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Stolb23 [73]
3 years ago
11

Please help me asap​

Mathematics
2 answers:
kotegsom [21]3 years ago
6 0

Answer:

-.51

-1/2

.42

11/20

omeli [17]3 years ago
3 0

Answer:

-.51, .1/2, .42, 11/20

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Engineers must consider the breadths of male heads when designing helmets. The company researchers have determined that the popu
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Answer:

The minimum head breadth that will fit the clientele is 4.4 inches.

The maximum head breadth that will fit the clientele is 7.8 inches.

Step-by-step explanation:

Let <em>X</em> = head breadths of men that is considered for the helmets.

The random variable <em>X</em> is normally distributed with mean, <em>μ</em> = 6.1 and standard deviation, <em>σ</em> = 1.

To compute the probability of a normal distribution we first need to convert the raw scores to <em>z</em>-scores using the formula:

z=\frac{x-\mu}{\sigma}

It is provided that the helmets will be designed to fit all men except those with head breadths that are in the smallest 4.3% or largest 4.3%.

Compute the minimum head breadth that will fit the clientele as follows:

P (X < x) = 0.043

⇒ P (Z < z) = 0.043

The value of <em>z</em> for this probability is:

<em>z</em> = -1.717

*Use a <em>z</em>-table.

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

z=\frac{x-\mu}{\sigma}\\-1.717=\frac{x-6.1}{1}\\x=6.1-(1.717\times 1)\\x=4.383\\x\approx4.4

Thus, the minimum head breadth that will fit the clientele is 4.4 inches.

Compute the maximum head breadth that will fit the clientele as follows:

P (X > x) = 0.043

⇒ P (Z > z) = 0.043

⇒ P (Z < z) = 1 - 0.043

⇒ P (Z < z) = 0.957

The value of <em>z</em> for this probability is:

<em>z</em> = 1.717

*Use a <em>z</em>-table.

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

z=\frac{x-\mu}{\sigma}\\1.717=\frac{x-6.1}{1}\\x=6.1+(1.717\times 1)\\x=7.817\\x\approx7.8

Thus, the maximum head breadth that will fit the clientele is 7.8 inches.

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~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill&\$200\\ P=\textit{original amount deposited}\dotfill & \$850\\ r=rate\to 7\%\to \frac{7}{100}\dotfill &0.07\\ t=years \end{cases} \\\\\\ 200=850(0.07)t\implies \cfrac{200}{850(0.07)}=t\implies \stackrel{\textit{about 3 years, 4 months and 10 days}}{3.36\approx t}

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