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Alexus [3.1K]
1 year ago
11

A $55 pair of shoes was discounted 15 percent, making the sale price $46.75. The discounted price was then discounted again by 1

0 percent. Explain how you would find the final price of the shoes.
Mathematics
2 answers:
BigorU [14]1 year ago
5 0
46.75 - 0.10(46.75) = 46.75 - 4.675 = 42.075 rounds to 42.08
ur simply taking ur discounted price of $ 46.75 and subtracting 10% of 46.75....leaving u with a final price of 42.08
Viefleur [7K]1 year ago
4 0

Answer:


Step-by-step explanation:

The final price will be the discounted price multiplied by 100% — 10%. I could also multiply the discounted price by 0.10 and subtract that amount from the discounted price to get the final price.

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According to a survey, 10% of Americans are afraid to fly. Suppose 1100 Americans are sampled. a) What is the probability that 1
JulijaS [17]

Answer:

a) 14.46%

b) 0.00%

c) 1.54%

Step-by-step explanation:

According to the survey, 10% of Americans are afraid to fly.

This means p=0.10 and q=1-0.10=0.90.

If 1100 Americans are sampled, then the sample size is n=1100 .

The mean of the distribution is \mu=np.

This means \mu=1100\times 0.10=110

The standard deviation is \sigma=\sqrt{npq}

We substitute the values to get:

\sigma=\sqrt{1100\times 0.1\times 0.9}=9.95

a) We want to find the probability that 121 or more Americans in the survey are afraid to fly.

We first apply the continuity correction factor to get: P(X\ge121)=P(X\:>\:121-0.5)\\P(X\ge121)=P(X\:>\:120.5)

We now convert to Z-scores to get:

P(X\:>\:120.5)=P(z\:>\:\frac{120.5-110}{9.95})=1.06\\

From the standard normal distribution table P(z>1.06)=0.1446

As a percentage, the probability is 14.46%

b) We want to find the probability that 165 or more Americans in the survey are afraid to fly.

We apply the CCF to get:

P(X\ge165)=P(X\:>\:165-0.5)\\P(X\ge165)=P(X\:>\:164.5)

We convert to z-scores:

P(X\:>\:164.5)=P(z\:>\:\frac{164.5-110}{9.95})=5.48

From the normal distribution, P(z>164.5)=0

c)  First, 8% of 1100 is 88.

We want to find the probability that 88 or less Americans in the survey are afraid to fly.

We apply the CCF to get:

P(X\le88)=P(X\:

We convert to z-scores:

P(X\:

From the normal distribution, P(z>\:-2.16)=0.0154

As a percentage, we get 1.54%

3 0
2 years ago
Why is customer awareness a synonym of customer responsibility?
LUCKY_DIMON [66]
Customer awareness is the same as customer responsibility. this is because a customer should always be aware of any deals, defectives, or flaws in the product. being “aware” means “having knowledge of a situation”. being aware of your purchases affects your responsibility and attentiveness. customers should always examine the closely before purchasing.
7 0
1 year ago
Hikers came across a part of a redwood stump. If the length of the chord is 8 feet, what was the diameter of the tree?
erastova [34]
So in the problem, the length of the chord there is the circumference of the tree. So in order to get the diameter of the tree, we must use the formula in getting the circumference of a circle that is stated as follows. 

C = 2pi *Radius

so first we need the get the radius of the tree which represent by this formula:

Radius = C /2pi = 8/6.2832 = 1.2732 ft 
Diameter = 2*radius = 2 * 1.27 32 = 2.5465 feet 

In summary, the diameter of the tree is 2.565 feet. 


7 0
1 year ago
A toy company produces rubber balls that have a radius of 1.7 cm. A sphere has a radius of 1.7 centimeters. What is the volume o
kow [346]

Answer: 1 ) 20.58 cm^3

2 ) $0.09

3 ) $ 0.41

Step-by-step explanation:

7 0
1 year ago
Read 2 more answers
The College Board SAT college entrance exam consists of three parts: math, writing and critical reading (The World Almanac 2012)
Eddi Din [679]

Answer:

Step-by-step explanation:

Hello!

You have the math and writing SAT scores of twelve students.

There are two variables of interest X₁: Math SAT score of a student. and X₂: Writing the SAT score of a student.

These two variables aren't independent since both of them represent data corresponding to the same student, meaning, that the math and writing scores belong to the same students and not to two separate groups of students.

This is an example of paired samples, to make the statistical test you have to establish a third variable, this variable will be the difference between the other two:

Xd= X₁-X₂

Xd: "Difference between the Math and Writing SAT scores of a student.

This variable has a normal distribution Xd~N(μd;σd²)

a. Using a .05 level of significance and test for a difference between the population mean for the math scores and the population mean for the writing scores? Enter negative values as negative numbers. Round your answer to two decimal places.

The parameter of interest is μd is the population mean of the difference between the math and writing SAT scores of the students.

The hypotheses are:

H₀: μd=0

H₁: μd≠0

α: 0.05

The test statistic is a t-student for dependent samples and it's rejection region is two-tailed.

t= \frac{Xd[bar]-Mud}{Sd/\sqrt{n} } = \frac{25-0}{37.05/\sqrt{12} } = 2.33

What is the p-value? Round your answer to four decimal places.

The p-value for this test is: 0.0394

The p-value is less than the level of significance, the decision is to reject the null hypothesis.

b. What is the point estimate of the difference between the mean scores for the two tests?

The sample mean for the variable "difference" is X[bar]d

You can calculate the point estimate of the sample mean of the variable Xd using two ways.

1) You calculate all the differences between the pairs of scores, add them and divide them by the sample size

X[bar]d= ∑di/n

∑di= 300

n=12

X[bar]d= 300/12= 25

2) You can calculate the sample mean for each variable and then calculate the difference between the two sample means

X[bar]₁= ∑X₁/n

∑X₁= 6168

X[bar]₁= 6168/12= 514

X[bar]₂= ∑X₂/n

∑X₂= 5868

X[bar]₂= 5868/12= 489

X[bar]d= X[bar]₁-X[bar]₂ = 514 - 489= 25

What are the estimates of the population mean scores for the two tests?

Math test X[bar]₁= 514

Writing test X[bar]₂= 489

Which test reports the higher mean score?

The Math test reports a higher mean score.

I hope it helps!

3 0
1 year ago
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