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rosijanka [135]
2 years ago
12

3. Jeff Ruiz wants to purchase chocolates. The price of a 2-ounce box is $1.10, a 16-ounce box is $8.98, and a

Mathematics
1 answer:
bekas [8.4K]2 years ago
5 0

Jeff should purchase 25 , 2 ounce box and it will cost him $27.5 .

<h3>What is an ounce ?</h3>

Ounce is defined as 1/16 part of a pound.

It is given that

price of a 2-ounce box is $1.10,

a 16-ounce box is $8.98

a 32-ounce box is $17.98.

As Jeff wants to purchase total 48 ounce chocolates , the box which is more cheaper has to be determined  

The price of 1 ounce in a 2 ounce box is $1.1 /2 = $ 0.55

The price of 1 ounce in a 16 ounce box is $8.98 /16 = $ 0.56

The price of 1 ounce in a 32 ounce box is $17.98 /32 = $ 0.56

Therefore Jeff should purchase 25 , 2 ounce box and it will cost him $27.5 .

To know more about Ounce

brainly.com/question/26950819

#SPJ1

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Answer:

Step-by-step explanation:

12 - 12x = -6x + 48

Add 12x to both sides. That eliminates the x term on the left side.

12 = 6x + 48

Subtract 48 from both sides. That eliminates the constant term on the right side.

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3 years ago
g what is the 95% confidence interval a researcher wishes to compare the average amount of time spent in extracurricular
Serggg [28]

Complete question is;

A researcher wished to compare the average amount of time spent in extracurricular activities by high school students in a suburban school district with that in a school district of a large city. The researcher obtained a simple random sample of 60 high school students in a large suburban school district and found the mean time spent in extracurricular activities per week to be six hours with a standard deviation of three hours. The researcher also obtained an independent simple random sample of 40 high school students in a large city school district and found the mean time spent in extracurricular activities per week to be four hours with a standard deviation of two hours. Let x¯1 and x¯2 represent the mean amount of time spent in extracurricular activities per week by the populations of all high school students in the suburban and city school districts, respectively. Assume two-sample t procedures are safe to use?

what is the 95% confidence interval a researcher wishes to compare the average amount of time spent in extracurricular?

Answer:

CI = (0.755, 3.245)

Step-by-step explanation:

For SRS of 60;

Mean: x1¯ = 6

Standard deviation: s1 = 3

For SRS of 40;

Mean: x2¯ = 4

Standard deviation; s2 = 2

Critical value for the confidence interval of 95% is: t = 1.96

Formula for the CI is;

CI = (x¯1 - x¯2) ± t√[(s1²/n1) + ((s2)²/n1)]

Plugging in the relevant values, we have:

CI = (6 - 4) ± 1.96√[(3²/60) + ((4)²/40)]

CI = 2 ± 1.96√[(3²/60) + ((4)²/40)]

CI = 2 ± 1.96√0.55

CI = 2 ± 1.245

CI = [(2 - 1.245), (2 + 1.245)]

CI = (0.755, 3.245)

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