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Naily [24]
3 years ago
5

Mrs. Dodson has a jar full of pennies and quarters. There are 120 coins in the jar worth a

Mathematics
1 answer:
Anton [14]3 years ago
5 0

<em><u>Answer: 10.</u></em>

<em><u /></em>

<em><u>Step-by-step explanation:</u></em>

<em><u>DIVISION. 120/12.</u></em>

<em><u>That is hard! Here's a trick.</u></em>

<em><u>12x9= 99=9=108. Or 12x6=66+6=72.</u></em>

<em><u>We will stick with 12x9,  that equals 108.</u></em>

<em><u>We should add another 12 until we get to 120. </u></em>

<em><u>108+12=120. That is 10 times. The trick was...</u></em>

<em><u>dividing by 12 is hard, so lets say 12x4. I can put 4 and 4 together.</u></em>

<em><u>44 then and another 4. That is 48. So 12x4=48.</u></em>

<em><u /></em>

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The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean of 1262 and a s
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Answer:

a) 1186

b) Between 1031 and 1493.

c) 160

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with mean of 1262 and a standard deviation of 118.

This means that \mu = 1262, \sigma = 118

a) Determine the 26th percentile for the number of chocolate chips in a bag. ​

This is X when Z has a p-value of 0.26, so X when Z = -0.643.

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

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X - 1262 = -0.643*118

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(b) Determine the number of chocolate chips in a bag that make up the middle 95% of bags.

Between the 50 - (95/2) = 2.5th percentile and the 50 + (95/2) = 97.5th percentile.

2.5th percentile:

X when Z has a p-value of 0.025, so X when Z = -1.96.

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

-1.96 = \frac{X - 1262}{118}

X - 1262 = -1.96*118

X = 1031

97.5th percentile:

X when Z has a p-value of 0.975, so X when Z = 1.96.

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

1.96 = \frac{X - 1262}{118}

X - 1262 = 1.96*118

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​(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip​ cookies?

Difference between the 75th percentile and the 25th percentile.

25th percentile:

X when Z has a p-value of 0.25, so X when Z = -0.675.

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

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X when Z has a p-value of 0.75, so X when Z = 0.675.

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

0.675 = \frac{X - 1262}{118}

X - 1262 = 0.675*118

X = 1342

IQR:

1342 - 1182 = 160

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