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Andrews
18 days ago
5

Kevin runs a farm stand that sells apples and raspberries. Each pound of apples sells for $3.50 and each pound of raspberries se

lls for $4. Kevin sold twice as many pounds of raspberries as pounds of apples and he made $115 altogether. Determine the number of pounds of apples sold and the number of pounds of raspberries sold.
Mathematics
1 answer:
babunello [11.3K]18 days ago
3 0

Answer:
Step-by-step explanation:
10 pounds of apples sold, and 20 pounds of raspberries sold
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A waiter receives a tip of $3.25. If this tip represents 20% of the price of the meal, what is the price of the meal?
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19 days ago
Tags are placed to the left leg and right leg of a bear in a forest. Let A1 be the event that the left leg tag is lost and the e
babunello [11366]

Answer:

0.75 = 75% chance that only one tag is lost, provided at least one tag is lost

Step-by-step explanation:

Independent events:

If A and B are independent events, then:

P(A \cap B) = P(A)*P(B)

Conditional probability:

P(B|A) = \frac{P(A \cap B)}{P(A)}

Here

P(B|A) refers to the probability of event B occurring, given that event A has occurred.

P(A \cap B) is the probability of both A and B occurring together.

P(A) is the probability of event A occurring.

In this scenario:

Event A: At least one tag is missing

Event B: Only one tag is missing.

Each tag has a 40% likelihood of being lost, which is equal to 0.4.

Probability of at least one tag missing:

The events can be considered as either no tags are missing or at least one is. Their probabilities sum to 1. Thus

p + P(A) = 1

p is the probability that none are lost. Each tag has a 60% = 0.6 chance of not being lost, and since they are independent,

p = 0.6*0.6 = 0.36

Then

P(A) = 1 - p = 1 - 0.36 = 0.64

Intersection:

The intersection of at least one lost (A) and exactly one lost (B) is precisely one lost.

Then

Probability of at least one lost:

The first being lost (0.4 chance) and the second not lost (0.6 chance)

Or

The first not being lost (0.6 chance) and the second lost (0.4 chance)

So

P(A \cap B) = 0.4*0.6 + 0.6*0.4 = 0.48

Calculate the probability that exactly one tag is lost, given that at least one tag is lost (round to two decimal places).

P(B|A) = \frac{0.48}{0.64} = 0.75

0.75 = 75% likelihood that precisely one tag is lost, assuming at least one tag is lost

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18 days ago
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