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3 years ago

Solve 5x-4=5(x-3)+11

1 answer:

7 0

Expanding the right side:-

5x - 4 = 5x - 15 + 11
5x - 4 = 5x - 4

left side = right side so there are infinitely many solutions

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What is the slope of the linear function shown in the graph? The options are in the picture!

frez [133]

Answer:

5/4

Step-by-step explanation:

start from 0,0 then go up 5 and right 4

5 0

3 years ago

Circumference of the circle

dimulka [17.4K]

Answer:

The formula of circumference in a circle is 3.14 times diameter.

Step-by-step explanation:

3.14×20=62.8

The circumference is 62.8

8 0

3 years ago

Make <br> q the subject of p=6q +7

Illusion [34]

Answer:

p= 6q+7

-7. -7

p-7 =6q

/6. /6

p-7/6

Step-by-step explanation:

if you want to make a specific term the subject, you need to isolate it on one side of the equation. You do this by 'getting rid' of anything on the term's side of the equation, starting with whatever's furthest from the term. (think BEMDAS but backwards)

p = 6q + 7

(take 7 from both sides to get rid of +7 on q's side)

p - 7 = 6q + 7 - 7

p - 7 = 6q(divide both sides by 6 to get rid of 6 on q's side)

4 0

3 years ago

-y>2x solving linear inequalities<br><img src="https://tex.z-dn.net/?f=%20-%20y%20%20%5Cgeqslant%202x%20%20%20%20" id="TexFor

topjm [15]

Im sorry i just wanna get points

6 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