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Alex17521 [72]
1 year ago
14

1. Miguel is playing a game in which a box contains four chips with numbers written on them. Two of the chips have the number 1,

one chip has the number 3, and the other chip has the number 5. Miguel must choose two chips, and if both chips have the same number, he wins $2. If the two chips he chooses have different numbers, he loses $1 (–$1). a. Let X = the amount of money Miguel will receive or owe. Fill out the missing values in the table. (Hint: The total possible outcomes are six because there are four chips and you are choosing two of them.) Xi 2 –1 P(xi) b. What is Miguel’s expected value from playing the game? c. Based on the expected value in the previous step, how much money should Miguel expect to win or lose each time he plays? d. What value should be assigned to choosing two chips with the number 1 to make the game fair? Explain your answer using a complete sentence and/or an equation.
Mathematics
2 answers:
joja [24]1 year ago
7 0

Answer: hi do you have the answers and work to the rest of this assignment (the student guide one from edge)? if u do can u let me know and i can give u my info to send it to me i would appreciate it so much!

Step-by-step explanation:

zzz [600]1 year ago
6 0

Answer:


Step-by-step explanation:

Given that Miguel is playing a game

The box contains 4 chips, 2 with number 1, and other two differntly numbered as 3 and 5.

OUt of these 4, 2 chips are drawn

P(drawing same number) = 2C2/4C2 =\frac{1}{6}

Prob (drawing differnt numbers) = 1-1/6 =\frac{5}{6}

Hence prob of winning 2 dollars = \frac{1}{6}

Prob of losing 1 dollar = \frac{5}{6}

b) Expected value = sum of prob x amount won

= \frac{1}{6}2+\frac{5}{6}(-1)=-\frac{1}{2}

c) Miguel can expect to lose 1/2 dollars for every game he plays

d) If it is to be a fair game expected value =0

i.e. let the amount assigned be s

Then \frac{1}{6}s-\frac{5}{6}=0\\s=5

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A business has $25,000 to spend on training sessions for its employees. It wants 45 of its employees to attend. The business wan
lilavasa [31]

Answer: x+y=45\\\\1000 x + 500y = $2500

Step-by-step explanation:

Let x =  Number of employees taking technology training

y= Number of employees taking customer service training

Given, The technology training costs $1,000 per person. The customer service training costs $500 per person.

Total cost = 1000 x + 500y

Since, Total cost = $25,000 and total employee to attend training= 45 .

That means , the required equations are:

x+y=45\\\\1000 x + 500y = $2500

5 0
2 years ago
the richmond park gamekeeper wanted to know how many deer were in the park . he caught 90 of them and put a small tag round a ho
Alenkasestr [34]

Answer:

a. Since 10 out of 70 had a tag we can write the ratio 10:70. We need to solve for x in 90:x. x = 630 so the answer is 630.

b. An assumption he could have made is that the number of deer that had a tag and the total amount of deer were directly proportional.

5 0
2 years ago
Given the general identity tan X =sin X/cos X , which equation relating the acute angles, A and C, of a right ∆ABC is true?
irakobra [83]

First, note that m\angle A+m\angle C=90^{\circ}. Then

m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.

Consider all options:

A.

\tan A=\dfrac{\sin A}{\sin C}

By the definition,

\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.

Now

\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.

Option A is true.

B.

\cos A=\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.

Then

\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}=\dfrac{\dfrac{AB}{BC}}{\dfrac{BC}{AC}}=\dfrac{AB\cdot AC}{BC^2}\neq \dfrac{AB}{AC}.

Option B is false.

3.

\sin C = \dfrac{\cos A}{\tan C}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

Now

\dfrac{\cos A}{\tan C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{AB}{BC}}=\dfrac{BC}{AC}\neq \sin C.

Option C is false.

D.

\cos A=\tan C.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

As you can see \cos A\neq \tan C and option D is not true.

E.

\sin C = \dfrac{\cos(90^{\circ}-C)}{\tan A}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos (90^{\circ}-C)=\cos A=\dfrac{AB}{AC},\\ \\\tan A=\dfrac{BC}{AB}.

Then

\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.

This option is false.

8 0
2 years ago
Read 2 more answers
A really bad carton of eggs contains spoiled eggs. An unsuspecting chef picks eggs at random for his ""Mega-Omelet Surprise."" F
Dima020 [189]

Answer:

(a) The probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b) The probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c) The probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

Step-by-step explanation:

The complete question is:

A really bad carton of 18 eggs contains 8 spoiled eggs. An unsuspecting chef picks 5 eggs at random for his “Mega-Omelet Surprise.” Find the probability that the number of unspoiled eggs among the 5 selected is

(a) exactly 5

(b) 2 or fewer

(c) more than 1.

Let <em>X</em> = number of unspoiled eggs in the bad carton of eggs.

Of the 18 eggs in the bad carton of eggs, 8 were spoiled eggs.

The probability of selecting an unspoiled egg is:

P(X)=p=\frac{10}{18}=0.556

A randomly selected egg is unspoiled or not is independent of the others.

It is provided that a chef picks 5 eggs at random.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.556.

The success is defined as the selection of an unspoiled egg.

The probability mass function of <em>X</em> is given by:

P(X=x)={5\choose x}(0.556)^{x}(1-0.556)^{5-x};\ x=0,1,2,3...

(a)

Compute the probability that of the 5 eggs selected exactly 5 are unspoiled as follows:

P(X=5)={5\choose 5}(0.556)^{5}(1-0.556)^{5-5}\\=1\times 0.05313\times 1\\=0.0531

Thus, the probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b)

Compute the probability that of the 5 eggs selected 2 or less are unspoiled as follows:

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

              =\sum\imits^{2}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=0.0173+0.1080+0.2706\\=0.3959

Thus, the probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c)

Compute the probability that of the 5 eggs selected more than 1 are unspoiled as follows:

P (X > 1) = 1 - P (X ≤ 1)

              = 1 - P (X = 0) - P (X = 1)

              =1-\sum\limits^{1}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=1-0.0173-0.1080\\=0.8747

Thus, the probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

6 0
1 year ago
Can you simplify before multiplying 14×25/27​
castortr0y [4]

Answer:

No.

Step-by-step explanation:

You can not simplify this expression before multiplying it, because no whole numbers go into both 14 and 27, and also no whole numbers go into both, 25 and 27. This is the correct answer to this question.

Hope this helps!!!

Kyle.

5 0
2 years ago
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