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zzz [600]
3 years ago
15

Sora was given two data sets, one without an outlier and one with an outlier. Data without an outlier: 108, 113, 105, 118, 124,

121, 109 Data with an outlier: 108, 113, 105, 118, 124, 121, 109, 61 How is the mean affected by an outlier? The outlier did not affect the mean at all. The outlier made the mean higher than all the other values. The outlier made the mean lower than all the other values. The mean is centered around all the other numbers in both sets of data.
Mathematics
2 answers:
gladu [14]3 years ago
8 0

The outlier (61) is at the low end of the data set, but doesn't affect the mean by a lot, so ...

The mean is centered among the other numbers in both sets of data.

_____

The mean without the outlier is 114. With the outlier, it is 107.4. The lower quartile is 108, so the mean does get moved outside the "box" of the box-and-whisker plot of the data set without the outlier.

Nikitich [7]3 years ago
6 0

Answer:

The approved answer is wrong the answer was it was lower than all the others. I took the test

Step-by-step explanation:

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To solve this we are going to use the formula for compounded interest: A=P(1+ \frac{r}{n})^{nt}
where 
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r is the interest rate in decimal form 
n is the number of times the interest is compounded per year
t is the time in years 

We know for our problem that P=1380, r= \frac{5}{100} =0.05, and t=3. Since the interest is compounded daily, it is compounded 365 times in year; therefore, n=365. Lets replace those values in our formula to find A:
A=P(1+ \frac{r}{n})^{nt}
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We can conclude the amount in Diane's after 3 years will be <span>$1,603.31</span>
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3 years ago
WhT is 6811.09 rounded to the nearest ten
Ede4ka [16]

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The reason the 0 moved up to a 1 is because the number in the hundredth place was 5 or over. If it were 4 or under, then it would have remained a 0.

8 0
3 years ago
According to a study conducted by the Gallup Organization, the the proportion of Americans who are afraid to fly is 0.10. A rand
notsponge [240]

Answer: 0.1457

Step-by-step explanation:

Let p be the population proportion.

Given: The proportion of Americans who are afraid to fly is 0.10.

i.e. p= 0.10

Sample size : n= 1100

Sample proportion of Americans who are afraid to fly =\hat{p}=\dfrac{121}{1100}=0.11

We assume that the population is normally distributed

Now, the probability that the sample proportion is more than 0.11:

P(\hat{p}>0.11)=P(\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}>\dfrac{0.11-0.10}{\sqrt{\dfrac{0.10(0.90)}{1100}}})\\\\=P(z>\dfrac{0.01}{0.0090453})\ \ \ [\because z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}} ]\\\\=P(z>1.1055)\\\\=1-P(z\leq1.055)\\\\=1-0.8543=0.1457\ \ \ [\text{using z-table}]

Hence, the probability that the sample proportion is more than 0.11 = 0.1457

3 0
3 years ago
Round 34,576 to the nearest thousand
VARVARA [1.3K]

Answer:

35000

Step-by-step explanation:

6 0
3 years ago
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