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borishaifa [10]
3 years ago
11

Need help ASAP !!! And show work please

Mathematics
1 answer:
sammy [17]3 years ago
6 0

Answer:

\sqrt{40} or 6.32

Step-by-step explanation:

We can find the distance between two points by using this formula

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

where the x and y values are derived from the given points

The points given are ( -2 , 5 ) and ( 4 , 3 )

So we plug in the x and y values of those coordinates into the distance formula

\sqrt{(4-(-2)^2+(3-5)^2} \\4-(-2)=6\\3-5=-2\\d=\sqrt{6^2+-2^2} \\6^2=36\\-2^2=4\\36+4=40\\d=\sqrt{40}

or 6.32

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Answer:

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Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
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a) The 90% confidence interval of the percentage of all mangoes on the truck that fail to meet the standards is: (7.55%, 12.45%).

b) The margin of error is: 2.45%.

c) The 90% confidence is the level of confidence that the true population percentage is in the interval.

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<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions has the bounds given by the rule presented as follows:

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

The margin of error is given by:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

The variables are listed as follows:

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  • z is the critical value.
  • n is the sample size.

The confidence level is of 90%, hence the critical value z is the value of Z that has a p-value of \frac{1+0.90}{2} = 0.95, so the critical value is z = 1.645.

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  • The upper bound is: 10 + 2.45 = 12.45%.

For a margin of error of 3% = 0.03, the needed sample size is obtained as follows:

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0.03\sqrt{n} = 1.645\sqrt{0.1(0.9)}

\sqrt{n} = \frac{1.645\sqrt{0.1(0.9)}}{0.03}

(\sqrt{n}})^2 = \left(\frac{1.645\sqrt{0.1(0.9)}}{0.03}\right)^2

n = 271 (rounded up).

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

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