Which statement, if true, would help prove that quadrilateral

25 POINTS!!! x = 4 6 12

shutvik [7]

Answer:

288

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Express (5a to the power 2) x (2b to the power 2) x (c) as a power of a product.

ella [17]

Answer:

50 a^2 b c^2

Step-by-step explanation:

I'm quite confused about your question, but I try my best. I have attached the explanation above. Hopefully this will help

4 0

3 years ago

Each time they Playa game there's a 90% chance that the blue comets soccer team will score the first goal. What is the probabili

aalyn [17]

Step-by-step explanation:

in 4 straight games ?

game 1 has a probability of 0.9 (= 90%).

for 2 games we have the probability of the first game, and now based on this we have to apply the probability of the second game.

in other words, the desired event has to happen in the first game AND then also in the second game.

since both probabilities are smaller than 1 (0.9), the total probability for 2 games must be less than the probability for 1 isolated game.

to combine such probabilities as "and" events, we have to multiply these probabilities.

so, the probability for 2 games is

0.9 × 0.9 = 0.9² = 0.81

and in the same way we get the probability to score the first goal in 4 games in a row :

0.9⁴ = 0.6561

8 0

2 years ago

Which of the following is likely to have the greatest variability? *

noname [10]

Answer:

G

Step-by-step explanation:

4 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