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mel-nik [20]
1 year ago
7

60% of the voters favor Ms. Stein. If 200 voters are chosen at random, what is the expected number of voters who indicate favori

ng Ms. Stein?The expected number of voters who indicate favoring Ms. Stein is
Mathematics
1 answer:
damaskus [11]1 year ago
6 0

Given,

The parcentage of voters in favr of Ms Stein is 60%.

The total number of voters are 200.

The expected number of voters who indicate favoring Ms. Stein is,

\text{Expected value = probability}\times sample\text{ size}

Substituting the values,

\begin{gathered} \text{Expected value=}\frac{\text{60}}{100}\times200 \\ =120 \end{gathered}

Hence, the expected number of voters who indicate favoring Ms. Stein is 120

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Of the entering class at a​ college, ​% attended public high​ school, ​% attended private high​ school, and ​% were home schoole
Veronika [31]

Answer:

(a) The probability that the student made the​ Dean's list is 0.1655.

(b) The probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c) The probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

Step-by-step explanation:

The complete question is:

Of the entering class at a college, 71% attended public high school, 21% attended private high school, and 8% were home schooled. Of those who attended public high school, 16% made the Dean's list, 19% of those who attended private high school made the Dean's list, and 15% of those who were home schooled made the Dean's list.

a) Find the probability that the student made the Dean's list.

b) Find the probability that the student came from a private high school, given that the student made the Dean's list.

c) Find the probability that the student was not home schooled, given that the student did not make the Dean's list.

Solution:

Denote the events as follows:

<em>A</em> = a student attended public high school

<em>B</em> = a student attended private high school

<em>C</em> = a student was home schooled

<em>D</em> = a student made the Dean's list

The provided information is as follows:

P (A) = 0.71

P (B) = 0.21

P (C) = 0.08

P (D|A) = 0.16

P (D|B) = 0.19

P (D|C) = 0.15

(a)

The law of total probability states that:

P(X)=\sum\limits_{i} P(X|Y_{i})\cdot P(Y_{i})

Compute the probability that the student made the​ Dean's list as follows:

P(D)=P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)

         =(0.16\times 0.71)+(0.19\times 0.21)+(0.15\times 0.08)\\=0.1136+0.0399+0.012\\=0.1655

Thus, the probability that the student made the​ Dean's list is 0.1655.

(b)

Compute the probability that the student came from a private high school, given that the student made the Dean's list as follows:

P(B|D)=\frac{P(D|B)P(B)}{P(D)}

             =\frac{0.21\times 0.19}{0.1655}\\\\=0.2410876\\\\\approx 0.2411

Thus, the probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c)

Compute the probability that the student was not home schooled, given that the student did not make the Dean's list as follows:

P(C^{c}|D^{c})=1-P(C|D^{c})

               =1-\frac{P(D^{c}|C)P(C)}{P(D^{c})}\\\\=1-\frac{(1-P(D|C))\times P(C)}{1-P(D)}\\\\=1-\frac{(1-0.15)\times 0.08}{(1-0.1655)}\\\\=1-0.0815\\\\=0.9185

Thus, the probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

3 0
2 years ago
Kate shakes a 2 liter bottle of soda, and 60 centiliters of the soda sprays out of the bottle. How much soda is left in the bott
dmitriy555 [2]
85 is the answer may be helpful
3 0
3 years ago
(13p5 + 6p - 12p2) - (-9p - p2 - 13p5)
anastassius [24]

Answer:

Step-by-step explanation:

13p⁵ + 6p - 12p² -(-9p - p² - 13p⁵) = 13p⁵ + 6p - 12p² + 9p + p² + 13p⁵

{Distribute  (-1) to the second expression}

= <u>13p⁵ + 13p⁵</u>    <u>-12p² + p²</u>     <u>+ 6p + 9p</u>

{Combine like terms}

= 26p⁵ - 11p² + 15p

7 0
2 years ago
Which is the correct answer?
vovangra [49]

B is the correct answer , it is incorrect.

5 0
3 years ago
Read 2 more answers
Sec squared 55 - tan squared 55
sergejj [24]

<u>Answer: </u>

sec squared 55 – tan squared 55  = 1

<u>Explanation:</u>

Given, sec square 55 – tan squared 55

We know that,

\sec \Theta=\frac{\text {hypotenuse}}{\text {base}}

And,

\tan \theta=\frac{\text { perpendicular }}{\text { base }}

where Ө is the angle

Substituting the values

\left(\frac{\text {hypotenuse}}{\text {base}}\right)^{2}-\left(\frac{\text { perpendicular }}{\text {base}}\right)^{2}

Solving,

\frac{(\text {hypotenuse})^{2}-(\text {perpendicular})^{2}}{(\text {base}) *(\text {base})}

According to Pythagoras theorem,

\text { (hypotenuse) }^{2}-\text { (perpendicular) }^{2}=(\text { base })^{2}

Putting this in the equation;

squared 55 - tan squared 55 =

\frac{(\text {hypotenuse})^{2}-(\text {perpendicular})^{2}}{(\text {base}) *(\text {base})}=\frac{(\text {base})^{2}}{(\text {base}) *(\text {base})}=1

Therefore, sec squared 55 – tan squared 55 = 1

6 0
3 years ago
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