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kherson [118]
4 years ago
15

Inside a bag of candy there are 16 lollipops, 12 pieces of chocolate, and 7 pieces of licorice. If you reach in the bag and grab

a piece of candy, what is the probability that it will be either a piece of chocolate or a lollipop?
Mathematics
2 answers:
FinnZ [79.3K]4 years ago
5 0
28/35. Add the total amounts of candy (35) and then add the amounts of chocolate or lollipops (16+12 = 28) and then place over 35.

28/35
denpristay [2]4 years ago
4 0
The probability of getting a piece of chocolate or a lollipop would be \frac{28}{35}. You can find this by first adding all of the values of the amount of chocolate, lollipops, and licorice. Then, add the value of the amount of chocolate and lollipops.
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never [62]

Answer:

-3/4

Step-by-step explanation:

Perpendicular slopes are reciprocal opposites of the original slope. This means that the reciprocal of 4/3 is 3/4, and the opposite of 3/4 is -3/4.

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What is 32986538537 plus 78.9 times 5
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164,932,693,079.5

Step-by-step explanation:

To start you must add 78.9 to 32,986,538,537 which will give you 32,986,538,615.9. Then you must multiply 5.

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8 0
4 years ago
A Travel Weekly International Air Transport Association survey asked business travelers about the purpose for their most recent
siniylev [52]

Answer:

(a) P (p' > 0.25) = 0.

(b) P (0.15 < p' < 0.20) = 0.7784.

(c) P (133 < X < 171) = 0.2205

Step-by-step explanation:

Let <em>p</em> = proportion of business trip that were for  internal company visit.

It is provided that 19% responded that it was for an internal company visit, i.e.

<em>p</em> = 0.19.

A sample of <em>n</em> = 950 business travelers are randomly selected.

According to the Central limit theorem, if from an unknown population large samples of sizes <em>n</em> > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

 \mu_{\hat p}=p

The standard deviation of this sampling distribution of sample proportion is:

\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}

Thus, the distribution of \hat p is N (<em>μ</em> = 0.19, <em>σ </em>= 0.013).

(a)

Compute the probability that more than 25% of the business travelers say that the reason for their most recent business trip was an internal company visit as follows:

P (p' > 0.25) = P (Z > 4.62) = 0

*Use a <em>z</em>-table for the probability.

Thus, the probability that more than 25% of the business travelers say that the reason for their most recent business trip was an internal company visit is 0.

(b)

Compute the probability that between 15% and 20% of the business travelers say that the reason for their most recent business trip was an internal company visit as follows:

P (0.15 < p' < 0.20) = P (-3.08 < Z < 0.77)

                               = P (Z < 0.77) - P (Z < -3.08)

                               = 0.7794 - 0.0010

                               = 0.7784

*Use a <em>z</em>-table for the probability.

Thus, the probability that between 15% and 20% of the business travelers say that the reason for their most recent business trip was an internal company visit is 0.7784.

(c)

Compute the proportion of <em>X</em> = 133  and <em>X</em> = 171 as follows:

p' = 133/950 = 0.14

p' = 171/950 = 0.18

Compute the value of P (0.14 < p' < 0.20) as follows:

P (0.14 < p' < 0.18) = P (-3.85< Z < -0.77)

                               = P (Z < -0.77) - P (Z < -3.85)

                               = 0.2206- 0.0001

                               = 0.2205

*Use a <em>z</em>-table for the probability.

Thus, the probability that between 133 and 171 of the business travelers say that the reason for their most recent business trip was an internal company visit is 0.2205.

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Answer:

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