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The daily revenues of a cafe near the university are approximately normally distributed. The owner recently collected a random s

lbvjy [14]

Answer:

The sample size to obtain the desired margin of error is 160.

Step-by-step explanation:

The Margin of Error is given as

$MOE=z_{crit}\times\dfrac{\sigma}{\sqrt{n}}$

Rearranging this equation in terms of n gives

$n=\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2$

Now the Margin of Error is reduced by 2 so the new M_2 is given as M/2 so the value of n_2 is calculated as

$n_2=\left[z_{crit}\times \dfrac{\sigma}{M_2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{\sigma}{M/2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{2\sigma}{M}\right]^2\\n_2=2^2\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4n$

As n is given as 40 so the new sample size is given as

$n_2=4n\\n_2=4*40\\n_2=160$

So the sample size to obtain the desired margin of error is 160.

4 0

4 years ago

Triangle KJL is inscribed in semicircle O shown below such that JK=6 and JL=11. Which of the following is closest to the radius

lakkis [162]

Uhh, answer choices?

In any case, since $\triangle KJL$ is inscribed in a semicircle, it is a right triangle (due to the inscribed angle theorem). Thus, by the Pythagorean theorem, $KL=\sqrt{6^2+11^2}=\sqrt{157}.$ Now, the circumradius of a right triangle (the radius of the circle passing through all three of its vertices) is simply half its hypotenuse, so $OJ=OK=OL=\frac{\sqrt{157}}{2}\approx \boxed{6.265}.$

7 0

3 years ago

What’s the answer ???

Aleonysh [2.5K]

Answer:

3rd

Step-by-step explanation:

7 0

3 years ago

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BC is parallel to DE. What is the length of CE? A) 2 1/3 B) 2 2/3 C) 3 1/3 D)3

Bas_tet [7]

Is there supposed to be a picture

3 0

3 years ago

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On a coordinate plane, triangles P Q R and S T U are shown. Triangle P Q R has points (4, 4), (negative 2, 0), (negative 2, 4).

mamaluj [8]

Answer:

The answer is below

Step-by-step explanation:

Two triangles are said to be similar if their corresponding angles are equal and the corresponding sides are in proportion.

The distance between two points on the coordinate plane is given as:

$Distance=\sqrt{(x_2-x_1}^2+(y_2-y_1)^2 \\\\Therefore\ in\ triangle\ PQR:\\\\|QR|=\sqrt{(-2-(-2))^2+(4-0)^2}=4\\\\|PQ|=\sqrt{(-2-4)^2+(0-4)^2}=\sqrt{52}=2\sqrt{13} \\\\|PR|=\sqrt{(-2-4)^2+(4-4)^2}=6$

In triangle STU:

$|ST|=\sqrt{(-1-2)^2+(-2-(-4))^2}=\sqrt{13}\\\\|SU|=\sqrt{(-1-2)^2+(-4-(-4))^2} =3\\\\|TU|=\sqrt{(-1-(-1))^2+(-4-(-2))^2}=2$

|QR| / |TU| = 4/2 = 2

|PR| / |SU| = 6/3 = 2

|PQ| / |ST| = 2√13 / √13 = 2

Hence:

|QR| / |TU| = |PR| / |SU| = |PQ| / |ST|

Therefore, △PQR and △STU are similar triangles since the ratio of their sides are in the same proportion.

6 0

3 years ago

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