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Artist 52 [7]
2 years ago
10

A study was designed to investigate the effects of two variables ​(1) a​ student's level of mathematical anxiety and​ (2) teachi

ng method on a​ student's achievement in a mathematics course. Students who had a low level of mathematical anxiety were taught using the traditional expository method. These students obtained a mean score of with a standard deviation of on a standardized test. Assuming a​ mound-shaped and symmetric​ distribution, in what range would approximately ​% of the students​ score?
Mathematics
1 answer:
kicyunya [14]2 years ago
8 0

Answer:

Solution using the Empirical Rule, which is explained here.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

The question is incomplete, but you solve this using the Empirical Rule.

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3 years ago
Victoria searched her house for quarters. She started with 17 quarters and found 6 more quarters. She then goes to the arcade. I
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Answer:

11 games

Step-by-step explanation:

17+6=23

23/2=11.5

Victoria can play 11 games. You'd usually round up, but in this case you can't play half a game so she'll play 11 games and have one quarter leftover.

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The table shown below represents a function. Which of the following values could not be used to complete the table?
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4 0
3 years ago
The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean of 1262 and a s
Andrew [12]

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}

-0.643 = \frac{X - 1262}{118}

X - 1262 = -0.643*118

X = 1186

(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

X = 1493

Between 1031 and 1493.

​(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}

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

X - 1262 = -0.675*118

X = 1182

75th percentile:

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

7 0
3 years ago
Betty is having invitations printed for a holiday party. It costs $3.70 to set up the printing machine and $1.55 per invitation.
PIT_PIT [208]
Y = 1.55x + 3.70
1.55 x 150 = 232.5
232.5 + 3.70 = 236.2

The answer is C 236.20 hope that helped.
4 0
3 years ago
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