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3 years ago

What is the common ratio? {250, 50, 10, 2...}

Mathematics

1 answer:

Allisa [31]3 years ago

6 0

Answer:

Step-by-step explanation:

The common ration can be found by dividing the 2nd number from the 1st number:

250 / 50

which equals 5

250 / 50 = 5

you can do this by any number, for example:

50 / 10 =5

10 / 2 = 5

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What is 2,117 divided by 3

sweet [91]

The answer is 705.666667. Rounded the answer is about 706. As a fraction it it 705 666667/1000000 (it cannot be simplified any further).
I hope this helps:)

6 0

2 years ago

Fred has 36 strawberries and 42 blueberries. He wants to use them to garnish desert so that each desert has the same number of b

DENIUS [597]

To do this problem you must find the GCF or the greatest common factor of 36 and 42. For example 36 can be made by 1 and 36 2 and 18 3 and 12 4 and 9 and 6 and 6. 42 can be made by 1 and 42 2 and 21 3 and 14 and 6 and 7. The highest common factor is 6. So, if you put 6 berries on all of the desserts, you will put 6 strawberries on 6 tarts and 6 blueberries on 7 tarts. That’s a total of 13 desserts

7 0

2 years ago

Consider three boxes with numbered balls in them. Box A con- tains six balls numbered 1, . . . , 6. Box B contains twelve balls

murzikaleks [220]

Answer:

a) 0.73684

b) 2/3

Step-by-step explanation:

part a)

$P ( A is 1 / exactly two balls are 1) = \frac{P ( A is 1 and that exactly two balls are 1)}{P (Exactly two balls are one)}$

Using conditional probability as above:

(A,B,C)

Cases for numerator when:

P( A is 1 and exactly two balls are 1) = P( 1, not 1, 1) + P(1, 1, not 1)

= $(\frac{1}{6}* \frac{11}{12}*\frac{1}{4}) + (\frac{1}{6}*\frac{1}{12}*\frac{3}{4}) = 0.048611111$

Cases for denominator when:

P( Exactly two balls are 1) = P( 1, not 1, 1) + P(1, 1, not 1) + P(not 1, 1 , 1)

$= (\frac{1}{6}* \frac{11}{12}*\frac{1}{4}) + (\frac{1}{6}*\frac{1}{12}*\frac{3}{4}) + (\frac{5}{6}*\frac{1}{12}*\frac{1}{4})= 0.0659722222$

Hence,

$P ( A is 1 / exactly two balls are 1) = \frac{P ( A is 1 and that exactly two balls are 1)}{P (Exactly two balls are one)} = \frac{0.048611111}{0.06597222} \\\\= 0.73684$

Part b

$P ( B = 12 / A+B+C = 21) = \frac{P ( B = 12 and A+B+C = 21)}{P (A+B+C = 21)}$

Cases for denominator when:

P ( A + B + C = 21) = P(5,12,4) + P(6,11,4) + P(6,12,3)

$= 3*P(5,12,4 ) =3* \frac{1}{6}*\frac{1}{12}*\frac{1}{4}\\\\= \frac{1}{96}$

Cases for numerator when:

P (B = 12 & A + B + C = 21) = P(5,12,4) + P(6,12,3)

$= 2*P(5,12,4 ) =2* \frac{1}{6}*\frac{1}{12}*\frac{1}{4}\\\\= \frac{1}{144}$

Hence,

$P ( B = 12 / A+B+C = 21) = \frac{\frac{1}{144} }{\frac{1}{96} }\\\\= \frac{2}{3}$

4 0

3 years ago

-QUICK PLEASE-

Zepler [3.9K]

Answer:

Option B. 0.75 gallons is the right answer.

Step-by-step explanation:

Let the load of dishwasher used in house A = x gallons

Let the load of dishwasher used in house B is = y gallons

and similarly we assume for house C = z gallons

Now from the given question

On Friday amount of water used in all three houses = 17.5 gallons

From the table we will make the equation

x + 2y + 2z = 17.5-----(1)

For Saturday

2x + 3y + 4z = 31.5 ----(2)

For Sunday

2x + 2y + 3z = 24.75 -----(3)

By subtracting equation 3 from 2

y + z = 31.5 - 24.75 = 6.75 ------(4)

Equation (1) multiplied by 2 - equation (2)

2×(x + 2y + 2z) - (2x + 3y + 4z) = 17.5×2 - 31.5

y = 35 - 31.5 = 3.5

By putting y = 3.5 in the equation (4)

z = 6.75 - 3.5 = 3.25

By putting y = 3.5 and z = 3.25 in the equation (1)

x + 7 + 6.5 = 17.5

x = 17.5 - 13.5 = 4

Now the difference between load of dishwasher that uses least water and the most.

x - z = 4 - 3.25 = 0.75 gallons

Option B is the right answer.

6 0

3 years ago

WILL GIVE FIRST ANSWER BRAINLIEST

DedPeter [7]

My answer is D.(5, -2)

4 0

3 years ago

Read 2 more answers