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2 years ago

Winona asked randomly selected students in

Mathematics

1 answer:

MA_775_DIABLO [31]2 years ago

3 0

Sample space=S=240
N=75
Y=165

$\\ \sf\longmapsto P(N)=\dfrac{75}{240}=\dfrac{25}{48}\approx 2$

$\\ \sf\longmapsto P(Y)=\dfrac{165}{240}$

$\\ \sf\longmapsto P(Y)=\dfrac{33}{48}$

$\\ \sf\longmapsto P(Y)=\dfrac{11}{16}$

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Solve System Of Equation −4x+11y=15 x=2y x= y=

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Answer:

y=5 x=110

Step-by-step explanation:

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The sum of three numbers is 105. The third number is 2 times the first. The first number is 5 less than the second. What are the

Radda [10]

Answer:

The second number = 30

The first number = 25

The third number = 50

Step-by-step explanation:

Let

The second number = x

The first number = x - 5

The third number = 2(x - 5)

Total = 105

105 = x + (x - 5) + 2(x - 5)

105 = x + x - 5 + 2x - 10

105 = 4x - 15

105 + 15 = 4x

120 = 4x

x = 120/4

x = 30

The second number = x

= 30

The first number = x - 5

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= 25

The third number = 2(x - 5)

= 2(30 - 5)

= 2(25)

= 50

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3 years ago

4/5 equivalent fraction?

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4/5 = 8/10 = 12/15 = 16/20 = 20/25
Hope it helps

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The volume of

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1/10 of an ounce is the answer

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4 years ago

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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

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