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anyanavicka [17]

3 years ago

9

This week, Sam spent 2 3/4 hours on his homework. This was 2 2/3 times less than last week. How many hours did he spend on homew

ork last week?
Show your work.

1 answer:

Semenov [28]3 years ago

3 0

Answer:

7.315 hours

Step-by-step explanation:

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The following data represent the monthly phone use, in minutes, of a customer enrolled in a fraud prevention program for the pas

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Answer:

The answer is "717.25 minutes".

Step-by-step explanation:

In this question first, we arrange the value in ascending order that are:

307,311,321,322,354,363,377,406,425,435,461,464,474,499,505,513,517,534,548

The median is always in an orderly position that is $= \frac{(n+1)}{2}$ .

$\to \frac{(n+1)}{2} = \frac{(20+1)}{2} = 10.5$ position orders

$10^{th} \ and \ 11^{th}$ place an average of observations so, the

$Average = \frac{(425 +435)}{2} = \frac{(860)}{2} =430$

Because as medium stands at 10.5 that median is as below 10 are greater than the level.

$Q_1$ often falls throughout the ordered BELOW average is $= \frac{(n+1)}{2}$ place.

$\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}$ ordered position

Average $5^{th} \ and \ 6^{th}$ location findings

$Q_1 = \frac{(354+363)}{2} = \frac{(717)}{2}= 358.5$

In $= \frac{(n+1)}{2}$ the ordered place, Q3 always falls ABOVE the median.

$\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}$ ordered At median ABOVE

Consequently $Q_3$ falls between $5^{th} \ and \ 6^{th}$ ABOVE the median position

$15^{th}\ and \ 16^{th}$ place average of observations

$\to Q_3 = \frac{(499+505)}{2}=\frac{(1004)}{2} = 502$

$\to IQR = Q_3 - Q_1= 502-358.5= 143.5$

$\to \text{Upper fence} = Q_3 + 1.5 \times IQR= 502 + 1.5 \times 143.5 = 717.25$

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some friends went to the local movie theater and bought four buckets of large popcorn and six boxes of candy. the total for the

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$4x+6y=46.50\\x+y=9.75\\\\4x+6y=46.50\\2x+3y=23.25$

$23.25

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Answer:

he answer is in the image

Step-by-step explanation:

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