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2 years ago

5

A flower vendor sells roses for 50 cents each. How much does she pay per flower is she makes $6.00 on every twenty dollars worth

sold?

1 answer:

Lelu [443]2 years ago

3 0

One rose is 50 cents, so 2 roses cost $1 ( 50 cents x 2).

2 roses per dollar x 20 dollars = 40 total roses sold.

When they sell $20 dollars they make $6, so that means they pay 20-6 = $14 dollars for the 40 roses.

$14 / 40 roses = 0.35 per rose.

She pays 35 cents per rose.

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A company makes a product and has no way to determine which ones are faulty until an unhappy customer returns it. Three percent

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<span>(x) - (0.03)(200) = 2.00
Let x be the </span><span>company charge to make the profit

solving for x gives 8$</span>

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2 years ago

Read 2 more answers

Alonso went to the market with \$55$55dollar sign, 55 to buy eggs and sugar. He knows he needs a package of 121212 eggs that cos

eduard

Answer:

2.75+11.50S $\leq$ 55

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Step-by-step explanation:

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2 years ago

The aquarium at Sea Critters Depot contains 140 fish. Eighty of these fish are green swordtails (44 female and 36 male) and 60 a

maxonik [38]

Given that:

Total number of fish = 140

Fish are green swordtails female = 44

Fish are green swordtails male = 36

Fish are orange swordtails female = 36

Fish are orange swordtails male = 24

Solution:

A. We have to find the probability that the selected fish is a green swordtail.

$\text{P(green swordtail)}=\dfrac{\text{Total green swordtail fish}}{\text{Total fish}}$

$\text{P(green swordtail)}=\dfrac{80}{140}$

$\text{P(green swordtail)}=\dfrac{4}{7}$

Therefore, the probability that the selected fish is a green swordtail is $\dfrac{4}{7}.$

B. We have to find the probability that the selected fish is male.

$\text{P(Male fish)}=\dfrac{\text{Total male fish}}{\text{Total fish}}$

$\text{P(Male fish)}=\dfrac{36+24}{140}$

$\text{P(Male fish)}=\dfrac{60}{140}$

$\text{P(Male fish)}=\dfrac{3}{7}$

Therefore, the probability that the selected fish is a male, is $\dfrac{3}{7}.$

C. We have to find the probability that the selected fish is a male green swordtail.

$\text{P(Male green swordtail)}=\dfrac{\text{Total male green swordtail fish}}{\text{Total fish}}$

$\text{P(Male green swordtail)}=\dfrac{36}{140}$

$\text{P(Male green swordtail)}=\dfrac{9}{35}$

Therefore, probability that the selected fish is a male green swordtail is $\dfrac{9}{35}.$

D.

We have to find the probability that the selected fish is either a male or a green swordtail.

$\text{P(Male or green swordtail)}=\dfrac{\text{Total male or green swordtail fish}}{\text{Total fish}}$

$\text{P(Male or green swordtail)}=\dfrac{44+36+24}{140}$

$\text{P(Male or green swordtail)}=\dfrac{96}{140}$

$\text{P(Male or green swordtail)}=\dfrac{24}{35}$

Therefore, the probability the selected fish is either a male or a green swordtail is $\dfrac{24}{35}.$

4 0

2 years ago

The length of time a full length movie runs from opening to credits is normally distributed with a mean of 1.9 hours and standar

Llana [10]

Answer:

a) The probability that a random movie is between 1.8 and 2.0 hours = 0.2586.

b) The probability that a random movie is longer than 2.3 hours is 0.0918.

c) The length of movie that is shorter than 94% of the movies is 1.4 hours

Step-by-step explanation:

In the above question, we would solve it using z score formula

z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation

a) A random movie is between 1.8 and 2.0 hours

z = (x-μ)/σ,

x1 = 1.8,

x2 = 2.0

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z1 = (1.8 - 1.9)/0.3

z1 = -1/0.3

z1 = -0.33333

Using the z score table

P(z1 = -0.33) = 0.3707

z2 = (2.0 - 1.9)/0.3

z1 = 1/0.3

z1 = 0.33333

p(z2 = 0.33) = 0.6293

= P(- 0.33 ≤ z ≤ 0.33)

= 0.6293 - 0.3707

= 0.2586

The probability that a random movie is between 1.8 and 2.0 hours = 0.2586

b) A movie is longer than 2.3 hours

z = (x-μ)/σ,

x1 = 2.3

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z = (2.3 - 1.9)/0.3

z = 4/0.3

z = 1.33333

P(z = 1.33) = 0.90824

P(x>2.3) = = 1 - 0.90824

= 0.091759

≈ 0.0918

The probability that a random movie is longer than 2.3 hours is 0.0918.

3) The length of movie that is shorter than 94% of the movies.

z = (x-μ)/σ

Probability (z ) = 94% = 0.94

Movie that is shorter than 0.94

= P(1 - 0.94) = P(0.06)

Finding the P (x< 0.06) = -1.555

≈ -1.56

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

-1.56 = (x - 1.9)/ 0.3

Cross multiply

-1.56 × 0.3 = x - 1.9

- 0.468 + 1.9 = x

= 1.432 hours

≈ 1.4 hours

Therefore, the length of movie that is shorter than 94% of the movies is 1.4 hours

5 0

1 year ago

The Census Bureau reports that 82% of Americans over the age of 25 are high school graduates. A survey of randomly selected resi

SVETLANKA909090 [29]

Answer:

a) Mean = 1030; Standard deviation = 12.38.

b) The county result is unusually high.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

(a) Find the mean and standard deviation for the number of high school graduates in groups of 1210 Americans over the age of 25.

This first question is a binomial propability distribution.

We have a sample of 1210 Amricans, so $n = 1210$ .

The mean of the sample is 1030.

The probability of a success is $\pi = \frac{1030}{1210} = 0.8512$ .

The standard deviation of the sample is $s = \sqrt{n\pi(1-\pi)} = \sqrt{1210*0.8512*0.1488} = 12.38$

(b) Is that county result of 1030 unusually high, or low, or neither?

The first step is find the zscore when $X = 1030$ .

Then we find the pvalue of this zscore.

If this pvalue is bigger than 0.95, the county result is unusually high.

If this pvalue is smaller than 0.05, the county result is unusually low.

Otherwise, it is neither.

The national mean is 82%. So,

$\mu = 0.82(1210) = 992.2$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1030 - 992.2}{12.38}$

$Z = 3.05$

$Z = 3.05$ has a pvalue of 0.9989.This means that the county result is unusually high.

4 0

2 years ago