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kumpel [21]
3 years ago
10

Please answer immediately.​

Mathematics
1 answer:
salantis [7]3 years ago
4 0

1. Angle L

2. A

3. Side PL

4. Side AR

5. A and K

6. B

7. Side Br

8. A

9. T

10. TA

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pashok25 [27]
The answer is 2 18/19
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1. The cost of a fish dinner for a restaurant is $7.95. If they sell it for $11.95, what is the percent markup?
kvv77 [185]

Answer:

the answer will be

50.3144 %

7 0
3 years ago
A local restaurant advertises that the mode cost of their most popular meals is $8. If the costs of their most popular meals are
hodyreva [135]

Answer: This kind of advertising is accurate.

Step-by-step explanation:

Since we have given that

Cost of their most popular meals are as follows:

\$7,\$8,\$8,\$12,\$13,\$15,\$17,\$18,\ and\ \$20

As we have given that the mode cost of their most popular meals is $8.

As we know that "Mode" refers to the most occurring number that we can found in the data.

Here, we can see that $8 occurs 2 times , which is the most occurring cost among them.

Hence, Mode = $8.

So, This kind of advertising is accurate.

8 0
3 years ago
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Someone please help me omg i am so lost and this is due tonight
Amiraneli [1.4K]

Answer:

45

Step-by-step explanation:

105+30 = 135

180-135= 45

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5 0
3 years ago
The Wall Street Journal reported that Walmart Stores Inc. is planning to lay off employees at its Sam's Club warehouse unit. App
tatyana61 [14]

Answer:

a) \bar X = 52

Median = 55

b) Q_1= \frac{44+44}{2}=44

Q_3= \frac{57+60}{2}=58.5

c) Range = Max -Min = 69-36=33

d) s^2 =100.1429

s= \sqrt{100.143}=10.0071

e) Lower = Q_1 -1.5 IQR = 44-1.5(58.5-44) = 22.25

Upper = Q_1 +1.5 IQR = 44+1.5(58.5-44) = 65.75

A possible outlier would be the value of 69 since its above the upper limti for the boxplot.

Step-by-step explanation:

For this case we have the following dataset:

55 56 44 43 44 56 60 62 57 45 36 38 50 69 65

A total of 15 observations

Part a

We calculate the mean with the following formula:

\bar X = \frac{\sum_{i=1}^{15} X_i}{15}

And for this case we got \bar X = 52

For the median we ust need to order the data on increasing way like this:

36, 38,43,44,44,45,50,55,56,56,57,60,62,65,69

Since the number of observations is an odd number the median would be on the 8 position from the dataset ordered on this case:

Median = 55

Part b

In order to calculate the Q1 we need to select the following data:

36, 38,43,44,44,45,50,55

And the Q1 would be the average between the 4 and 5 positions like this:

Q_1= \frac{44+44}{2}=44

And for the Q3 we select these values:

55,56,56,57,60,62,65,69

And the Q3 would be the average between the 4 and 5 positions like this:

Q_3= \frac{57+60}{2}=58.5

Part c

The Range is defined as:

Range = Max -Min = 69-36=33

Part d

In order to calculate the sample variance we can use the following formula:

s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}

And if we replace we got:

s^2 =100.1429

And the deviation is just the square root of the variance:

s= \sqrt{100.143}=10.0071

Part e

For this case we need to find the lower and upper limits for the boxplot given by:

Lower = Q_1 -1.5 IQR = 44-1.5(58.5-44) = 22.25

Upper = Q_1 +1.5 IQR = 44+1.5(58.5-44) = 65.75

A possible outlier would be the value of 69 since its above the upper limti for the boxplot.

4 0
4 years ago
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