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frutty [35]
3 years ago
10

If Alex has 4 times as many dimes as nickels and they have a combined value of 270 cents, how many of

Mathematics
1 answer:
bija089 [108]3 years ago
5 0

Answer:

Nickel has 54 cents and Alex has 216 cents.

Step-by-step explanation:

Hope it helps u :)

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d.2/3

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48,62,81,94,27 to their nearest tens​
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the name Joe is very common at a school in one out of every ten students go by the name. If there are 15 students in one class,
kumpel [21]

Using the binomial distribution, it is found that there is a 0.7941 = 79.41% probability that at least one of them is named Joe.

For each student, there are only two possible outcomes, either they are named Joe, or they are not. The probability of a student being named Joe is independent of any other student, hence, the <em>binomial distribution</em> is used to solve this question.

<h3>Binomial probability distribution </h3>

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

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

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • One in ten students are named Joe, hence p = \frac{1}{10} = 0.1.
  • There are 15 students in the class, hence n = 15.

The probability that at least one of them is named Joe is:

P(X \geq 1) = 1 - P(X = 0)

In which:

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

P(X = 0) = C_{15,0}.(0.1)^{0}.(0.9)^{15} = 0.2059

Then:

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.2059 = 0.7941

0.7941 = 79.41% probability that at least one of them is named Joe.

To learn more about the binomial distribution, you can take a look at brainly.com/question/24863377

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A sandbox has an area of 32 square feet, and
Ksivusya [100]
The width is 8 because 4x8=32
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3 years ago
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