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Gemiola [76]
2 years ago
10

A company has a $150 budget to provide lunch for its 20 employees. The options are to provide either roast beef sandwiches, whic

h cost $5 apiece, or tuna sandwiches, which also cost $5 apiece. The company also wants to use the entire budget. Suppose r represents the number of roast beef sandwiches it provides and t represents the number of tuna sandwiches. Which statement is correct?
Mathematics
2 answers:
Shkiper50 [21]2 years ago
5 0

Answer:

i know im vv late but i dont want anyone to get it wrong. anyway if your on e2020 i believe the correct answer is

D. The company cannot provide lunch for all 20 employees and use the entire budget because there is no solution to the system of equations mc019-r+t=20 and 5r+5t=150

Step-by-step explanation:

Stella [2.4K]2 years ago
3 0

Answer:

D

Step-by-step explanation:

The company cannot provide lunch for all 20 employees and use the entire budget because there is no solution to the system of equations r + t = 20 and 5 r + 5 t = 150.

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A hospital finds that 22% of its accounts are at least 1 month in arrears. A random sample of 425 accounts was taken. What is th
GenaCL600 [577]

Answer:

8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears

Step-by-step explanation:

For each account, there are only two possible outcomes. Either they are at least 1 month in arrears, or they are not. The probability of an account being at least 1 month in arrears is independent from other accounts. So the binomial probability distribution is used to solve this question.

However, we are working with a large sample. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

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

Normal probability distribution

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}

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.22, n = 425

So

\mu = E(X) = np = 425*0.22 = 93.5

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{425*0.22*0.78} = 8.54

What is the probability that fewer than 82 accounts in the sample were at least 1 month in arrears

This probability is the pvalue of Z when X = 82. So

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

Z = \frac{82 - 93.5}{8.54}

Z = -1.35

Z = -1.35 has a pvalue of 0.0885.

8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears

8 0
2 years ago
The difference of two numbers is 20 and their product is 125, what is the answer?
brilliants [131]

find prime factors to get the inbetween numbers for product, and choose the one that has a difference of 20, and it should be 5 and 25, both requirements are met.

5 0
2 years ago
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The graph shows a fir tree's diameter growth over a 50 year period.
Brilliant_brown [7]

Answer:

The answer is B.) The diameter of the fir tree when planted and 20 inches

Step-by-step explanation:

The y-intercept is the initial diameter of the fir tree (10 inches). At the end of 50 years, the tree's diameter is 30 inches. Therefore, 30 − 10 = 20 inches of growth occurred over the 50 year period. The diameter of the fir tree when planted. The fir tree's diameter was 10 inches when it was planted.

6 0
2 years ago
Answer the following question. Click on symbol to choose correct answer. Given: R = {(x, y): y = -x^2} What is the range of R? y
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Answer: c

I answered c and got a 100 so it has to be it

3 0
2 years ago
According to the WHO MONICA Project the mean blood pressure for people in China is 128 mmHg with a standard deviation of 23 mmHg
Sati [7]

Answer:

a) Mean blood pressure for people in China.

b) 38.21% probability that a person in China has blood pressure of 135 mmHg or more.

c) 71.30% probability that a person in China has blood pressure of 141 mmHg or less.

d) 8.51% probability that a person in China has blood pressure between 120 and 125 mmHg.

e) Since Z when X = 135 is less than two standard deviations from the mean, it is not unusual for a person in China to have a blood pressure of 135 mmHg

f) 157.44mmHg

Step-by-step explanation:

When the distribution is normal, we use 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}

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If X is two standard deviations from the mean or more, it is considered unusual.

In this question:

\mu = 128, \sigma = 23

a.) State the random variable.

Mean blood pressure for people in China.

b.) Find the probability that a person in China has blood pressure of 135 mmHg or more.

This is 1 subtracted by the pvalue of Z when X = 135.

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

Z = \frac{135 - 128}{23}

Z = 0.3

Z = 0.3 has a pvalue of 0.6179

1 - 0.6179 = 0.3821

38.21% probability that a person in China has blood pressure of 135 mmHg or more.

c.) Find the probability that a person in China has blood pressure of 141 mmHg or less.

This is the pvalue of Z when X = 141.

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

Z = \frac{141 - 128}{23}

Z = 0.565

Z = 0.565 has a pvalue of 0.7140

71.30% probability that a person in China has blood pressure of 141 mmHg or less.

d.)Find the probability that a person in China has blood pressure between 120 and 125 mmHg.

This is the pvalue of Z when X = 125 subtracted by the pvalue of Z when X = 120. So

X = 125

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

Z = \frac{125 - 128}{23}

Z = -0.13

Z = -0.13 has a pvalue of 0.4483

X = 120

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

Z = \frac{120 - 128}{23}

Z = -0.35

Z = -0.35 has a pvalue of 0.3632

0.4483 - 0.3632 = 0.0851

8.51% probability that a person in China has blood pressure between 120 and 125 mmHg.

e.) Is it unusual for a person in China to have a blood pressure of 135 mmHg? Why or why not?

From b), when X = 135, Z = 0.3

Since Z when X = 135 is less than two standard deviations from the mean, it is not unusual for a person in China to have a blood pressure of 135 mmHg.

f.) What blood pressure do 90% of all people in China have less than?

This is the 90th percentile, which is X when Z has a pvalue of 0.28. So X when Z = 1.28. Then

X = 120

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

1.28 = \frac{X - 128}{23}

X - 128 = 1.28*23

X = 157.44

So

157.44mmHg

6 0
2 years ago
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