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2 years ago

Mathew walks 5/8 mile to the bus stop each morning.How far will he walk in 5 days

Mathematics

1 answer:

LenaWriter [7]2 years ago

7 0

Answer:

3 1/8 miles I think

Step-by-step explanation:

5/8 x 5/1 = 25/8

Divide 25 by 8 and you get 3 1/8

Add label of miles

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Avery had $25.56 in her wallet. If she bought lunch with dollars from her wallet, how much money did she have in her wallet afte

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$25.56

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3 years ago

Without calculating the sum, determine whether each answer will be rational or irrational.

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Irrational

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Step-by-step explanation:

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3 years ago

Help!!!Tony bought a new car at a cost of $20,000. The value of the car decreases by approximately 12% each year.

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Step-by-step explanation:

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3 years ago

Car colors preference change over the years and according to the particular model that the customer selects. In a recent year, 1

Margarita [4]

Answer:

99.05% probabilitiy that at most six cars are black.

Step-by-step explanation:

For each luxury car, there are only two possible outcomes. Either it is black, or it is not black. The probability of a luxury car being black is independent from other luxury cars. 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.

In this problem we have that:

$p = 0.1, n = 25$

If 25 cars of that year and type are randomly selected, find the probabilitiy that at most six cars are black.

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

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

$P(X = 0) = C_{25,0}.(0.1)^{0}.(0.9)^{25} = 0.0718$

$P(X = 1) = C_{25,1}.(0.1)^{1}.(0.9)^{24} = 0.1994$

$P(X = 2) = C_{25,2}.(0.1)^{2}.(0.9)^{23} = 0.2659$

$P(X = 3) = C_{25,3}.(0.1)^{3}.(0.9)^{22} = 0.2265$

$P(X = 4) = C_{25,4}.(0.1)^{4}.(0.9)^{21} = 0.1384$

$P(X = 5) = C_{25,5}.(0.1)^{5}.(0.9)^{20} = 0.0646$

$P(X = 6) = C_{25,6}.(0.1)^{6}.(0.9)^{19} = 0.0239$

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0718 + 0.1994 + 0.2659 + 0.2265 + 0.1384 + 0.0646 + 0.0239 = 0.9905$

99.05% probabilitiy that at most six cars are black.

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3 years ago