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lys-0071 [83]
3 years ago
6

HELP PLEASE RN ASAP!!!!!

Mathematics
2 answers:
Sergeeva-Olga [200]3 years ago
5 0

Answer:

1/9^15

Step-by-step explanation:

PIT_PIT [208]3 years ago
4 0

Answer:

0.05555555555555555555

Step-by-step explanation:

9-3 is 6 and 9x12 is 108

6/108 is 0.0555555555555555555555555555555555

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A leaking pond loses 18 gallons of water in 35 hours. How many gallons of water will it lose in 24 hours?
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Answer: 12gallons

Step-by-step explanation:

18/35= 0.5 gallons per hour

0.5x24= 12 gallons

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Please help!<br> Anwser photo below
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Step-by-step explanation:

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A small car company sold 65.000 cars last year. Ninety-five
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Answer:61750 cars

Step-by-step explanation:

95% of 65000

95/100 x 65000

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3 0
3 years ago
Suppose a spider was able to create one thread that would extend from the top-right back corner of a room to the bottom-left fro
NNADVOKAT [17]

Answer:

d^2 = 30^2 + 12^2

e^2 = d^2 + 8^2

e^2 = 30^2 + 12^2 + 8^2

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7 0
4 years ago
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Scott and Letitia are brother and sister. After dinner, they have to do the dishes, with one washing and the other drying. They
ehidna [41]

Answer:

The probability that Scott will wash is 2.5

Step-by-step explanation:

Given

Let the events be: P = Purple and G = Green

P = 2

G = 3

Required

The probability of Scott washing the dishes

If Scott washes the dishes, then it means he picks two spoons of the same color handle.

So, we have to calculate the probability of picking the same handle. i.e.

P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)

This gives:

P(G_1\ and\ G_2) = P(G_1) * P(G_2)

P(G_1\ and\ G_2) = \frac{n(G)}{Total} * \frac{n(G)-1}{Total - 1}

P(G_1\ and\ G_2) = \frac{3}{5} * \frac{3-1}{5- 1}

P(G_1\ and\ G_2) = \frac{3}{5} * \frac{2}{4}

P(G_1\ and\ G_2) = \frac{3}{10}

P(P_1\ and\ P_2) = P(P_1) * P(P_2)

P(P_1\ and\ P_2) = \frac{n(P)}{Total} * \frac{n(P)-1}{Total - 1}

P(P_1\ and\ P_2) = \frac{2}{5} * \frac{2-1}{5- 1}

P(P_1\ and\ P_2) = \frac{2}{5} * \frac{1}{4}

P(P_1\ and\ P_2) = \frac{1}{10}

<em>Note that: 1 is subtracted because it is a probability without replacement</em>

So, we have:

P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)

P(Same) = \frac{3}{10} + \frac{1}{10}

P(Same) = \frac{3+1}{10}

P(Same) = \frac{4}{10}

P(Same) = \frac{2}{5}

8 0
3 years ago
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