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spayn [35]
2 years ago
8

Jane paid $40 for an item after she received a 20% discount. Jane's friend says this means that the original price of the item w

as $48. Determine if Jane's friend is correct about the original price of the item.
Mathematics
2 answers:
atroni [7]2 years ago
6 0

Answer:

Jane's friend's assumption about the original price was incorrect.

Step-by-step explanation:

A price of $40 after being given a 20% discount would end up being $50 instead of $48. This is because 20% of $50 is $10, and that is what was deducted from the price to make it $40.

Hope I helped! ☺

Alika [10]2 years ago
3 0

Answer:

she is incorrect because if she was given a 20% discount the amount she would have to pay is 38.4

Step-by-step explanation:

48 x .2 = 9.6

48 - 9.6 = 38.4 not 40

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Answer:

7,7, and 7

Step-by-step explanation:

If she vase is cubical we know that all 3 dimensions are the same. This means that x^3 = 343. So we know the answer is the cube root of 343, which is 7

3 0
3 years ago
A grocery store’s receipts show that Sunday customer purchases have a skewed distribution with a mean of 27$ and a standard devi
34kurt

Answer:

(a) The probability that the store’s revenues were at least $9,000 is 0.0233.

(b) The revenue of the store on the worst 1% of such days is $7,631.57.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.  

Then, the mean of the distribution of the sum of values of X is given by,  

 \mu_{X}=n\mu

And the standard deviation of the distribution of the sum of values of X is given by,  

\sigma_{X}=\sqrt{n}\sigma

It is provided that:

\mu=\$27\\\sigma=\$18\\n=310

As the sample size is quite large, i.e. <em>n</em> = 310 > 30, the central limit theorem can be applied to approximate the sampling distribution of the store’s revenues for Sundays by a normal distribution.

(a)

Compute the probability that the store’s revenues were at least $9,000 as follows:

P(S\geq 9000)=P(\frac{S-\mu_{X}}{\sigma_{X}}\geq \frac{9000-(27\times310)}{\sqrt{310}\times 18})\\\\=P(Z\geq 1.99)\\\\=1-P(Z

Thus, the probability that the store’s revenues were at least $9,000 is 0.0233.

(b)

Let <em>s</em> denote the revenue of the store on the worst 1% of such days.

Then, P (S < s) = 0.01.

The corresponding <em>z-</em>value is, -2.33.

Compute the value of <em>s</em> as follows:

z=\frac{s-\mu_{X}}{\sigma_{X}}\\\\-2.33=\frac{s-8370}{316.923}\\\\s=8370-(2.33\times 316.923)\\\\s=7631.56941\\\\s\approx \$7,631.57

Thus, the revenue of the store on the worst 1% of such days is $7,631.57.

5 0
2 years ago
Help me solve for x and y .
Olegator [25]

Answer:

Step-by-step explanation:

7 0
2 years ago
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2(3x + 1) - (x - 5) = 42
lbvjy [14]

Answer:

6x+2-1x+5=42

Step-by-step explanation:

6 0
3 years ago
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Draw a picture of the standard normal curve and shade the area that corresponds to the requested probabilities. Then use the sta
elena-14-01-66 [18.8K]

Answer:

a)P [ z > 1,38 ] = 0,08379

b) P [ 1,233 < z < 2,43 ]  = 0,1012

c)  P [ z > -2,43 ]  = 0,99245

Step-by-step explanation:

a) P [ z > 1,38 ] = 1 -  P [ z < 1,38 ]

From z-table  P [ z < 1,38 ] = 0,91621

P [ z > 1,38 ] = 1 - 0,91621

P [ z > 1,38 ] = 0,08379

b)  P [ 1,233 - 2,43 ]  must be  P [ 1,233 < z < 2,43 ]

P [ 1,233 < z < 2,43 ]  = P [ z < 2,43 ] - P [ z > 1,233 ]

P [ z < 2,43 ]  = 0,99245

P [ z > 1,233 ] = 0,89125    ( approximated value  without interpolation)

Then

P [ 1,233 < z < 2,43 ]  = 0,99245 - 0,89125

P [ 1,233 < z < 2,43 ]  = 0,1012

c) P [ z > -2,43 ]

Fom z-table

P [ z > -2,43 ] = 1 - P [ z < -2,43 ]

P [ z > -2,43 ] = 1 - 0,00755

P [ z > -2,43 ]  = 0,99245

8 0
2 years ago
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