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pentagon [3]
3 years ago
13

Jessie received the following scores on her math tests this year. (45, 65, 70, 80, 85, 100) Suppose the teacher removes the lowe

st and highest scores. (65, 70, 80, 85) What are the interquartile ranges of Jessie’s original scores and her new scores?
Mathematics
2 answers:
notsponge [240]3 years ago
7 0
Original scores :
45,65,70,80,85,100
IQR = 85 - 65 = 20

new scores :
65,70,80,85
IQR = (80 + 85)/2 - (65 + 70)/2 = 82.5 - 67.5 = 15
vesna_86 [32]3 years ago
7 0

Answer:

Interquartile ranges of Jessie’s original scores = 20

Interquartile ranges of Jessie’s new scores = 15

Step-by-step explanation:

Interquartile range IQR = Higher quartile (Q3) - lower quartile (Q1)

Where Q3 is the mid-value of the second half of a dataset and

Q1 is the mid-value of the first half of a dataset.

Original score = 45, 65, 70, 80, 85, 100

Q1 of original score:

First half of original score = 45, 65, 70

Therefore Q1 = 65

Q3 of original score:

Second half of original score = 80, 85, 100

Therefore Q3 = 85

:. Interquartile range IQR of original score = Q3 of original score - Q1 of original score

:. IQR = 85 - 65 = 20

New score = 65, 70, 80, 85

Q1 of new score:

First half of new score = 65, 70

Therefore Q1 = (65+70)/2 = 67.5

Q3 of new score:

Second half of new score = 80, 85

Therefore Q3 = (80 + 85)/2 = 82.5

:. Interquartile range IQR of new score = Q3 of new score - Q1 of new score

:. IQR = 82.5 - 67.5 = 15

Therefore:

Interquartile ranges of Jessie’s original scores = 20

Interquartile ranges of Jessie’s new scores = 15

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Answer: Rounded: 236 seconds

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5 0
3 years ago
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IrinaK [193]

Answer:

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The 30th term is -47

Step-by-step explanation:

∵ f(n) = 11 and g(n) = -2(n - 1) = -2n + 2

∴ f(n) + g(n) = 11 + -2n + 2 = 13 - 2n

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∵ n = 1 ⇒ 13 - 2(1) = 11

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∵ n = 4 ⇒ 13 - 2(4) = 13 - 8 = 5

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∴ The rule of the arithmetic sequence is 13 - 2n

∴ The 30th term = 13 - 2(30) = -47

5 0
2 years ago
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A teacher was interested in knowing the amount of physical activity that his students were engaged in daily. He randomly sampled
klasskru [66]

Answer:

The standard error of the mean is 4.5.

Step-by-step explanation:

As we don't know the standard deviation of the population, we can estimate the standard error of the mean from the standard deviation of the sample as:

\sigma_{\bar{x}}\approx\frac{s}{\sqrt{n}}

The sample is [30mins, 40 mins, 60 mins, 80 mins, 20 mins, 85 mins]. The size of the sample is n=6.

The mean of the sample is:

\bar{x}=\frac{1}n} \sum x_i =\frac{30+40+60+80+20+85}{6}=52.5

The standard deviation of the sample is calculated as:

s=\sqrt{\frac{1}{n-1}\sum (x_i-\bar x)^2} \\\\ s=\sqrt{\frac{1}{5}\cdot ((30-52.5)^2+(40-52.5)^2+(60-52.5)^2+(80-52.5)^2+(20-52.5)^2+(85-52.5)^2}\\\\s=\sqrt{\frac{1}{5} *3587.5}=\sqrt{717.5}=26.8

Then, we can calculate the standard error of the mean as:

\sigma_{\bar{x}}\approx\frac{s}{\sqrt{n}}=\frac{26.8}{6}= 4.5

6 0
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A recent survey determined the IQ score of a random selection of residents of Alaska. The accompanying relative frequency distri
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Answer:

The class width is 20

Step-by-step explanation:

In a frequency or a relative frequency distribution the class width is calculated as the difference between the lower or upper class limits of consecutive classes. A point to note is that all the categories or classes usually have the same class width.

We use the first two classes to calculate the class width by using their respective upper limits;

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4 0
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It was greater than 18.96 whether interest was compounded daily or monthly

Step-by-step explanation:

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