19780726 [428]3 years ago
3 0
The answer is 128

hope that helps!
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Step-by-step explanation:

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What equals 28 in times tables?
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The following gives 28 in multiplication tables (for positive integers):

1x28
2x14
4x7
7x4
14x2

Of course you can go into -28x-1 etc. but I feel this is adequate. If you'd like more just say.
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The graph shows the functions f(x), p(x), and g(x):
Sidana [21]
The function f(x) is:

f(x)=x

This is because the line f(x) passes through the points (-1,-1), (0,0), (1,1) etc.

The function p(x) is:

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Whereby (m) is the slope and (C) is a constant.

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p(x)=-4/3x -7

Now, where p(x)=g(x), x=-6.

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p(x) and g(x) meet at (-6, 1) and the solution to p(x)=g(x) is x=-6.
3 0
3 years ago
If triangle ABC congruent triangle MNO, then what corresponding angle is congruent to angle N?
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B
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7 0
3 years ago
Read 2 more answers
Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro
Vinvika [58]

Answer:

40.1% probability that he will miss at least one of them

Step-by-step explanation:

For each target, there are only two possible outcomes. Either he hits it, or he does not. The probability of hitting a target is independent of other targets. 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.

0.95 probaiblity of hitting a target

This means that p = 0.95

10 targets

This means that n = 10

What is the probability that he will miss at least one of them?

Either he hits all the targets, or he misses at least one of them. The sum of the probabilities of these events is decimal 1. So

P(X = 10) + P(X < 10) = 1

We want P(X < 10). So

P(X < 10) = 1 - P(X = 10)

In which

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

P(X = 10) = C_{10,10}.(0.95)^{10}.(0.05)^{0} = 0.5987

P(X < 10) = 1 - P(X = 10) = 1 - 0.5987 = 0.401

40.1% probability that he will miss at least one of them

7 0
3 years ago
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