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Anuta_ua [19.1K]
2 years ago
8

Paolo paid $ 28 $ 28 for a hat that was originally priced at $ 35 $ 35 . By what percent was the hat discounted?

Mathematics
1 answer:
devlian [24]2 years ago
5 0

The original price of the hat is $35. Paolo paid $28 for the hat.

So discount for the hat is $(35-28) = $7

$7 is discounted from $35. We have to find the percentage of the discount.

To find the percentage we have to divide the discount amount by the original price and then have to multiply it by 100.

So the discount percentage =

((7/35)×100) %

We can simplify 7/35 by dividing 7 and 35 both by 7. So we will get 1/5 after simplifying.

((1/5) ×100) %

(100/5)% = 20%

So the hat was discounted by 20%.

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The equation |x − 8| = 3 represents the minimum and maximum percent of people in a survey who are undecided about an issue. What
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A savings and loan association needs information concerning the checking account balances of its local customers. A random sampl
Citrus2011 [14]

Answer:

a) 98% confidence interval for the true mean checking account balance for local customers.

(453.586 , 874.693)

b)   95% confidence interval for the standard deviation.

(214.91 , 441.53)

Step-by-step explanation:

Given a size of sample 'n' =14

given mean of the sample x⁻ = $664.14

standard deviation of the sample 'S' = $297.29.

a)

<u>98% of confidence intervals</u>

<u></u>(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } , x^{-}+ t_{\alpha }\frac{S}{\sqrt{n} } )<u></u>

The degrees of freedom γ=n-1 =14-1 =13

t₁₃ = 2.650 at 98% of confidence level of signification.

(664.14- 2.650\frac{297.29}{\sqrt{14} } , 664.14+ 2.650\frac{297.29}{\sqrt{14} } )

on calculation, we get

(664.14-210.553 , 664.14+210.553)

(453.586 , 874.693)

98% confidence interval for the true mean checking account balance for local customers.

(453.586 , 874.693)

<u>95% of confidence intervals</u>

({s\sqrt{\frac{n-1}{X^{2} _{(\frac{\alpha }{2} ,n-1) } } } ,s\sqrt{\frac{n-1}{X^2_{\frac{1-\alpha }{2},n-1 } } }  )

The degrees of freedom γ=n-1 =14-1 =13

X^2_{0.05,13} =22.36     (check table)

X^2_{0.95,13} = 5.892    (check table)

(297.29. (\sqrt{\frac{14-1}{X^2_{0.05,13}  } } ),297.29(\sqrt{\frac{14-1}{X^2_{0.95,13} } } )

(297.29. (\sqrt{\frac{14-1}{22.36  } } ),297.29(\sqrt{\frac{14-1}{5.892 } } )

(214.91 , 441.53)

95% confidence interval for the standard deviation.

(214.91 , 441.53)

7 0
2 years ago
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