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1 year ago

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The regular selling price of an item is $261. For special year-end sale the price is at a markdown of 20%. Find the discount pri

ce.

1 answer:

MrMuchimi1 year ago

7 0

Answer:

208.8 or 209 for the discount price

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See You Later Based on a Harris Interactive poll, 20% of adults believe in reincarnation. Assume that six adults are randomly se

REY [17]

Answer:

a) There is a 0.15% probability that exactly five of the selected adults believe in reincarnation.

b) 0.0064% probability that all of the selected adults believe in reincarnation.

c) There is a 0.1564% probability that at least five of the selected adults believe in reincarnation.

d) Since $P(X \geq 5) < 0.05$ , 5 is a significantly high number of adults who believe in reincarnation in this sample.

Step-by-step explanation:

For each of the adults selected, there are only two possible outcomes. Either they believe in reincarnation, or they do not. 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.

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

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

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

And p is the probability of X happening.

In this problem we have that:

$n = 6, p = 0.2$

a. What is the probability that exactly five of the selected adults believe in reincarnation?

This is P(X = 5).

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

$P(X = 5) = C_{6,5}.(0.2)^{5}.(0.8)^{1} = 0.0015$

There is a 0.15% probability that exactly five of the selected adults believe in reincarnation.

b. What is the probability that all of the selected adults believe in reincarnation?

This is P(X = 6).

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

$P(X = 6) = C_{6,6}.(0.2)^{6}.(0.8)^{0} = 0.000064$

There is a 0.0064% probability that all of the selected adults believe in reincarnation.

c. What is the probability that at least five of the selected adults believe in reincarnation?

This is

$P(X \geq 5) = P(X = 5) + P(X = 6) = 0.0015 + 0.000064 = 0.001564$

There is a 0.1564% probability that at least five of the selected adults believe in reincarnation.

d. If six adults are randomly selected, is five a significantly high number who believe in reincarnation?

5 is significantly high if $P(X \geq 5) < 0.05$

We have that

$P(X \geq 5) = P(X = 5) + P(X = 6) = 0.0015 + 0.000064 = 0.001564 < 0.05$

Since $P(X \geq 5) < 0.05$ , 5 is a significantly high number of adults who believe in reincarnation in this sample.

5 0

1 year ago

contractor wishes to build 9 houses, each different in design. In how many ways can he place these houses on a street if 6 lots

Tju [1.3M]

The houses can be placed in 362,880 ways.

<u>Step-by-step explanation:</u>

The 9 houses are each in different design.

The each lot can place any of the 9 houses.

The 1st lot can place anyone house of all the 9 houses.
The 2nd lot can place one of remaining 8 houses.
The 3rd lot can place one of remaining 7 houses.

Similarly, the process gets repeated until the last house is placed on a lot.

<u>From the above steps, it can be determined that :</u>

The number of ways to place the 9 houses in 9 lots = 9!

⇒ 9×8×7×6×5×4×3×2×1

⇒ 362880 ways.

Therefore, the houses can be placed in 362880 ways.

5 0

2 years ago

Find the fifth roots of 243(cos 300° + i sin 300°).

Stels [109]

In looking for the fifth roots, we will use De Moivre's theorem.
The formula for this problem is z^5 = 243 (cos 300 degress + i sin 300 degrees)

Where you'll also need the following data:
300/5 = 60
360/5 = 72
- you'll input these two after the cos and i sin (60+k*72) where k = 0,1,2,3,4

solution:

z^5 = 243 (cos 300 degrees + i sin 300 degrees)
z= 243^1/5 (cos 300 degrees + i sin 300 degrees)
z= 3 (cos (60 + k*72) degrees) + (i sin (60 + k*72) degrees)

so the following are the roots:

3 (cos 60 degrees + i sin 60 degrees)
3 (cos 132 degrees + i sin 132 degrees)
3 (cos 204 degrees + i sin 204 degrees)
3 (cos 276 degrees + i sin 276 degrees)
3 (cos 348 degrees + i sin 348 degrees)

7 0

2 years ago

The diagram shows two different nature trails in a state park. The solid line shows the Dogwood Trail. The dashed line shows the

Ad libitum [116K]

Lets find the missing sides first
big triangle, we apply the Pythagorean theorem a^2+b^2=c^2
a^=13^2-5^=169-25=144, Square root of a=square root of 144=12
a=12

smaller triangle
5^2-3^2=b^2
25-16=9
b=4
Elm trail is 13km+5km=18km
Dogwood trail=12+5+4+3=24km

correct choices are 2 and 5

4 0

2 years ago

Read 2 more answers

The graph shows the distribution of the number of text messages young adults send per day. The distribution is approximately Nor

Mars2501 [29]

<h3>Answer: C) 81.5%</h3>

This value is approximate.

====================================================

Explanation:

We have a normal distribution with these parameters

mu = 128 = population mean
sigma = 30 = population standard deviation

The goal is to find the area under the curve from x = 68 to x = 158, where x is the number of text messages sent per day. So effectively, we want to find P(68 < x < 158).

Let's convert the score x = 68 to its corresponding z score

z = (x-mu)/sigma

z = (68-128)/30

z = -60/30

z = -2

This tells us that the score x = 68 is exactly two standard deviations below the mean mu = 128.

Repeat for x = 158

z = (x-mu)/sigma

z = (158-128)/30

z = 30/30

z = 1

This value is exactly one standard deviation above the mean

-------------------------------------------

The problem of finding P(68 < x < 158) can be rephrased into P(-2 < z < 1)

We do this because we can then use the Empirical rule as shown in the diagram below.

We'll focus on the regions between z = -2 and z = 1. This consists of the blue 13.5% on the left, and the two pink 34% portions. So we will say 13.5% + 34% + 34% = 81.5%

Approximately 81.5% of the the population sends between 68 and 158 text messages per day. This value is approximate because the percentages listed in the Empirical rule below are approximate.

5 0

1 year ago