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Genrish500 [490]
3 years ago
10

Sam is hired for a 20-day period. On days that he works, he earns $60. For each day that he does not work, $30 is subtracted fro

m his earnings. At the end of the 20-day period, he received $660. How many days did he not work?A) 2B) 4C) 6D) 3
Mathematics
1 answer:
Alenkinab [10]3 years ago
5 0

Answer:

He did not work for 6 days

Step-by-step explanation:

Sam is hired for a 20-day period.

Let x be the no. of days he did not work

So, No. of days he worked = 20-x

he earns per working day = $60

So, he earns for (20-x) days = 60(20-x)

For each day that he does not work, $30 is subtracted from his earnings.

So, Amount deducted for x days = 30x

So, His earning for 20 days =60(20-x)-30x

We are given that At the end of the 20-day period, he received $660

So, 60(20-x)-30x=660

1200-60x-30x=660

1200-90x=660

1200-660=90x

540=90x

\frac{540}{90}=x

6=x

So, he did not work for 6 days

So, Option C is true

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Answer:

a. 0.691

b. 0.382

c. 0.933

d. $88.490

e. $58.168

f. 5th percentile: $42.103

95th percentile: $107.897

Step-by-step explanation:

We have, for the purchase amounts by customers, a normal distribution with mean $75 and standard deviation of $20.

a. This can be calculated using the z-score:

z=\dfrac{X-\mu}{\sigma}=\dfrac{85-75}{20}=\dfrac{10}{20}=0.5\\\\\\P(X

The probability that a randomly selected customer spends less than $85 at this store is 0.691.

b. We have to calculate the z-scores for both values:

z_1=\dfrac{X_1-\mu}{\sigma}=\dfrac{65-75}{20}=\dfrac{-10}{20}=-0.5\\\\\\z_2=\dfrac{X_2-\mu}{\sigma}=\dfrac{85-75}{20}=\dfrac{10}{20}=0.5\\\\\\\\P(65

The probability that a randomly selected customer spends between $65 and $85 at this store is 0.382.

c. We recalculate the z-score for X=45.

z=\dfrac{X-\mu}{\sigma}=\dfrac{45-75}{20}=\dfrac{-30}{20}=-1.5\\\\\\P(X>45)=P(z>-1.5)=0.933

The probability that a randomly selected customer spends more than $45 at this store is 0.933.

d. In this case, first we have to calculate the z-score that satisfies P(z<z*)=0.75, and then calculate the X* that corresponds to that z-score z*.

Looking in a standard normal distribution table, we have that:

P(z

Then, we can calculate X as:

X^*=\mu+z^*\cdot\sigma=75+0.67449\cdot 20=75+13.4898=88.490

75% of the customers will not spend more than $88.49.

e. In this case, first we have to calculate the z-score that satisfies P(z>z*)=0.8, and then calculate the X* that corresponds to that z-score z*.

Looking in a standard normal distribution table, we have that:

P(z>-0.84162)=0.80

Then, we can calculate X as:

X^*=\mu+z^*\cdot\sigma=75+(-0.84162)\cdot 20=75-16.8324=58.168

80% of the customers will spend more than $58.17.

f. We have to calculate the two points that are equidistant from the mean such that 90% of all customer purchases are between these values.

In terms of the z-score, we can express this as:

P(|z|

The value for z* is ±1.64485.

We can now calculate the values for X as:

X_1=\mu+z_1\cdot\sigma=75+(-1.64485)\cdot 20=75-32.897=42.103\\\\\\X_2=\mu+z_2\cdot\sigma=75+1.64485\cdot 20=75+32.897=107.897

5th percentile: $42.103

95th percentile: $107.897

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