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lesya692 [45]
2 years ago
14

What is the area...?​

Mathematics
1 answer:
vesna_86 [32]2 years ago
4 0

here is your answer

HOPE IT HELPS

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A pair of shoes is on sale for 20% off the regular price of $72.50. How much will the customer pay with the 20% markdown?
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72.50/100 X 80 = 58

total cost with mark down $58

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ANSWER FAST PLZ 25 POINTS!!!!!!!!!!!!!!!
liberstina [14]

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44?

Step-by-step explanation:

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What is the Difference Between a 16-Ounce Brick and a Carpenter?
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In a randomly selected sample of 1169 men ages 35–44, the mean total cholesterol level was 210 milligrams per deciliter with a s
Aneli [31]

Answer:

The highest total cholesterol level a man in this 35–44 age group can have and be in the lowest 10% is 160.59 milligrams per deciliter.

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

\mu = 210, \sigma = 38.6

Find the highest total cholesterol level a man in this 35–44 age group can have and be in the lowest 10%.

This is the 10th percentile, which is X when Z has a pvalue of 0.1. So X when Z = -1.28.

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

-1.28 = \frac{X - 210}{38.6}

X - 210 = -1.28*38.6

X = 160.59

The highest total cholesterol level a man in this 35–44 age group can have and be in the lowest 10% is 160.59 milligrams per deciliter.

5 0
3 years ago
Past records indicate that the probability of online retail orders that turn out to be fraudulent is 0.08. Suppose that, on a gi
Sunny_sXe [5.5K]

Answer:

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

Step-by-step explanation:

We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.08.

The probability of k online retail orders that turn out to be fraudulent in the sample is:

P(x=k)=\dbinom{n}{k}p^k(1-p)^{n-k}=\dbinom{20}{k}\cdot0.08^k\cdot0.92^{20-k}

We have to calculate the probability that 2 or more online retail orders that turn out to be fraudulent. This can be calculated as:

P(x\geq2)=1-[P(x=0)+P(x=1)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.08^{0}\cdot0.92^{20}=1\cdot1\cdot0.189=0.189\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.08^{1}\cdot0.92^{19}=20\cdot0.08\cdot0.205=0.328\\\\\\\\P(x\geq2)=1-[0.189+0.328]\\\\P(x\geq2)=1-0.517=0.483

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

5 0
3 years ago
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