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dangina [55]
4 years ago
9

Joe has created a budget and has allocated $250 per month for food. at the end of the month he finds that he actually spent $340

on food. this is:
Mathematics
1 answer:
Lilit [14]4 years ago
5 0
So he is over budget by $90 and he needs to allocate his bills to allow for it

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A rectangular lot is 115 yards long and 70 yards wide give the length and width of another rectangle lot that has the same perim
s2008m [1.1K]

Answer:165 length 20 width

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4 years ago
First come first serve<br> 1+1 equals what <br> you will get 40 points
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Answer: 1+1 = 2 :)

Step-by-step explanation:

7 0
3 years ago
Of the marbles in a bag, 4 are yellow, 5 are blue, and 4 are red. Sandra will randomly choose one
Tom [10]

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where is part b? Part A is 4/13

Step-by-step explanation:

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3 0
3 years ago
A random sample of 20 purchases showed the amounts in the table (in $). The mean is $49.57 and the standard deviation is $20.28.
son4ous [18]

Answer:

The 98​% confidence interval for the mean purchases of all​ customers is ($37.40, $61.74).

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.98}{2} = 0.01

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.01 = 0.99, so z = 2.325

Now, find M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 2.325*\frac{20.28}{\sqrt{20}} = 12.17

The lower end of the interval is the mean subtracted by M. So it is 49.57 - 12.17 = $37.40.

The upper end of the interval is the mean added to M. So it is 49.57 + 12.17 = $61.74.

The 98​% confidence interval for the mean purchases of all​ customers is ($37.40, $61.74).

3 0
3 years ago
The entry tickets at a community fair cost $5 for children and $10 for adults. On a certain day 1000 people we entered in the fa
Hitman42 [59]
580 kids
420 adults


hope this helps

brainliest plz
6 0
3 years ago
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