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Sedbober [7]
3 years ago
7

Gavin thought he was making a good investment when he heard about an online grocery company. So he bought 500 shares in the comp

any at $25 a share. The company ended up going bankrupt, and when Gavin sold his shares he only made $150 in total. What was the percent decrease in the share price? A)1.2%
B) 3%
C) 82.3%
D) 98.8%
Mathematics
1 answer:
kirill [66]3 years ago
5 0
The answer would be D) 98.8% because if the initial price is $25 per share, then Gavin would have spent $12,500. None of the other answers make sense since if it was just 3% or 82.3%, then it would be too much to equal $150 in return. If you were to put C, then the shares would equal $4.43 each, and If you bought 500 shares, then it would equal $2,212.5 which would be way more than $150. In the other hand, D is correct because then only 1 share would equal $0.3 and that multiplied by 500 equals precisely $150.

Hope this helps!
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A bag contains 4 blue marbles and 2 yellow marbles. Two marbles are randomly chosen (the first marble is NOT replaced before dra
VMariaS [17]

Answer:

0.4 = 40% probability that both marbles are blue.

0.0667 = 6.67% probability that both marbles are yellow.

53.33% probability of one blue and then one yellow

If you are told that both selected marbles are the same color, 0.8571 = 85.71% probability that both are blue

Step-by-step explanation:

To solve this question, we need to understand conditional probability(for the final question) and the hypergeometric distribution(for the first three, because the balls are chosen without being replaced).

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

Conditional Probability

We use the conditional probability formula to solve this question. It is

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

In which

P(B|A) is the probability of event B happening, given that A happened.

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

P(A) is the probability of A happening.

What is the probability that both marbles are blue?

4 + 2 = 6 total marbles, which means that N = 6

4 blue, which means that k = 4

Sample of 2, which means that n = 2

This is P(X = 2). So

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

P(X = 2) = h(2,6,2,4) = \frac{C_{4,2}*C_{2,0}}{C_{6,2}} = 0.4

0.4 = 40% probability that both marbles are blue

What is the probability that both marbles are yellow?

4 + 2 = 6 total marbles, which means that N = 6

2 yellow, which means that k = 2

Sample of 2, which means that n = 2

This is P(X = 2). So

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

P(X = 2) = h(2,6,2,2) = \frac{C_{2,2}*C_{4,0}}{C_{6,2}} = 0.0667

0.0667 = 6.67% probability that both marbles are yellow.

What is the probability of one blue and then one yellow?

Total is 100%.

Can be:

Both blue(40%)

Both yellow(6.67%)

One blue and one yellow(x%). So

40 + 6.67 + x = 100

x = 100 - 46.67

x = 53.33

53.33% probability of one blue and then one yellow.

If you are told that both selected marbles are the same color, what is the probability that both are blue?

Conditional probability.

Event A: Both same color

Event B: Both blue

Probability of both being same color:

Both blue(40%)

Both yellow(6.67%)

This means that P(A) = 0.4 + 0.0667 = 0.4667

Probability of both being the same color and blue:

40% probability that both are blue, which means that P(A \cap B) = 0.4

Desired probability:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4}{0.4667} = 0.8571

If you are told that both selected marbles are the same color, 0.8571 = 85.71% probability that both are blue

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