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zysi [14]
3 years ago
14

Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

Mathematics
1 answer:
jekas [21]3 years ago
6 0

Answer:

Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16% Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%. Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%. Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%

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PLEASE HELPPPPP. URGEENNTT!!! I DONT KNOW HOW TO SOLVE THIS
Oksi-84 [34.3K]

Answer:

A = 58.7 degrees

B = 66.9 degrees

C = 34.1 degrees

Step-by-step explanation:

<u><em>For <A:</em></u>

Tan A = \frac{opposite}{adjacent}

Tan A = \frac{23}{14}

Tan A = 1.6

A = Tan^{-1} 1.6

A = 58.7 degrees

<u>For <B:</u>

Sin B = \frac{opposite}{hypotenuse}

Sin B = \frac{23}{25}

Sin B = 0.92

B = Sin^{-1} 0.92

B = 66.9 degrees

<em><u>For <C:</u></em>

Sin C = \frac{opposite}{hypotenuse}

Sin C = \frac{14}{25}

Sin C = 0.56

C = Sin^{-1}0.56

C = 34.1 degrees

6 0
3 years ago
Lori had $88. She earned $29 walking dogs. After she spent money for concert tickets, Lori has $74 remaining.
BaLLatris [955]

Answer: $43

Step-by-step explanation:

total 88 + 29 = 117

117-74 = cost of concert tickets

= $43

4 0
3 years ago
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If one bag of topsoil covers 10 square feet, how many bags are needed to cover this flowr bed?
Nataly [62]
3 bags are needed to cover the flower bed.
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Please help me as soon as possible
Anika [276]

Answer:

A

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A is the correct answer

5 0
3 years ago
Does anyone know the answer to this!?
PolarNik [594]

Answer:

|A_y|=\left|\begin{array}{cc}12&7\\ \\17&-51\end{array}\right|

Step-by-step explanation:

You can use these Cramer's formulas to solve for x and y:

x=\dfrac{|A_x|}{|A|},\\ \\y=\dfrac{|A_y|}{|A|},

where

|A|=\left|\begin{array}{cc}12&-13\\ \\17&-22\end{array}\right|\\ \\ \\|A_x|=\left|\begin{array}{cc}7&-13\\ \\-51&-22\end{array}\right|\\ \\ \\|A_y|=\left|\begin{array}{cc}12&7\\ \\17&-51\end{array}\right|

So,

|A_y|=\left|\begin{array}{cc}12&7\\ \\17&-51\end{array}\right|

4 0
3 years ago
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