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svet-max [94.6K]

3 years ago

10

If King sets the price at $2.50 per doll how many disappointed customers will each store have during this week?.explain your pro

cess in complete sentences

1 answer:

Anastasy [175]3 years ago

5 0

Answer: about TWO

Step-by-step explanation:

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Visa Card USA studied how frequently young consumers, ages 18 to 24, use plastic (debit and credit) cards in making purchases (A

umka2103 [35]

Answer:

Given the consumer is 18 to 24 years old, there is 37% probability that he uses a plastic card.

Step-by-step explanation:

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem, we have:

What is the probability of the consumer using a plastic card, given that the consumer is 18 to 24 years old.

The problem states that the probability that a consumer uses a plastic card when making a purchase is .37, so $P(B) = 0.37$

P(A/B) is the probability that the consumer being 18 to 24 years old, given that he uses a plastic card. The problem states that this probability is .19. So $P(A/B) = 0.19$

P(A) is the probability that the consumer is 18 to 24 years old. There is a .81 probability that the consumer is more than 24 years old. So there is a .19 probability that he is 18 to 24 years old. So $P(A) = 0.19$

The probability is

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.37*0.19}{0.19} = 0.37$

Given the consumer is 18 to 24 years old, there is 37% probability that he uses a plastic card.

4 0

4 years ago

The function y= 5^x passes through the point

wlad13 [49]

The correct answer is c :)
the point is described by two variables: x and y. it is written like that: (x, y). so the first variable in brackets is the x variable. the other is y. so pick every point a-d, put these values into the equation, solve it and check if the values before and after = are the same. if so - you are correct!
look:
c) (1, 5) so x=1, y=5
equation: y=5^x
putting values: 5=5^1 and this is true!

4 0

3 years ago

In the figure below, Find AB Help

Tema [17]

we have:

CD ÷ CA = DE ÷ AB

=> 6 ÷ 11 = 12 ÷ AB

=> AB = 12 × 6 ÷ 11 = 72/11 ≈ 6,5

Answer: 72/11 ≈ 6,5

P/s: CA = CD + AD = 6 + 5 = 11

Ok done. Thank to me :>

4 0

2 years ago

Please solve for n the following equation..............-2 = -n + 8

Zielflug [23.3K]

Answer:

n=-5

Step-by-step explanation:

4 0

3 years ago

Suppose you toss a fair coin 10 times, let X denote the number of heads. (a) What is the probability that X=5? (b) What is the p

zubka84 [21]

Answer: The required answers are

(a) 0.25, (b) 0.62, (c) 6.

Step-by-step explanation: Given that we toss a fair coin 10 times and X denote the number of heads.

We are to find

(a) the probability that X=5

(b) the probability that X greater or equal than 5

(c) the minimum value of a such that P(X ≤ a) > 0.8.

We know that the probability of getting r heads out of n tosses in a toss of coin is given by the formula of binomial distribution as follows :

$P(X=r)=^nC_r\left(\dfrac{1}{2}\right)^r\left(\dfrac{1}{2}\right)^{n-r}.$

(a) The probability of getting 5 heads is given by

$P(X=5)\\\\\\=^{10}C_5\left(\dfrac{1}{2}\right)^5\left(\dfrac{1}{2}\right)^{10-5}\\\\\\=\dfrac{10!}{5!(10-5)!}\dfrac{1}{2^{10}}\\\\\\=0.24609\\\\\sim0.25.$

(b) The probability of getting 5 or more than 5 heads is

$P(X\geq 5)\\\\=P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)\\\\=^{10}C_5\left(\dfrac{1}{2}\right)^5\left(\dfrac{1}{2}\right)^{10-5}+^{10}C_6\left(\dfrac{1}{2}\right)^6\left(\dfrac{1}{2}\right)^{10-6}+^{10}C_7\left(\dfrac{1}{2}\right)^7\left(\dfrac{1}{2}\right)^{10-7}+^{10}C_8\left(\dfrac{1}{2}\right)^8\left(\dfrac{1}{2}\right)^{10-8}+^{10}C_9\left(\dfrac{1}{2}\right)^9\left(\dfrac{1}{2}\right)^{10-9}+^{10}C_{10}\left(\dfrac{1}{2}\right)^{10}\left(\dfrac{1}{2}\right)^{10-10}\\\\\\=0.24609+0.20507+0.11718+0.04394+0.0097+0.00097\\\\=0.62295\\\\\sim 0.62.$

(c) Proceeding as in parts (a) and (b), we see that

if a = 10, then

$P(X\leq 0)=0.00097,\\\\P(X\leq 1)=0.01067,\\\\P(X\leq 2)=0.05461,\\\\P(X\leq 3)=0.17179,\\\\P(X\leq 4)=0.37686,\\\\P(X\leq 5)=0.62295,\\\\P(X\leq 6)=0.82802.$

Therefore, the minimum value of a is 6.

Hence, all the questions are answered.

3 0

3 years ago