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Snezhnost [94]
3 years ago
15

Can someone help me please ?

Mathematics
1 answer:
Oliga [24]3 years ago
7 0
I think that it might be "c" but I am not sure but hey it is worth a shot sorry if I am wrong!
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Estimate each percent 19% of $53
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19 percent of 53 is 10.07
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Ms. Turner is buying deli meat to make sandwiches for a class picnic. She buys 14 pounds of roast beef for $51.10. She also buys
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RB: $51.10/14=$3.65 per lb
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Ham: $3.80 per lb

Answer: Turkey is the LEAST expensive
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On monday kayla borrowed $15 from his mom on Tuesday he borrowed another $10 what is the equation to show how much money kayla b
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Answer:

$35

Step-by-step explanation:

Monday: $15

Tuesday: $10

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3 years ago
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The average demand for rental skis on winter Saturdays at a particular area is 150 pairs, which has been quite stable over time.
pentagon [3]

Answer:

a) 15.87% probability that demand will exceed this supply on any winter Saturday

b) The shop needs to have 197 pairs of skis in stock.

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 = 150, \sigma = 20

(a) The rental shop stocks 170 pairs of skis. What is the probability that demand will exceed this supply on any winter Saturday?

This probability is 1 subtracted by the pvalue of Z when X = 150. So

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

Z = \frac{170 - 150}{20}

Z = 1

Z = 1 has a pvalue of 0.8413

1 - 0.8413 = 0.1587

15.87% probability that demand will exceed this supply on any winter Saturday

(b) How many pairs of skis in stock does the shop have to have to make the probability in question (a) less than .01?

This is X when Z has a pvalue of 0.99.

So X when Z = 2.33

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

2.33 = \frac{X - 150}{20}

X - 150 = 2.33*20

X = 196.6

Rounding up, since the number of pairs is a discrete number

The shop needs to have 197 pairs of skis in stock.

6 0
3 years ago
8x subtract 2 equals negative 9 plus 7x
ohaa [14]
The answer to this equation is x = 11 You can usually get answer to questions like these at https://www.mathpapa.com/algebra-calculator.html
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