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3 years ago

A case holds 48 boxes of candy and costs $24. How much would 2 % cases of candy cost?

Mathematics

1 answer:

Tatiana [17]3 years ago

5 0

Answer:

48 cents

Step-by-step explanation:

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Find and simplify the difference quotient of the form<img src="https://tex.z-dn.net/?f=%5Cfra%5Cfrac%7Bf%28x%29-f%28a%29%7D%7Bx-

Ilia_Sergeevich [38]

Answer: 3x + 13

Step-by-step explanation:

$f(x)=3x^2+x\\f(a)=3a^2+a\\f(4)=3*4^2+4=52\\\\f(x)-f(4)=3x^2+x-52=(x-4)(3x+13)\\\\\dfrac{f(x)-f(4)}{x-4} =\dfrac{3x^2+x-52}{x-4} \\\\if\ x \neq 4\ then \\\\\dfrac{f(x)-f(4)}{x-4} =\dfrac{(x-4)(3x+13)}{x-4} \\\\=3x+13\\$

6 0

3 years ago

Read 2 more answers

1.Where is the least change in the amount of money earned?

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30 or 3 and 4 weeks hope this helps

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3 years ago

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Simplify the expression using the distributive property. –4(–3x + 11) 1. Distribute the –4 through the parentheses: -4(–3x) + (-

gayaneshka [121]

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3 years ago

A bag contains 7 pieces of paper numbered 1 to 7. P(2)=. Is

pochemuha

Answer:

$P (2) =\frac{1}{7}$ Theoretical probability

Step-by-step explanation:

The theoretical probability is defined as:

$P = \frac{number\ of\ desired\ results}{number\ of\ possible\ results}$

In this case we look for the probability of taking a 2 out of the bag. As there is only one paper with the number 2 in the bag then:

number of desired results = 1

The amount of paper in the bag is equal to 7, so:

number of possible results = 7

Thus:

$P (2) =\frac{1}{7}$

This is a theoretical probability, since we do not need to perform the experiment to calculate the probability.

To calculate the experimental probability we must perform the following experiment:

Take a paper out of the bag, record the number obtained and then return the paper to the bag.

Now repeat this experiment n times. (Perform n trials)

So:

$P (2) = \frac{number\ of\ times\ you\ obtained\ the\ number\ 2}{number\ of\ trials\ performed}$

To calculate a theoretical probability you always need to perform an experiment with n trials.

5 0

3 years ago

A recent study suggested that 70% of all eligible voters will vote in the next presidential election. Suppose 20 eligible voters

natita [175]

Answer:

0.0479 = 4.79% probability that fewer than 11 of them will vote

Step-by-step explanation:

For each voter, there are only two possible outcomes. Either they will vote, or they will not. The probability of a voter voting is independent of any other voter, which means that the binomial probability distribution is used 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.

70% of all eligible voters will vote in the next presidential election.

This means that $p = 0.7$

20 eligible voters were randomly selected from the population of all eligible voters.

This means that $n = 20$

What is the probability that fewer than 11 of them will vote?

This is:

$P(X < 11) = P(X = 10) + P(X = 9) + P(X = 8) + P(X = 7) + P(X = 6) + P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0)$

In which

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

$P(X = 10) = C_{20,10}.(0.7)^{10}.(0.3)^{10} = 0.0308$

$P(X = 9) = C_{20,9}.(0.7)^{9}.(0.3)^{11} = 0.0120$

$P(X = 8) = C_{20,8}.(0.7)^{8}.(0.3)^{12} = 0.0039$

$P(X = 7) = C_{20,7}.(0.7)^{7}.(0.3)^{13} = 0.0010$

$P(X = 6) = C_{20,10}.(0.7)^{6}.(0.3)^{12} = 0.0002$

$P(X = 5) = C_{20,5}.(0.7)^{5}.(0.3)^{15} \approx 0$

The probability of 5 or less voting is very close to 0, so they will not affect the outcome. Then

$P(X < 11) = P(X = 10) + P(X = 9) + P(X = 8) + P(X = 7) + P(X = 6) + P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0) = 0.0308 + 0.0120 + 0.0039 + 0.0010 + 0.0002 = 0.0479$

0.0479 = 4.79% probability that fewer than 11 of them will vote

8 0

3 years ago