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babunello [35]
3 years ago
11

Avery’s piggy bank has 300 nickels, 450 pennies, and 150 dimes. She randomly picks three coins. Each time she picks a coin, she

makes a note of it and puts it back into the piggy bank before picking the next coin.
The table of randomly generated coin selections simulates this scenario. Each letter represents the first letter of the associated coin.

NPP NDN NNP PDP NPP
PDN DPN NNN PDN NPP
PNN NPP NPP NND DNP
PNN NNP PPP PNP NNN
PPP DPP PPD PDD PDP
PDP PPP NPP DNP PNN
PDN NNN PPP DPN DPD
NND NPP PDP PNP DND
PPN DPN PNP NPN NDN
PPP DPP PPN NDP PPP


The estimated probability that exactly two of the three coins Avery randomly picked are nickels is .

The estimated probability that exactly one of the three coins Avery randomly picked is a dime is .

The estimated probability that all three coins Avery randomly picked are pennies is .

The estimated probability that exactly two of the three coins Avery randomly picked are nickels is the estimated probability that exactly one of the three coins Avery randomly picked is a dime.
Mathematics
1 answer:
Mashutka [201]3 years ago
4 0

Answer:

(1) .20 (2) .40 (3) .12 (4) Less than

Step-by-step explanation:

You have to look at the table.  There are 5 columns with 10 rows.  5x10=50

Then simply count the boxes that have the correct number of currency for instance, if they are asking for EXACTLY 1 dime then you rule out the ones that have 2 or 3 dimes and only the count the ones that have a single dime.   So you count PDN but you would not count PDD.  There are 20 boxes that have a single dime in them.  20 out of the 50 boxes. 20/50=.40 (answer 2)

The estimated probability that exactly two of the three coins Avery randomly picked are nickels is . 20

The estimated probability that exactly one of the three coins Avery randomly picked is a dime is . 40

The estimated probability that all three coins Avery randomly picked are pennies is . 12

The answer to #1 is .20 or 20% and the answer to #2 is .40 or 40%.  20% is less than 40% so...

The estimated probability that exactly two of the three coins Avery randomly picked are nickels is  LESS THAN the estimated probability that exactly one of the three coins Avery randomly picked is a dime.

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Answer:

336\:\mathrm{ft^2}

Step-by-step explanation:

The total surface area simply consists of three rectangles and two triangles. To find the total surface of the figure, find the sum of each of the shapes that make up the outside of the figure:

2 triangles: 2 x 1/2 x 6 x 8 = 48 ft^2

First rectangle: 6 x 12 = 72 ft^2

Second rectangle: 8 x 12 = 96 ft^2

Third rectangle: 10 x 12 = 120 ft^2

*Note: One of the dimensions is in yards, so make sure to convert it to ft (1 yd = 3ft)

Thus, the total surface area of the figure is 48+72+96+120 = 336 ft^2.

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3 years ago
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t (23) ≤ 97 - 23

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If x + y = 5 and at the same time 2x - y = 7. find the value of x
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Probabilities with possible states of nature: s1, s2, and s3. Suppose that you are given a decision situation with three possibl
amm1812

Answer:

1. P(s_1|I)=\frac{1}{11}

2. P(s_2|I)=\frac{8}{11}

3. P(s_3|I)=\frac{2}{11}

Step-by-step explanation:

Given information:

P(s_1)=0.1, P(s_2)=0.6, P(s_3)=0.3

P(I|s_1)=0.15,P(I|s_2)=0.2,P(I|s_3)=0.1

(1)

We need to find the value of P(s₁|I).

P(s_1|I)=\frac{P(I|s_1)P(s_1)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}

P(s_1|I)=\frac{(0.15)(0.1)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}

P(s_1|I)=\frac{0.015}{0.015+0.12+0.03}

P(s_1|I)=\frac{0.015}{0.165}

P(s_1|I)=\frac{1}{11}

Therefore the value of P(s₁|I) is \frac{1}{11}.

(2)

We need to find the value of P(s₂|I).

P(s_2|I)=\frac{P(I|s_2)P(s_2)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}

P(s_2|I)=\frac{(0.2)(0.6)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}

P(s_2|I)=\frac{0.12}{0.015+0.12+0.03}

P(s_2|I)=\frac{0.12}{0.165}

P(s_2|I)=\frac{8}{11}

Therefore the value of P(s₂|I) is \frac{8}{11}.

(3)

We need to find the value of P(s₃|I).

P(s_3|I)=\frac{P(I|s_3)P(s_3)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}

P(s_3|I)=\frac{(0.1)(0.3)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}

P(s_3|I)=\frac{0.03}{0.015+0.12+0.03}

P(s_3|I)=\frac{0.03}{0.165}

P(s_3|I)=\frac{2}{11}

Therefore the value of P(s₃|I) is \frac{2}{11}.

4 0
2 years ago
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