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3 years ago

Subtract the fraction write your answer in simplest form 6/8 - 3/10

Mathematics

1 answer:

jok3333 [9.3K]3 years ago

4 0

Answer:

9/10 or 0.45 in decimal form

Step-by-step explanation:

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Veronikas five test scores are 59 80 95 and 93 if the outlier of 59 is removed what is the mean absolute deviation of the remain

Anit [1.1K]

Complete Question

Veronikas four test scores are 59, 80, 95, 88 and 93 if the outlier of 59 is removed what is the mean absolute deviation of the remaining four test scores?

Answer:

Step-by-step explanation:

We have the four test scores

80, 95, 88 and 93

Step 1

We find the mean of the 4 test scores

= 80 + 95 + 88 + 93/4

= 356 / 4

Step 2

The formula for Mean Absolute Deviation =

Summation( x - Mean)/n

Hence,

|(80 - 89 )+( 95 - 89) + (88 - 89) + (93 - 89)|/5

= 9 + 6 +1 + 4/4

= 20/4

= 5

8 0

3 years ago

Find the volume for the problem

dolphi86 [110]

The answer is 523.6 yards squared

4 0

3 years ago

Helpp I'm struggling on both of these questions !! I'm in my last questions!!

kari74 [83]

For question one 1) >

7 0

3 years ago

Read 2 more answers

Use the graph of △ABC with midsegments DE, EF and DF. Show that EF is parallel to AC and that EF=1/2 AC

Vinvika [58]

According to the midsegment theorem, the midsegments are parallel to

and half the length of the opposite side.

The completed statement are as follows;

Because the slopes of $\overline{EF}$ and $\overline{AC}$ are both -4, $\overline{EF}$ ║ $\overline{AC}$

EF = $\underline{\sqrt{17}}$ and AC = $\underline{2 \cdot \sqrt{17} }$

Because $\underline{\sqrt{17} }$ = $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17} }$ , $EF = \frac{1}{2} \cdot AC$

Reasons:

From the given graph of ΔABC, we have;

Coordinates of the points A, B, and C are; A(-5, 2), B(1, -2), and C(-3, -6)

The coordinates of the point D and E on $\mathbf{\overline{DE}}$ are; D(-4, -2), and E(-2, 0)

The coordinates of the point F is; F(-1, -4)

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\displaystyle Slope \ of \ line \ \overline{AC} = \mathbf{\frac{(-6) - 2}{-3 - (-5)} = \frac{-8}{2}} = -4$

$\displaystyle Slope \ of \ line \ \overline{EF} = \frac{0 - (-4)}{-2 - (-1)} = \frac{4}{-1} = -4$

$Length \ of \ segment,\ l = \sqrt{\left (x_{2}-x_{1} \right )^{2}+\left (y_{2}-y_{1} \right )^{2}}$

Length of EF = √((-1 - (-2))² + (-4 - 0)²) = √(17)

Length of AC = √((-3 - (-5))² + (-6 - 2)²) = √(4 × 17) = 2·√(17)

Therefore, we have;

Because the slope of $\mathbf{\overline {EF}}$ and $\mathbf{\overline {AC}}$ are both , -4, $\overline {EF}$ ║ $\overline {AC}$ . EF = $\underline{\sqrt{17}}$ , and AC

= $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17}}$ ,. Because $\underline{\sqrt{17}}$ = $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17}}$ , $EF =\mathbf{ \frac{1}{2} AC}$

Learn more about midsegment theorem of a triangle here:

brainly.com/question/7423948

8 0

3 years ago

4+(-7) what is the solution

MrMuchimi

Answer:

-3

Step-by-step explanation:

8 0

4 years ago

Read 2 more answers