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kicyunya [14]
4 years ago
11

A) [40,100] B- [30,100] C- [16,100] D - [11,100]

Mathematics
2 answers:
Allisa [31]4 years ago
7 0

Hello from MrBillDoesMath!

Answer:

Choice C   [16,100]


Discussion:

Let "x" be the score needed on the 5th exam to achieve an average score >= 60.  Then

(50 + 90 + 66 + 78 + x)/5 >= 60  =>        (multiply both sides by 5)

(50 + 90 +66 + 78) + x >= 60*5 = 300 =>

(284) + x >= 300 =>

x >= 300 - 284 = 16.

As x is a percent, x <= 100, so the interval is [16, 100], which is Choice C.


Thank you,

MrB

liraira [26]4 years ago
7 0
Mthe correct answer is C
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3 years ago
Which of the following is NOT a requirement of testing a claim about two population means when 1 and 2 are unknown and not assum
andrey2020 [161]

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3 years ago
Here is some information about 26 houses.a,b and c are all different numbers.Number of bedrooms:1,2,3,4,5.Number of houses:7,a,b
kkurt [141]

Answer:

[a,b,c]=[4,2,5] or [a,b,c]=[2,4,5]

Step-by-step explanation:

Given

\begin{array}{cccccc}{Bedroom} & {1} & {2} & {3} & {4} & {5} \ \\ {Houses} & {7} & {a} & {b} & {c} & {8} \ \end{array}

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Median = 3.5

Required

Find a, b and c

Median is calculated as:

Median = \frac{n+1}{2}

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Median = \frac{27}{2}

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This means that the median is the average of the 13th and 14th item.

\begin{array}{cccccc}{Bedroom} & {1} & {2} & {3} & {4} & {5} \ \\ {Houses} & {7} & {a} & {b} & {c} & {8} \ \end{array}

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Since a, b and c are different, then

a and b cannot be 5 because c = 5

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a

So, possible sets are:

[a,b]=[2,4]

[a,b]=[4,2]

Include c, we have:

[a,b,c]=[4,2,5] or [a,b,c]=[2,4,5]

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3 years ago
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