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zubka84 [21]
3 years ago
8

5.23y+3·(2.7−1.6y) simplify

Mathematics
1 answer:
vampirchik [111]3 years ago
5 0

Answer:

0.43y + 8.1

Step-by-step explanation:

5.23y + 8.1 - 4.8y

5.23y - 4.8y + 8.1

0.43y +8.1

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In a random sample of 18 families, the average weekly food expense was $95.60 with a sample standard deviation of $22.50. Determ
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Answer:

t-distribution should be used to construct a confidence interval.

Step-by-step explanation:

We are given that a random sample of 18 families, the average weekly food expense was $95.60 with a sample standard deviation of $22.50.

We have to determine whether a normal distribution (Z values) or a t- distribution should be used or whether neither of these can be used to construct a confidence interval.

<em><u>Since in this question we are provided with;</u></em>

Sample average weekly food expense, \bar X = $95.60

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