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nalin [4]
2 years ago
11

The five members of the Traynor family each buy train tickets. During the train ride, each family member buys a boxed lunch for

$6.50. If the total cost of the trip is $248.50, what is the price of each train ticket? Enter your answer in the box.
Mathematics
2 answers:
Mandarinka [93]2 years ago
8 0
Each ticket is $43.20.......
Yuri [45]2 years ago
5 0
So 5(6.50) = 32.5
248.5 - 32.5 = 216
216/5 = 43.2
each ticket is $43.20
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garri49 [273]
It equals 96. hope that helped
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The distance between Cincinnati, Ohio, and charlotte, North Carolina, is about 336 miles.The distance between Cincinnati and Chi
bija089 [108]
Cincinnati Ohio à<span> Charlote North CA = a total of 336 miles
Cincinnati </span>à<span> Chicago Illinois = a total of 247 miles
Perry drove from Charlote to Chicago by passing Cincinati. Find the distance she drove.
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=> 336 miles + 247 Miles
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Penny drove for a total of 583 miles from Charlotte to Cincinnati to Chicago.

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8 0
2 years ago
A painter needs 5 gallons of paint to finish a house. He has 3 quarts and 1 pint. How much more paint does he need?
NARA [144]

Answer:One pint and 16 quarts or one pint and four gallons

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3 0
2 years ago
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A dog groomer is offering 20% off a full groom for one dog, and 15% off for each additional dog. You decide to take your two fur
jarptica [38.1K]

Answer:the total discounted price for two dogs to be groomed is 22.75

Step-by-step explanation:

A dog groomer is offering 20% off a full groom for one dog, and 15% off for each additional dog. Normally, a full groom costs $65 for each dog.

If you decide to take your two furry friends to be groomed,

the discount on the first one would be

20/100 × 65 = 0.2 × 65 = 13

the discount on the second one would be

10/100 × 65 = 0.15 × 65 = 9.75

the total discounted price for two dogs to be groomed would be

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3 0
2 years ago
A 2012 Gallup survey of a random sample of 1014 American adults indicates that American families spend on average $151 per week
stiv31 [10]

Answer:

a.[144;158]$

b. The parameter estimated is the population mean of the weekly expenses on the food of American families.

Step-by-step explanation:

Hello!

The study variable for this problem is "Weekly expenses on food spent by an American family"($)

For a sample n=1014 American families the corresponding sample mean is x[bar]=$151

It is stated that for the 95%CI the margin of error is ±$7

a.

The confidence interval that estimates the population mean is constructed as the "estimator ± margin of error" for example, under a Standard Normal distribution the formula for the confidence interval is:

[x[bar]±Z_{\alpha/2}*(δ/√n)]

For the sample taken it is:

[151±7]

The inferior limit is 144$

The superior limit is 158$

b.

The parameter estimated is the population mean of the weekly expenses on the food of American families.

The confidence level of an interval is the probability under which it is built. This probability indicates  that if they build 100 confidence intervals, we expect 95 to contain the value of the population mean we are trying to estimate.

I hope you have a SUPER day!

3 0
2 years ago
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