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podryga [215]
1 year ago
9

A standard deck of 52 cards contains four suits: clubs, spades, hearts, and diamonds. Each deck contains an equal number of card

s in each suit. Rochelle chooses a card from the deck, records the suit, and replaces the card. Her results are shown in the table.
Cards
Suit
Observed Frequency
Clubs
29
Spades
13
Hearts
15
Diamonds
23

How does the experimental probability of choosing a heart compare with the theoretical probability of choosing a heart?

A) The theoretical probability of choosing a heart is StartFraction 1 over 16 EndFraction greater than the experimental probability of choosing a heart.
B) The experimental probability of choosing a heart is StartFraction 1 over 16 EndFraction greater than the theoretical probability of choosing a heart.
C) The theoretical probability of choosing a heart is StartFraction 1 over 26 EndFraction greater than the experimental probability of choosing a heart.
D) The experimental probability of choosing a heart is StartFraction 1 over 26 EndFraction greater than the theoretical probability of choosing a heart.
Mathematics
1 answer:
Artist 52 [7]1 year ago
7 0

Answer:

A) The theoretical probability of choosing a heart is 1/16 greater than the experimental probability of choosing a hear

Step-by-step explanation:

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A customer visiting the suit department of a certain store will purchase a suit with probability .22, a shirt with probability .
BigorU [14]

Answer:

a) The probability that he doesnt but any items is 0.49

b) He buys exactly 1 of those items with probability 0.28

Step-by-step explanation:

lets call su the event that the customer purchases a suit, sh the event that teh customer purchases a shirt and t the event that the customer purchases a tie.

Remembe that for events A, B and C we have that

P(A U B) = P(A) + P(B) - P(A ∩ B)

P(A U B U C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)

Also, we are given that

P(su) = 0.22

P(sh) = 0.3

p(t) = 0.28

p(su ∩ sh) =  0.11

P(su ∩ t) = 0.14

P(sh ∩ t) = 0.1

P(sh ∩ t ∩ su) = 0.06

The event that he doesnt buy any item has as complementary event su ∪ sh ∪ t, therefore

P( he doesnt but any items) = 1-P(su U sh U t) =

1-( P(su) + p(sh) + p(t) - P(su ∩ sh) - p(su∩t) - p(sh∩t) + p(su∩sh∩t) ) =

1-(0.22+0.30+0.28-0.11-0.14-0.1+0.06) = 1-0.51 = 0.49

b) The probability that he buys at least 2 items is equal to

p(su ∩ t) + p(su ∩ sh) + p(sh ∩ t) -2 p(su ∩ t ∩ sh) (because we are counting the triple intersection 3 times, so we need to remove it twice)

This number is

0.14+0.11+0.1-2*0.06 = 0.23

Thus, the probability that he buys exactly one item can be computed by substracting from one the probability of the complementary event : she buys 2 or more or non items

P(he buys exactly one item) = 1- ( p(he buys none items) + p(he buys at least 2) ) = 1- 0.49-0.23 = 0.28

7 0
2 years ago
Read 2 more answers
A study investigated the job satisfaction of teachers allowed to choose supplementary curriculum for their classes versus teache
ludmilkaskok [199]
The answer is b. the data shows that the authors connot make a determination either way with this data.
8 0
1 year ago
Suppose that only 20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions whe
Lina20 [59]

Answer:

a) 91.33% probability that at most 6 will come to a complete stop

b) 10.91% probability that exactly 6 will come to a complete stop.

c) 19.58% probability that at least 6 will come to a complete stop

d) 4 of the next 20 drivers do you expect to come to a complete stop

Step-by-step explanation:

For each driver, there are only two possible outcomes. Either they will come to a complete stop, or they will not. The probability of a driver coming to a complete stop is independent of other drivers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible.

This means that p = 0.2

20 drivers

This means that n = 20

a. at most 6 will come to a complete stop?

P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115

P(X = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576

P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.1369

P(X = 3) = C_{20,3}.(0.2)^{3}.(0.8)^{17} = 0.2054

P(X = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182

P(X = 5) = C_{20,5}.(0.2)^{5}.(0.8)^{15} = 0.1746

P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091

P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 + 0.1091 = 0.9133

91.33% probability that at most 6 will come to a complete stop

b. Exactly 6 will come to a complete stop?

P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091

10.91% probability that exactly 6 will come to a complete stop.

c. At least 6 will come to a complete stop?

Either less than 6 will come to a complete stop, or at least 6 will. The sum of the probabilities of these events is decimal 1. So

P(X < 6) + P(X \geq 6) = 1

We want P(X \geq 6). So

P(X \geq 6) = 1 - P(X < 6)

In which

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 = 0.8042

P(X \geq 6) = 1 - P(X < 6) = 1 - 0.8042 = 0.1958

19.58% probability that at least 6 will come to a complete stop

d. How many of the next 20 drivers do you expect to come to a complete stop?

The expected value of the binomial distribution is

E(X) = np = 20*0.2 = 4

4 of the next 20 drivers do you expect to come to a complete stop

4 0
2 years ago
How long will it take before twenty percent of our 1,000-gram sample of uranium-235 has decayed? [See Section 6.6 Example 13] Th
horsena [70]

Answer:

From what I can see of the problem, you CANNOT solve for the half-life of U 235 AND then solve for the length of time to determine the 20% decay.

U-235 half-life is 704,000,000 years. (Wikipedia)

The elapsed time formula = half-life * [log (Beginning Amount / ending amount) / log 2]

elapsed time = 7.04 x 10^8 * [log (100 % / 80%) / log 2]

elapsed time =7.04 x 10^8 *  [log (1.25) / .30103]

elapsed time =7.04 x 10^8 *  [0.096910 / .30103]

elapsed time = 7.04 x 10^8 *  0.321928047

elapsed time = 226,637,000 years

Step-by-step explanation:

8 0
1 year ago
The price of a recipe-book was reduced from $24.50 to$17.95. If p represents the percent decrease in price of the book, which pr
LuckyWell [14K]

Answer:

The required proportion is \frac{p}{100}=\frac{24.50-17.95}{24.50}.

Step-by-step explanation:

Consider the provided information.

The price of a recipe-book was reduced from $24.50 to$17.95.

The original price is 24.50

Reduced price is 17.95.

Let p represents the percent decrease in price of the book.

\text{Percent decrease}=\frac{\text{Original price-Reduced price}}{\text{Original price}}\times 100

Now substitute the respective values in the above formula.

p=\frac{24.50-17.95}{24.50}\times 100

\frac{p}{100}=\frac{24.50-17.95}{24.50}

Hence, the required proportion is \frac{p}{100}=\frac{24.50-17.95}{24.50}.

4 0
2 years ago
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