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Ray Of Light [21]

3 years ago

12

Maya and Sally were the two finalists in a singing competition. The person with the most votes from the audience was chosen as t

he winner. Sally received 25% of the total votes and lost by 62,500 votes. No person in the audience was allowed to vote more than once. The equation below represents this situation, where y is the total number of votes cast and x is the number of votes Maya received:
x − 62,500 = 0.25y

If a total of 125,000 people voted, how many votes did Maya receive? Do not enter comma for placeholder values.

1 answer:

Alex787 [66]3 years ago

5 0

x-62500=0.25y

x-62500=31250

x=93750

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Let X be the number of the cars being repaired at a repair shop. We have the following information:

worty [1.4K]

Answer:

(a) Sample Space

$S = \{0,1,2,3\}$

(b) PMF

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}$

(c) CDF

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}$

Step-by-step explanation:

Solving (a): The sample space

From the question, we understand that at most 3 cars will be repaired.

This implies that, the number of cars will be 0, 1, 2 or 3

So, the sample space is:

$S = \{0,1,2,3\}$

Solving (b): The PMF

From the question, we have:

$P(2) = P(1)$

$P(0) = P(3)$

$P(1\ or\ 2) = 0.5 * P(0\ or\ 3)$

$P(1\ or\ 2) = 0.5 * P(0\ or\ 3)$ can be represented as:

$P(1) + P(2) = 0.5[P(0) + P(3)]$

Substitute $P(2) = P(1)$ and $P(0) = P(3)$

$P(1) + P(1) = 0.5[P(0) + P(0)]$

$2P(1) = 0.5[2P(0)]$

$2P(1) = P(0)$

$P(0)= 2P(1)$

Also note that:

$P(0) + P(1) + P(2) + P(3) = 1$

Substitute $P(2) = P(1)$ and $P(0) = P(3)$

$P(0) + P(1) + P(1) + P(0) = 1$

$2P(1) + 2P(0) = 1$

Substitute $P(0)= 2P(1)$

$2P(1) + 2*2P(1) = 1$

$2P(1) + 4P(1) = 1$

$6P(1) = 1$

Solve for P(1)

$P(1) = \frac{1}{6}$

To calculate others, we have:

$P(2) = P(1)$

$P(2) = P(1) = \frac{1}{6}$

$P(0)= 2P(1)$

$P(0) =2 * \frac{1}{6}$ $P(0) =\frac{1}{3}$

$P(3) = P(0) =\frac{1}{3}$

Hence, the PMF is:

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}$

<em>See attachment (1) for histogram</em>

Solving (c): The CDF ; F(x)

This is calculated as:

$F(x) = P(X \le x) =\sum\limit^{3}_{x_i \le x} P(x_i)$

For x = 0;

We have:

$P(X \le 0) = P(0)$

$P(X \le 0) = 1/3$

For x = 1

$P(X \le 1) = P(0) + P(1)$

$P(X \le 1) = 1/3 + 1/6$

$P(X \le 1) = 1/2$

For x = 2

$P(X \le 2) = P(0) + P(1) + P(2)$

$P(X \le 2) = 1/3 + 1/6 + 1/6$

$P(X \le 2) = 2/3$

For x = 3

$P(X \le 3) = P(0) + P(1) + P(2) + P(3)$

$P(X \le 3) = 1/3 + 1/6 + 1/6 + 1/3$

$P(X \le 3) = 1$

Hence, the CDF is:

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}$

<em>See attachment (2) for histogram</em>

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2 years ago

on a 140 question test a student answered 84 correctly . what percent of the problems did the student work correctly

riadik2000 [5.3K]

Answer:

it would be a 60%

Step-by-step explanation:

Divide 84 by 140, and you get .6

Hope this helps!!

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2 years ago

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At a certain clinic, 2 percent of all pap smears show signs of abnormality. (a) What is the expected number of pap smears that m

eimsori [14]

Answer:

The expected amount of pap smears that must be inspected before the first abnormal one is found is 50

Step-by-step explanation:

If 2 percent of all pap smears show signs of abnormality, then the probability that a pap smear is abnormal is 0.02.

Let X be the amount of pap smears needed before the first abnormal case is found. X has geometric ditribution with parameter p = 0.02. The mean of X, in other words, the expected amount of cases that must be inspected before the first abnormal one appears, is 1/p = 1/0.02 = 50.

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2 years ago

yuradex [85]

Answer:

<h3>A. two more than seven times x</h3>

7 × x +2
7x+2

<h3>B. the product of ten and f increased by one</h3>

10 × f + 1
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r ÷ 11

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2 years ago

Samantha solved an equation, as shown below:

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Answer:

Step-by-step explanation:

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3 years ago