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zalisa [80]
3 years ago
6

Suppose that shoe sizes of American women have a bell-shaped distribution with a mean of 8.33 and a standard deviation of 1.5. U

sing
the empirical rule, what percentage of American women have shoe sizes that are no more than 11.33? Please do not round your answer.
Mathematics
1 answer:
never [62]3 years ago
8 0

"97.5" will be the American women's % having their shoe sizes that are no more than 11.33.

Given values are:

  • Mean = 8.33
  • Standard deviation = 1.5

Now,

= 8.33 + 2\times 1.5

= 11.33 (2 above Standard deviation)

→ The data values are between 8.33 and 11.33,

= \frac{95}{2}

= 47.5 (%)

→ The data value below 11.33,

= 50 (%)

hence,

→ The percentage of American women will be:

= 50+47.5

= 97.5 (%)

Thus the above answer is appropriate.

Learn more about empirical rule here:

brainly.com/question/18529739

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A jeweler wants to make 14 grams of an alloy that is precisely 75% gold.. The jeweler has alloys that are 25% gold, 50% gold, &a
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Given that the jeweler has alloys that are 25% gold, 50% gold, and 82% gold.

As he wants to make 14 grams of an alloy by adding two different alloys that is precisely 75% gold, so one alloy must have a percentage of gold more than 75%.

One alloy is 82% gold and, the second can be chosen between 25% gold, 50% gold, so there are two cases.

Case 1: 82% gold + 50% gold

Let x grams of 82% gold and y  grams of 50% gold added to make x+y=14 grams of 75% gold, so

75% of 14 = 82% of x + 50% of y

\Rightarrow 75/100 \times 14 = 82/100 \times x + 50/100 \times y \\\\

\Rightarrow 75/100 \times 14 = 82/100 \times x + 50/100 \times (14-x)  [as x+y=14]

\Rightarrow 75 \times 14 = 82 \times x + 50 \times (14-x)  \\\\\Rightarrow 75 \times 14 = 82 \times x + 50 \times14-50\times x \\\\\Rightarrow 75 \times 14 = 32 \times x + 50 \times14 \\\\\Rightarrow 32 \times x =75 \times 14 - 50 \times14 \\\\

\Rightarrow x =(25 \times 14)/32=10.9375 grams

and y = 14-x= 14-10.9375=3.0625 grams.

Hence, 10.9375 grams of 82% gold and 3.0625  grams of 50% gold added to make 14 grams of 75% gold.

Case 2: 82% gold + 25% gold

Let x grams of 82% gold and y  grams of 25% gold added to make x+y=14 grams of 75% gold, so

75% of 14 = 82% of x + 25% of y

\Rightarrow 75/100 \times 14 = 82/100 \times x + 25/100 \times y \\\\\Rightarrow 75/100 \times 14 = 82/100 \times x + 25/100 \times (14-x) \\\\ \Rightarrow 75 \times 14 = 82 \times x + 25 \times (14-x)  \\\\\Rightarrow 75 \times 14 = 82 \times x + 25 \times14-25\times x \\\\\Rightarrow 75 \times 14 = 57 \times x + 25 \times14 \\\\\Rightarrow 57 \times x =75 \times 14 - 25 \times14 \\\\

\Rightarrow x =(50 \times 14)/57=12.28 grams

and y = 14-x= 14-12.28=1.72 grams.

Hence, 12.28 grams of 82% gold and 1.72  grams of 50% gold added to make 14 grams of 75% gold.

3 0
2 years ago
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