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2 years ago

Whoever helps with number 5 gets brainliest answer and 10 points :)

Mathematics

2 answers:

nata0808 [166]2 years ago

6 0

Count:5Sum:341Mean:68.2Median:62Mode:48, 40, 93, 62, 98

arlik [135]2 years ago

6 0

5.
Mean = 68
Median= 92
Mode= no mode
Range = 58

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Endpoints of segment MN have coordinates (0, 0), (5, 1). The endpoints of segment AB have coordinates (1 1/22 , 2 1/4 ) and (−2

Nana76 [90]

Answer: $k=18\dfrac{8}{11}$ .

Step-by-step explanation:

If a line passing through two points, then

$Slope=\dfrac{y_2-y_1}{x_2-x_1}$

Endpoints of segment MN have coordinates (0, 0) and (5, 1).

Slope of MN $=\dfrac{1-0}{5-0}=\dfrac{1}{5}$

The endpoints of segment AB have coordinates $\left(1\dfrac{1}{22} , 2\dfrac{1}{4}\right)$ and $\left(-2\dfrac{1}{4} , k\right)$ .

$A=\left(1\dfrac{1}{22} , 2\dfrac{1}{4}\right)=\left(\dfrac{23}{22} ,\dfrac{9}{4}\right)$

$B=\left(-2\dfrac{1}{4} , k\right)=\left(-\dfrac{9}{4} , k\right)$ .

Slope of AB $=\dfrac{k-\frac{9}{4}}{-\frac{9}{4}-\dfrac{23}{22}}$

$=\dfrac{\frac{4k-9}{4}}{\frac{-99-46}{44}}$

$=\dfrac{4k-9}{4}\times \dfrac{44}{-145}$

$=4k-9\times \dfrac{11}{-145}$

$=\dfrac{44k-99}{-145}$

Product of slopes of two perpendicular segments is -1.

Slope of MN × Slope of AB = -1

$\dfrac{1}{5}\times \dfrac{44k-99}{-145}=-1$

$\dfrac{44k-99}{-725}=-1$

$44k-99=725$

$44k=725+99$

$k=\dfrac{824}{44}$

$k=\dfrac{206}{11}$

$k=18\dfrac{8}{11}$

Therefore, the value of k is $k=18\dfrac{8}{11}$ .

8 0

3 years ago

Read 2 more answers

The graph of the piecewise function f(x) is shown.

iris [78.8K]

1. The range of a function is the set of all values that f can produce for all the x-es in the domain.

2. If we are given the graph, in order to find the range, we project the graph into the y axis. Informally, we draw the "shadow" of the graph into the y axis as in the FIGURE atached.

3. The range is D || {y | −5 ≤ y ≤ −1}