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Ede4ka [16]
3 years ago
6

Suzanne bought a sweater at the sale price of $29.99. the original cost of the seater was $49.99. what percent represents the di

scount they suzanne received when buying the sweater?
a: 40% discount
b: 60% discount
c: 50% discount
d: none of the above

NEED HELP ASAP MASTERY GRADE
Mathematics
2 answers:
Nikitich [7]3 years ago
6 0

Answer:

= 40%

Step-by-step explanation:

Original price = 100%

Discount = Op - Np = $49.99 - $29.99 = $20

If $49.99 = 100%

What about $20 = ?

= (20 x 100) ÷ 49.99

= 2000 ÷ 49.99

= 40.008

= 40%

blondinia [14]3 years ago
3 0
The reduction in price is:
37.5
%

Original price
→
$40
Reduced to

→
$25

Writing this as a fraction of the original price we have
25/40

But this represents the final price.
What we need is the amount of reduction


The price has been reduced by

$40−$25=$15

So as a fraction of the original price the reduction is:
15/40

Converting this to a percentage

Let the unknown value be
x

15/40=x/100


Multiply both sides by 100
15/40×100 = x × 100/100


But 100/100= 1

x = 15/40 × 100 = 37 1/2→37.5 giving: 37.5/100 = 37.5 × 1/100 = 37.5%
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Answer:

6.26-2.01\frac{2.47}{\sqrt{50}}=5.56    

6.26+2.01\frac{2.47}{\sqrt{50}}=6.96    

So on this case the 95% confidence interval would be given by (5.56;6.96)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

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The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=6.26

The sample deviation calculated s=2.47

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=50-1=49

Assuming a Confidence of 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,49)".And we see that t_{\alpha/2}=2.01

Now we have everything in order to replace into formula (1):

6.26-2.01\frac{2.47}{\sqrt{50}}=5.56    

6.26+2.01\frac{2.47}{\sqrt{50}}=6.96    

So on this case the 95% confidence interval would be given by (5.56;6.96)    

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Answer:

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Step-by-step explanation:

Let's assume

tank can hold 'x' gallons when it is full

we are given

The gas gauge in a car shows it has 40% of a tank of gas

So, gas in car is

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