Shelley read 40 business proposals between 6 am and 11 am. On average how many proposals did she read per hour

Mathematics

1 answer:

UkoKoshka [18]2 years ago

6 0

Answer: 8 proposals per hour

Step-by-step explanation: 40/5 = 8

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The $100(1-\alpha)\%$ confidence interval of a standard deviation is given by:

$\sqrt{ \frac{(n-1)s^2}{\chi^2_{1- \frac{\alpha}{2} } }} \leq\sigma\leq\sqrt{ \frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2} } }}$

Given a sample size of 31 charge sales, the degree of freedom is 30 and for 90% confidence interval,

$\chi^2_{1- \frac{\alpha}{2} }=43.773 \\ \\ \chi^2_{\frac{\alpha}{2} }=18.493$

Therefore, the 90% confidence interval for the standard deviation is given by

$\sqrt{ \frac{(31-1)50^2}{43.773 }} \leq\sigma\leq\sqrt{ \frac{(31-1)50^2}{18.493 }} \\ \\ \Rightarrow\sqrt{ \frac{(30)2500}{43.773 }} \leq\sigma\leq\sqrt{ \frac{(30)2500}{18.493 }} \\ \\ \Rightarrow\sqrt{ \frac{75000}{43.773 }} \leq\sigma\leq\sqrt{ \frac{75000}{18.493 }} \\ \\ \Rightarrow \sqrt{1713.38} \leq\sigma\leq \sqrt{4055.59} \\ \\ \Rightarrow41.4\leq\sigma\leq63.7$

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Wittaler [7]

P(y)=4/15
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(I’m assuming this is without replacement since you are the starburst, so there is one less of the total.)

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A factory makes TV's and video machines. Checking for faults regularly, on average, 5 out of every 300 TV's are faulty and 8 out

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I need help please asap

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0, 1/13, 49/100, 1/2, 11/20, 8/9, 1

Step-by-step explanation:

You know the smallest is 0, largest is 1

the middle is 1/2

49/100 is slightly smaller than 1/2 or 50/100

11/20 is slightly larger than 1/2 or 10/20

1/13 is slightly larger than 0

8/9 is slightly smaller than 1

please give thanks :)

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