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Allushta [10]
2 years ago
9

COFFEE SHOP Chenoa wants to stop for coffee on her way to school. The distance from Chenoa’s house to the coffee shop is 3 miles

more than twice the distance from the coffee shop to Chenoa’s school. The total distance from Chenoa’s house to her school is 5 times the distance from the coffee shop to her school.
brainlist+ty+5 stars+points!

Mathematics
2 answers:
insens350 [35]2 years ago
6 0

Chenoa wants to stop for coffee on her way to school. The distance from Chenoa's house to the coffee shop is 3 miles more than twice the distance from the coffee shop to Chenoa's school. The total distance from Chenoa's house to her school is 5 times the distance from the coffee shop to her school.

a. What is the distance from Chenoa's house to the coffee shop? Write your answer as a decimal, if necessary.

b. What assumptions did you make when solving this problem?

ipn [44]2 years ago
5 0

Answer:

Chenoa wants to stop for coffee on her way to school. The distance from Chenoa's house to the coffee shop is 3 miles more than twice the distance from the coffee shop to Chenoa's school. The total distance from Chenoa's house to her school is 5 times the distance from the coffee shop to her school.

a. What is the distance from Chenoa's house to the coffee shop? Write your answer as a decimal, if necessary.

b. What assumptions did you make when solving this problem?

Explanation

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Nemecek Brothers make a single product on two separate production lines, A and B. Its labor force is equivalent to 1000 hours pe
frez [133]

Answer:

(a) The inequality for the number of items, x, produced by the labor, is given as follows;

250 ≤ x ≤ 600

(b) The inequality for the cost, C is $1,000 ≤ C ≤ $3,000

Step-by-step explanation:

The total time available for production = 1000 hours per week

The time it takes to produce an item on line A = 1 hour

The time it takes to produce an item on line B = 4 hour

Therefore, with both lines working simultaneously, the time it takes to produce 5 items = 4 hours

The number of items produced per the weekly labor = 1000/4 × 5 = 1,250 items

The minimum number of items that can be produced is when only line B is working which produces 1 item per 4 hours, with the weekly number of items = 1000/4 × 1 = 250 items

Therefore, the number of items, x, produced per week with the available labor is given as follows;

250 ≤ x ≤ 1250

Which is revised to 250 ≤ x ≤ 600 as shown in the following answer

(b) The cost of producing a single item on line A = $5

The cost of producing a single item on line B = $4

The total available amount for operating cost = $3,000

Therefore, given that we can have either one item each from lines A and B with a total possible item

When the minimum number of possible items is produced by line B, we have;

Cost = 250 × 4 = $1,000

When the maximum number of items possible, 1,250, is produced, whereby we have 250 items produced from line B and 1,000 items produced from line A, the total cost becomes;

Total cost = 250 × 4 + 1000 × 5 = 6,000

Whereby available weekly outlay = $3000, the maximum that can be produced from line A alone is therefore;

$3,000/$5 = 600 items = The maximum number of items that can be produced

The inequality for the cost, C, becomes;

$1,000 ≤ C ≤ $3,000

The time to produce the maximum 600 items on line A alone is given as follows;

1 hour/item × 600 items = 600 hours

The inequality for the number of items, x, produced by the labor, is therefore, given as follows;

250 ≤ x ≤ 600

8 0
3 years ago
The College Boards, which are administered each year to many thousands of high school students, are scored so as to yield a mean
Marysya12 [62]

Answer:

a) 15.87% of the scores are expected to be greater than 600.

b) 2.28% of the scores are expected to be greater than 700.

c) 30.85% of the scores are expected to be less than 450.

d) 53.28% of the scores are expected to be between 450 and 600.

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 = 500, \sigma = 100

a. Greater than 600

This is 1 subtracted by the pvalue of Z when X = 600. So

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

Z = \frac{600 - 500}{100}

Z = 1

Z = 1 has a pvalue of 0.8413.

1 - 0.8413 = 0.1587

15.87% of the scores are expected to be greater than 600.

b. Greater than 700

This is 1 subtracted by the pvalue of Z when X = 700. So

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

Z = \frac{700 - 500}{100}

Z = 2

Z = 2 has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% of the scores are expected to be greater than 700.

c. Less than 450

Pvalue of Z when X = 450. So

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

Z = \frac{450 - 500}{100}

Z = -0.5

Z = -0.5 has a pvalue of 0.3085.

30.85% of the scores are expected to be less than 450.

d. Between 450 and 600

pvalue of Z when X = 600 subtracted by the pvalue of Z when X = 450. So

X = 600

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

Z = \frac{600 - 500}{100}

Z = 1

Z = 1 has a pvalue of 0.8413.

X = 450

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

Z = \frac{450 - 500}{100}

Z = -0.5

Z = -0.5 has a pvalue of 0.3085.

0.8413 - 0.3085 = 0.5328

53.28% of the scores are expected to be between 450 and 600.

6 0
2 years ago
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