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3 years ago

Carrie is buying candy for a high school funndraiser. She buys 1000 bars for 350 what is the unit cost per candy bar

Mathematics

1 answer:

dem82 [27]3 years ago

3 0

Answer:

The cost is $0.35 per candy bar.

Step-by-step explanation:

To find the unit cost per candy bar, start with the overall cost and divide by the number of units (bars).

$350/1000 bars = $0.35 per candy bar.

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The Frosty Ice-Cream Shop sells sundaes for $2 and banana splits for $3. On a hot summer day, the shop sold twice as sundaes tha

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Answer:

i dont want to put the whole example cz like ur school might catch u so u gonna have to do the explainin part srry:( but the answear is 156$

Step-by-step explanation:

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3 years ago

An average person in the United States throws away 4.5 × 10² g of trash per day. In 2013, there were about 3.2×1063.2×10^6 peopl

satela [25.4K]

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3 years ago

45 is what percent of 60

STALIN [3.7K]

Answer:

75%

Step-by-step explanation:

$\frac{x}{100} = \frac{45}{60}$

45 x 100 = 4,500

4,500 divided by 60 = 75

7 0

3 years ago

Past records indicate that the probability of online retail orders

Tcecarenko [31]

Answer:

a) Mean = 1.6, standard deviation = 1.21

b) 18.87% probability that zero online retail orders will turn out to be fraudulent.

c) 32.82% probability that one online retail order will turn out to be fraudulent.

d) 48.31% probability that two or more online retail orders will turn out to be fraudulent.

Step-by-step explanation:

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.

The mean of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

In this problem, we have that:

$p = 0.08, n = 20$

a. What are the mean and standard deviation of the number of online retail orders that turn out to be fraudulent?

Mean

$E(X) = np = 20*0.08 = 1.6$

Standard deviation

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{20*0.08*0.92} = 1.21$

b. What is the probability that zero online retail orders will turn out to be fraudulent?

This is P(X = 0).

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

$P(X = 0) = C_{20,0}.(0.08)^{0}.(0.92)^{20} = 0.1887$

18.87% probability that zero online retail orders will turn out to be fraudulent.

c. What is the probability that one online retail order will turn out to be fraudulent?

This is P(X = 1).

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

$P(X = 1) = C_{20,1}.(0.08)^{1}.(0.92)^{19} = 0.3282$

32.82% probability that one online retail order will turn out to be fraudulent.

d. What is the probability that two or more online retail orders will turn out to be fraudulent?

Either one or less is fraudulent, or two or more are. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 1) + P(X \geq 2) = 1$

We want $P(X \geq 2)$

$P(X \geq 2) = 1 - P(X \leq 1)$

In which

$P(X \leq 1) = P(X = 0) + P(X = 1)$

From itens b and c

$P(X \leq 1) = 0.1887 + 0.3282 = 0.5169$

$P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.5169 = 0.4831$

48.31% probability that two or more online retail orders will turn out to be fraudulent.

4 0

4 years ago

Please help please A.68 B.39 C.73 D.107

lisov135 [29]

39 + 68 + x = 180
107 + x = 180
x = 73

your answer is actually C. 73

7 0

3 years ago