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svp [43]
3 years ago
10

A survey was taken of students in math classes to find out how many hours per day students spend on social media. The survey res

ults for the first-, second-, and third-period classes are as follows:
First period: 2, 4, 3, 1, 0, 2, 1, 3, 1, 4, 9, 2, 4, 3, 0

Second period: 3, 2, 3, 1, 3, 4, 2, 4, 3, 1, 0, 2, 3, 1, 2

Third period: 4, 5, 3, 4, 2, 3, 4, 1, 8, 2, 3, 1, 0, 2, 1, 3

Which is the best measure of center for third period and why?

A: Interquartile range, because there is 1 outlier that affects the center

B: Standard deviation, because there are no outliers that affect the center

C: Mean, because there are no outliers that affect the center

D: Median, because there is 1 outlier that affects the center
Mathematics
2 answers:
MissTica3 years ago
6 0

Answer: D : Median, because there is 1 outlier that affects the center

Step-by-step explanation:

Mean and median terms are use to measure the central tendency in a data set.

Mean is the best measure of center if there is no outlier present in the data otherwise median is the best measure of center if there is outlier present because outliers effects the means value of a data but not the median value.

In the third period, there is an outlier present as 8 that affects the center, then the best measure of center for third period must be Median.

Kamila [148]3 years ago
6 0

Answer:

I reckon,the correct option is option C,Mean, because there are no outliers that affect the center.

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To save money, a local charity organization wants to target its mailing requests for donations to individuals who are most suppo
mars1129 [50]

Answer:

The 80% confidence interval for difference between two means is (0.85, 1.55).

Step-by-step explanation:

The (1 - <em>α</em>) % confidence interval for difference between two means is:

CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2,(n_{1}+n_{2}-2)}\times SE_{\bar x_{1}-\bar x_{2}}

Given:

\bar x_{1}=M_{1}=6.1\\\bar x_{2}=M_{2}=4.9\\SE_{\bar x_{1}-\bar x_{2}}=0.25

Confidence level = 80%

t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.20/2, (5+5-2)}=t_{0.10,8}=1.397

*Use a <em>t</em>-table for the critical value.

Compute the 80% confidence interval for difference between two means as follows:

CI=(6.1-4.9)\pm 1.397\times 0.25\\=1.2\pm 0.34925\\=(0.85075, 1.54925)\\\approx(0.85, 1.55)

Thus, the 80% confidence interval for difference between two means is (0.85, 1.55).

3 0
3 years ago
Identify the transformation that was applied to this letter.
Shtirlitz [24]
Identify the transformation that was applied to this letter.

D. Rotation about the origin
8 0
2 years ago
Find the angle measure of PQR
sineoko [7]

Answer:

<em>64 degrees</em>

Step-by-step explanation:

When you have the SRQ angle, you can easily solve for PQR. Luckily, we're given that QRP and PRS are 32 and 84, respectively. Adding this together gives you 116. 116*2 = 232 and 360-232 = 128, and 128/2 = 64.

<em>Please Mark Brainliest if this helps!</em>

8 0
3 years ago
X - 5(x + 1) = 3x + 2<br> answer , but show work pls.
Nitella [24]

Answer: x = -1

Step-by-step explanation:

Starting:

x-5(x+1)=3x+2

Simplify the left side:

-4x-5=3x+2

Move all the terms with an x to the left:

-7x-5=2

Move all the terms without an x to the right:

-7x=7

Divide both sides by -7:

x = -1

3 0
3 years ago
What is the scale factor if the scale is 10 inches = 1 foot?
klasskru [66]

10 inches to 12 inches

10:12

5:6

<em>Have a Great Day</em>

5 0
3 years ago
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