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3 years ago

Raul has $40

Mathematics

2 answers:

Oksi-84 [34.3K]3 years ago

8 0

The answer to the problem is b

expeople1 [14]3 years ago

8 0

Let, amount of money he can spend on the present before tax=p

The tax on present of cost p =6% of p=0.06p

Amount of money he spent on present after tax= p +0.06p=1.06p

the amount he can spent should be less than 40

So, 1.06p≤40

To get the value of p, let us divide by 1.06 on both sides

$\frac{1.06}{1.06}p \leq\frac{40}{1.6}$

$\frac{1}{1}p \leq 37.73$

p≤37.73 or 37.73≥p

Answer:Option (b)

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95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

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We are given that the tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm 2 and a standard deviation of 2.15.

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P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $74.28-1.96 \times {\frac{2.15}{\sqrt{50} } }$ , $74.28+1.96 \times {\frac{2.15}{\sqrt{50} } }$ ]

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