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3 years ago

12

Pls help and show workings

1 answer:

Alexeev081 [22]3 years ago

4 0

Answer:

22

Step-by-step explanation:

you would take 14 plus 8 and that would give you your answer. I've done this before. I'm positive its 22

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Divisiones que den de residuo 300

Stolb23 [73]

Answer:

900÷3

Step-by-step explanation:

creo que es esto :)

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3 years ago

phoebe needs to use 3/4 cup to make one batch of cookies how many batches can she make with 12 cups of sugar?

sattari [20]

$\bf \begin{array}{ccll}\stackrel{sugar}{cups}&\stackrel{cookies}{batches}\\\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\\frac{3}{4}&1\\\\12&b\end{array}\implies \cfrac{\quad \frac{3}{4}\quad }{12}=\cfrac{1}{b}\implies \cfrac{\quad \frac{3}{4}\quad }{\frac{12}{1}}=\cfrac{1}{b}\\\\\\\cfrac{3}{4}\cdot \cfrac{1}{12}=\cfrac{1}{b}\implies \cfrac{3}{48}=\cfrac{1}{b}\implies 3b=48\implies b=\cfrac{48}{3}\implies b=16$

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3 years ago

Read 2 more answers

Solve pls brainliest

Olenka [21]

Answer:

b

Step-by-step explanation:

I Think B is the right answer

5 0

2 years ago

Evaluate the geometric series

Schach [20]

Your multiplying by -4 each time

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3 0

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:

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

7 0

3 years ago