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3 years ago

The endpoints of EF are E(1,3) and F(-2,8). Find EF.

Mathematics

1 answer:

Phantasy [73]3 years ago

8 0

Answer:

EF = $\sqrt{34}$ ≈ 5.83

Step-by-step explanation:

Calculate EF using the distance formula

d = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

with (x₁, y₁ ) = E(1, 3) and (x₂, y₂ ) = F(- 2, 8)

EF = $\sqrt{(-2-1)^2+(8-3)^2}$

= $\sqrt{(-3)^2+5^2}$

= $\sqrt{9+25}$

= $\sqrt{34}$ ≈ 5.83 ( to 2 dec. places )

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Suppose twenty-two communities have an average of = 123.6 reported cases of larceny per year. assume that σ is known to be 36.8

Delvig [45]

We are given the following data:

Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22

We are to find the confidence intervals for 90%, 95% and 98% confidence level.

Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.

Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.645* \frac{36.8}{ \sqrt{22}}=110.69$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.645* \frac{36.8}{ \sqrt{22}}=136.51$

Thus the 90% confidence interval will be (110.69, 136.51)

Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.96* \frac{36.8}{ \sqrt{22}}=108.22$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.96* \frac{36.8}{ \sqrt{22}}=138.98$

Thus the 95% confidence interval will be (108.22, 138.98)

Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-2.327* \frac{36.8}{ \sqrt{22}}=105.34$
Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+2.327* \frac{36.8}{ \sqrt{22}}=141.86$

Thus the 98% confidence interval will be (105.34, 141.86)

Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)

As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.

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3 years ago

What is the solution to question number 33?

Oksana_A [137]

If the width of the garden is w, then the maximum area available is a square, giving you w² as the dimensions. So:
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3 years ago

1 gallon of batter is poured equally into 4 bowls. How many gallons of batter are in each bowl?

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Answer:

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Step-by-step explanation:

This is because you are spliting one thing into four parts so each bowl would have 1/4 of the gallon.

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Answer: r = C/(2pi)

This is the same as writing $r = \frac{C}{2\pi}$

=========================================================

Explanation:

pi is some number (approximately 3.14) which means 2pi is also some number (roughly 6.28)

Saying c = 2pi*r means we have 2pi times r, and the result is the circumference c. To isolate r, we'll undo the multiplication. We'll undo it by dividing both sides by 2pi like so

C = 2pi*r

C/(2pi) = r

r = C/(2pi)

in which we can write it like $r = \frac{C}{2\pi}$

So whatever C is, we divide it over 2pi (aka roughly 6.28) to get the radius.

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Extra Info (Optional Section):

As an example, let's say the circumference of the circle is 628 feet. This means the distance around the circle is 628 feet. So C = 628 would lead to...

r = C/(2pi)

r = C/(6.28)

r = 628/(6.28)

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So the radius would be roughly 100 feet.

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