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svet-max [94.6K]
3 years ago
9

Natalie's group brought in pizza but is buying drinks at the concession stand medium sodas are a dollar and $.25 how many medium

s sodas can Natalie's group by with $20
Mathematics
1 answer:
Phantasy [73]3 years ago
8 0

Keywords:

<em>Medium sodas, buy, dollars, divide </em>

For this case we must find the amount of medium sodas that Natalie's group can buy, taking into account that they have 20 dollars and that each medium soda costs 1.25 dollars. To solve, we must divide:

Let "x" be the number of medium sodas you can buy, then:

x = \frac {20} {1.25}\\x = 16

So, Natalie's group can buy 16 medium sodas with 20 dollars

Answer:

16 medium sodas

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At the start of a tennis match Ivan had 28/32 quart of water in his water bottle. He drank water from the bottle during the matc
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5/8 of a quart.

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A quiz-show contestant is presented with two questions, question 1 and question 2, and she can choose which question to answer f
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Answer:

The contestant should try and answer question 2 first to maximize the expected reward.

Step-by-step explanation:

Let the probability of getting question 1 right = P(A) = 0.60

Probability of not getting question 1 = P(A') = 1 - P(A) = 1 - 0.60 = 0.40

Let the probability of getting question 2 right be = P(B) = 0.80

Probability of not getting question 2 = P(B') = 1 - P(B) = 1 - 0.80 = 0.20

To obtain the better option using the expected value method.

E(X) = Σ xᵢpᵢ

where pᵢ = each probability.

xᵢ = cash reward for each probability.

There are two ways to go about this.

Approach 1

If the contestant attempts question 1 first.

The possible probabilities include

1) The contestant misses the question 1 and cannot answer question 2 = P(A') = 0.40; cash reward associated = $0

2) The contestant gets the question 1 and misses question 2 = P(A n B') = P(A) × P(B') = 0.6 × 0.2 = 0.12; cash reward associated with this probability = $200

3) The contestant gets the question 1 and gets the question 2 too = P(A n B) = P(A) × P(B) = 0.6 × 0.8 = 0.48; cash reward associated with this probability = $300

Expected reward for this approach

E(X) = (0.4×0) + (0.12×200) + (0.48×300) = $168

Approach 2

If the contestant attempts question 2 first.

The possible probabilities include

1) The contestant misses the question 2 and cannot answer question 1 = P(B') = 0.20; cash reward associated = $0

2) The contestant gets the question 2 and misses question 1 = P(A' n B) = P(A') × P(B) = 0.4 × 0.8 = 0.32; cash reward associated with this probability = $100

3) The contestant gets the question 2 and gets the question 1 too = P(A n B) = P(A) × P(B) = 0.6 × 0.8 = 0.48; cash reward associated with this probability = $300

Expected reward for this approach

E(X) = (0.2×0) + (0.32×100) + (0.48×300) = $176

Approach 2 is the better approach to follow as it has a higher expected reward.

The contestant should try and answer question 2 first to maximize the expected reward.

Hope this helps!!!

3 0
3 years ago
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