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3 years ago

Huw is paid a weekly wage. Every week he: saves 1/5 of his wage,

Mathematics

2 answers:

elena55 [62]3 years ago

6 0

Answer: The answer is seven.

Nataly [62]3 years ago

6 0

The problems answer is 7

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Sum of the ages of A and B is 55 years. If the ratio of their ages is 6:5. find their ages...

lys-0071 [83]

Answer:

A's age = 6x = 6 × 5 = 30 years.

B's age = 5x = 5 × 5 = 25 years

Step-by-step explanation:

here's the solution : -

let their ages be 6x and 5x

and we know sum of their ages = 55

So, 6x + 5x = 55

=》11x = 55

=》x = 55 ÷ 11

=》x = 5

A's age = 6x = 6 × 5 = 30 years.

B's age = 5x = 5 × 5 = 25 years

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3 years ago

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Becky needs the regular shade of purple for her artwork. Which ratio of red paint to blue paint will she use? How many ounces of

lord [1]

Answer: c

Step-by-step explanation: I think

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3 years ago

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Which expression can be used to find 57% of 230? ⊥ω⊥

ella [17]

Answer:

131.10 (Answer C)

Step-by-step explanation:

Convert 57% to 0.57 and then multiply 230 by 0.57:

0.57(230) = 131.10

This is Answer C

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3 years ago

X + 3 = y <br><br> solve for x

alexira [117]

Subtract 3 from each side to isolate the x and you get x=y-3

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3 years ago

An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What

galina1969 [7]

Answer:

Part A:

The probability that all of the balls selected are white:

$P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\ P(A)=\frac{5}{66}=0.075757576$

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

$P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516$

Step-by-step explanation:

A is the event all balls are white.

D_i is the dice outcome.

Sine the die is fair:

$P(D_i)=\frac{1}{6}$ for i∈{1,2,3,4,5,6}

In case of 10 black and 5 white balls:

$P(A|D_1)=\frac{5_{C}_1}{15_{C}_1} =\frac{5}{15}=\frac{1}{3}$

$P(A|D_2)=\frac{5_{C}_2}{15_{C}_2} =\frac{10}{105}=\frac{2}{21}$

$P(A|D_3)=\frac{5_{C}_3}{15_{C}_3} =\frac{10}{455}=\frac{2}{91}$

$P(A|D_4)=\frac{5_{C}_4}{15_{C}_4} =\frac{5}{1365}=\frac{1}{273}$

$P(A|D_5)=\frac{5_{C}_5}{15_{C}_5} =\frac{1}{3003}=\frac{1}{3003}$

$P(A|D_6)=\frac{5_{C}_6}{15_{C}_6} =0$

Part A:

The probability that all of the balls selected are white:

$P(A)=\sum^6_{i=1} P(A|D_i)P(D_i)$

$P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\ P(A)=\frac{5}{66}=0.075757576$

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

We have to find $P(D_3|A)$

The data required is calculated above:

$P(D_3|A)=\frac{P(A|D_3)P(D_3)}{P(A)}\\ P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516$

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3 years ago