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Fudgin [204]
3 years ago
7

Leo solved the equation b = 12 a−1−−−−−√3 for a, but made an error. His work is shown. Complete the sentences that follow.

Mathematics
1 answer:
kiruha [24]3 years ago
8 0

The value of a in terms of b is given as a = 8b^3

Given the equation solved by Leo expressed as b=\frac{1}{2}\sqrt[3]{a-1}

We are to solve the equation for the variable "a"

Given;

b=\frac{1}{2}\sqrt[3]{a-1}

Cross multiply

2b=\sqrt[3]{a-1}

Cube both sides of the equation:

(2b)^3=(\sqrt[3]{a-1})^3 \\8b^3=a-1

Add 1 to both sides of the equation:

8b^3+1=a-1+1\\8b^3=a\\Swap\\a=8b^3

Hence the value of a in terms of b is given as a = 8b^3

Learn more on subject of formula here: brainly.com/question/657646

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a) P=0.091

b) If there are half of each taste, picking 3 vainilla in a row has a rather improbable chance (9%), but it is still possible that there are 6 of each taste.

c) The probability of picking 4 vainilla in a row, if there are half of each taste, is P=0.030.

This is a very improbable case, so if this happens we would have reasons to think that there are more than half vainilla candies in the box.

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P(x=k)=\dfrac{\binom{K}{k}\cdot \binom{N-K}{n-k}}{\binom{N}{n}}

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n is the number of draws (3 in point a, 4 in point c),

k is the number of observed successes (3 in point a, 4 in point c),

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b) If there are half of each taste, picking 3 vainilla in a row has a rather improbable chance (9%), but is possible.

c) In the case k=4, we have:

P(x=3)=\dfrac{\binom{6}{4}\cdot \binom{6}{0}}{\binom{12}{4}}=\dfrac{15\cdot 1}{495}=0.030

This is a very improbable case, so we would have reasons to think that there are more than half vainilla candies in the box.

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