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3 years ago

Find the midpoint and distance between (5,1) and (-15,11)

Mathematics

1 answer:

liq [111]3 years ago

4 0

Answer:

look i think is -5.005 I don't really know but I hope you get it right. Have a good day

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HI I NEVER KNEW HOW TO DO THIS BUT I DO SO CAN YALL REALLY HELP ME PLEASE

Aliun [14]

Answer:

A. 4/15 and 0.26 repeating

Step-by-step explanation:

As we see in the long division, 4/15 is equal to 0.26666666666666666666666 (or. 0.26 repeating).

4/15

4 doesn't go into 15, so add a zero.

4.0/15

40 goes into 15 twice

40 - 30 is 10. 10 is less than 15 so add a 0.

4.00

100 goes into 15 6 times.

100 - 90 is 10. 10 is less than 15, so add a 0.

The 100 into 15 process repeats, so there's your answer.

8 0

3 years ago

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A home repair service company earned $990 performing annual inspections. They charged $55 for each inspection. How many inspecti

OlgaM077 [116]

They performed 18 home inspections

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3 years ago

Read 2 more answers

Plssss help me I gotta turn the rest of my other assignments my 11:59 pm <br> T-T

Ad libitum [116K]

B hope this helps good luck

3 0

3 years ago

A square and a circle intersect so that each side of the square contains a chord of the circle equal in length to the radius of

krek1111 [17]

Answer:

Step-by-step explanation:

it is given that Square contains a chord of of the circle equal to the radius thus from diagram

$QR=chord =radius =R$

If Chord is equal to radius then triangle PQR is an equilateral Triangle

Thus $QO=\frac{R}{2}=RO$

In triangle PQO applying Pythagoras theorem

$(PQ)^2=(PO)^2+(QO)^2$

$PO=\sqrt{(PQ)^2-(QO)^2}$

$PO=\sqrt{R^2-\frac{R^2}{4}}$

$PO=\frac{\sqrt{3}}{2}R$

Thus length of Side of square $=2PO=\sqrt{3}R$

Area of square $=(\sqrt{3}R)^2=3R^2$

Area of Circle $=\pi R^2$

Ratio of square to the circle $=\frac{3R^2}{\pi R^2}=\frac{3}{\pi }$

5 0

3 years ago

Suppose that two openings on an appellate court bench are to be filled from current municipal court judges. The municipal court

Ksju [112]

Answer:

$(a)\dfrac{92}{117}$

$(b)\dfrac{8}{39}$

$(c)\dfrac{25}{117}$

Step-by-step explanation:

Number of Men, n(M)=24

Number of Women, n(W)=3

Total Sample, n(S)=24+3=27

Since you cannot appoint the same person twice, the probabilities are <u>without replacement.</u>

(a)Probability that both appointees are men.

$P(MM)=\dfrac{24}{27}X \dfrac{23}{26}=\dfrac{552}{702}\\=\dfrac{92}{117}$

(b)Probability that one man and one woman are appointed.

To find the probability that one man and one woman are appointed, this could happen in two ways.

A man is appointed first and a woman is appointed next.
A woman is appointed first and a man is appointed next.

P(One man and one woman are appointed) $=P(MW)+P(WM)$

$=(\dfrac{24}{27}X \dfrac{3}{26})+(\dfrac{3}{27}X \dfrac{24}{26})\\=\dfrac{72}{702}+\dfrac{72}{702}\\=\dfrac{144}{702}\\=\dfrac{8}{39}$

(c)Probability that at least one woman is appointed.

The probability that at least one woman is appointed can occur in three ways.

A man is appointed first and a woman is appointed next.
A woman is appointed first and a man is appointed next.
Two women are appointed

P(at least one woman is appointed) $=P(MW)+P(WM)+P(WW)$

$P(WW)=\dfrac{3}{27}X \dfrac{2}{26}=\dfrac{6}{702}$

In Part B, $P(MW)+P(WM)=\frac{8}{39}$

Therefore:

$P(MW)+P(WM)+P(WW)=\dfrac{8}{39}+\dfrac{6}{702}\\$P(at least one woman is appointed)=\dfrac{25}{117}$

5 0

3 years ago