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2 years ago

Part E

Mathematics

1 answer:

Ronch [10]2 years ago

3 0

The option which makes more sense for Jake to choose is option (1) He can accept the tie, and the game is over.

<h3>How to determine the best option to choose?</h3>

The complete question is added as an attachment

From the complete question, the probability in part D is

Probability = 33.5%

The options to choose one from are:

Option 1: He can accept the tie, and the game is over.
Option 2: He can make one more throw. Jake wins if he earns at least 30 points on the throw, but he loses if he earns less than 30 points.

The calculated probability is less than 0.5.

This means that Jake is more likely to lose than win

So, the best option for Jake is option (A) walk away

Hence, the option which makes more sense for Jake to choose is option (1) He can accept the tie, and the game is over.

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Step-by-step explanation:

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What is the measure of space inside a suitcase?

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2 years ago

Read 2 more answers

In the expansion (ax+by)^7, the coefficients of the first two terms are 128 and -224, respectively. Find the values of a and b

madam [21]

Answer:

a = 2, b = 3.5

Step-by-step explanation:

Expanding $(ax+by)^7$ using Binomial expansion, we have that:

$(ax+by)^7$ =

$(ax)^7(by)^0 + (ax)^6(by)^1 + (ax)^5(by)^2 + (ax)^4(by)^3 + (ax)^3(by)^4 + (ax)^2(by)^5 + (ax)^1(by)^6 + (ax)^0(by)^7$

$= (a)^7(x)^7+ (a)^6(x)^6(b)(y) + (a)^5(x)^5(b)^2(y)^2 + (a)^4(x)^4(b)^3(y)^3 + (a)^3(x)^3(b)^4(y)^4 + (a)^2(x)^2(b)^5(y)^5 + (a)(x)(b)^6(y)^6 + (b)^7(y)^7\\\\\\= (a)^7(x)^7+ (a)^6(b)(x)^6(y) + (a)^5(b)^2(x)^5(y)^2 + (a)^4(b)^3(x)^4(y)^3 + (a)^3(b)^4(x)^3(y)^4 + (a)^2(b)^5(x)^2(y)^5 + (a)(b)^6(x)(y)^6 + (b)^7(y)^7$

We have that the coefficients of the first two terms are 128 and -224.

For the first term:

=> $a^7 = 128$

=> $a = \sqrt[7]{128}\\ \\\\a = 2$

For the second term:

$a^6b = -224$

$b = \frac{-224}{a^6}$

$b = \frac{-224}{2^6} \\\\\\b = \frac{-224}{64} \\\\\\b = 3.5$

Therefore, a = 2, b = 3.5

5 0

3 years ago

A small regional carrier accepted 16 reservations for a particular flight with 12 seats. 8 reservations went to regular customer

wolverine [178]

Answer:

a) 32.04% probability that overbooking occurs.

b) 40.79% probability that the flight has empty seats.

Step-by-step explanation:

For each booked passenger, there are only two possible outcomes. Either they arrive for the flight, or they do not arrive. The probability of a passenger arriving is independent of other passengers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$