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3 years ago

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polygon EFGH is similar to polygon JKLM, and EF and JK are corresponding sides. you know the perimeter of each polygon and you k

now the measure of GH. what can you find?

1 answer:

Savatey [412]3 years ago

8 0

This means you can now find the measure of LM :)

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you are using division to determine how much whole wheat flour to use in a bread recipe.is an estimated answer good enough? plea

Pavel [41]

Yes, it is. if you end up getting 4.85239854285696357835482938339158989............................... you wouldn't want to measure out exactly that much.

5 0

3 years ago

Read 2 more answers

The life in hours of a 75-watt light bulb is known to be normally distributed with standard deviation σ = 25 hours. A random sam

irina [24]

Answer: a) (1008.34,1019.658) b) (1009.24,1018.76)

Step-by-step explanation:

Since we have given that

n = 75

mean = 1014 hours

Standard deviation = 25 hours

At 95% two sided , z = 1.96

So, confidence interval would be

$\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=1014\pm 1.96\dfrac{25}{\sqrt{75}}\\\\=1014\pm 5.658\\\\=(1014-5.658,1014+5.658)\\\\=(1008.34,1019.658)$

(b) Construct a 95% lower confidence bound on the mean life.

z = 1.65

So, confidence interval would be

$\bar{x}\pm z\times \dfrac{\sigma}{\sqrt{n}}\\\\=1014\pm 1.65\times \dfrac{25}{\sqrt{75}}\\\\=1014\pm 4.76\\\\=(1014-4.76,1014+4.76)\\\\=(1009.24,1018.76)$

Hence, a) (1008.34,1019.658) b) (1009.24,1018.76)

4 0

3 years ago

a. Fill in the midpoint of each class in the column provided. b. Enter the midpoints in L1 and the frequencies in L2, and use 1-

Tresset [83]

Answer:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

$M = \frac{1}{2}(Lower + Upper)$

Where

$Lower \to$ Lower class interval

$Upper \to$ Upper class interval

So, we have:

Class 63-65:

$M = \frac{1}{2}(63 + 65) = 64$

Class 66 - 68:

$M = \frac{1}{2}(66 + 68) = 67$

When the computation is completed, the frequency distribution will be:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

$\bar x = 70.2903$

$\sigma = 3.5795$

<em>See attachment for result of 1-VarStats</em>

8 0

3 years ago

Which term of ap 30,27,24is0

son4ous [18]

Answer:

11th term is 0

Step-by-step explanation:

30, 27 , 24 ,......0

a = first term = 30

Common difference = second term - first term = 27 - 30 = -3

nth term = a+(n-1)*d

a + (n-1)d = 0

30 + (n - 1) *(-3) = 0

30 + n*(-3) -1*(-3) = 0

30 - 3n + 3 = 0

-3n + 33 = 0

-3n = -33

n = -33/-3

n = 11

8 0

3 years ago

What's the probability of landing on green or purple?

Lena [83]

Answer:

green by searching me on is green

4 0

3 years ago

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