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3 years ago

Max finds the product 3/7 *16

2 answers:

4 0

$\\
\text{We need to find the product, }\frac{3}{7}\times 16\\
\\
\text{to multiply it, first we write the 16 in fraction form as }\frac{16}{1}\\
\\
\text{so we now the product becomes}\\
\\
\frac{3}{7}\times \frac{16}{1}\\
\\
\text{Now to multiply the fractions, we multiply the numerators together}\\
\text{and the denominators together. so we have}\\
\\
\frac{3}{7}\times \frac{16}{1}=\frac{3\times 16}{7\times 1}\\
\\
=\frac{48}{7}$

Hence the product is 48/7

Verdich [7]3 years ago

3 0

Max finds that the answer is 6.857 lol

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Step-by-step explanation:

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3 years ago

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2 years ago

One of the relatively uncommon instances in which researchers know the population standard deviation is in the case of the Intel

Phoenix [80]

Answer:

90% confidence interval for the mean IQ score in this community is [106.58 , 109.42].

Step-by-step explanation:

We are given that in general, the average IQ score in large, diverse populations is 100 and the standard deviation is 15.

Suppose that your sample of 300 members of your community gives you a mean IQ score of 108.

Firstly, the pivotal quantity for 90% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean IQ score = 108

$\sigma$ = population standard deviation = 15

n = sample of members = 300

$\mu$ = population mean IQ score

Here for constructing 90% confidence interval we have used One-sample z test statistics because we know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X -\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $108-1.645 \times {\frac{15}{\sqrt{300} } }$ , $108+1.645 \times {\frac{15}{\sqrt{300} } }$ ]

= [106.58 , 109.42]

Therefore, we can be 90% confident that the mean IQ score in this community lies between 106.58 and 109.42.

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3 years ago