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ddd [48]
4 years ago
6

57/9 to a mixed number

Mathematics
2 answers:
Yuki888 [10]4 years ago
5 0
All you have to do is divide your numbers out; 9 goes into 57 six times. Subtract your product 57-54 = 3. You now have 6 3/9 BUT you must simplify 3/9 to 1/3. Final answer: 6 1/3.
Alex4 years ago
4 0
U have to see how many times 9 goes into 57... 9 goes into 57 six times. 9 times 6 is 54, so u take 57 and subtract it by 54. 57 minus 54 is 3, so u would put three lver ur demoninator. The answer would be 6 3/9. Im really bad at explaining
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Suppose Felipe has 9.24 pounds of seed if it takes 6.6 pounds of seed to plant on acre of grass how many acres of grass can be p
Lyrx [107]

Answer:

1.4 acres of grass can be planted.

Step-by-step explanation:

He has 9.24 pounds of seeds.

If one acre of grass needs 6.6 pounds of seeds we can divide 9.24 pounds by 6.6 pounds to find how many acres of grass he can plant.

9.24 lbs × \frac{1acre}{6.6 lbs} = 1.4 acres

4 0
3 years ago
A 52-card deck is thoroughly shuffled and you are dealt a hand of 13 cards. (a) If you have at least one ace, what is the probab
jasenka [17]

Answer:

a) 0.371

b) 0.561

Step-by-step explanation:

We can answer both questions using conditional probability.

(a) We need to calculate the probability of obtaining two aces given that you obtained at least one. Let's call <em>A</em> the random variable that determines how many Aces you have. A is a discrete variable that can take any integer value from 0 to 4. We need to calculate

P(A \geq 2 | A \geq 1) = P(A\geq 2 \cap A \geq 1) / P(A \geq 1)

Since having 2 or more aces implies having at least one, the event A \geq 2 \cap A \geq 1 is equal to the event A \geq 2. Therefore, we can rewrite the previous expression as follows

P(A \geq 2) / P(A \geq 1)

We can calculate each of the probabilities by substracting from one the probability of its complementary event, which  are easier to compute

P(A \geq 2) = 1 - P((A \geq 2)^c) = 1 - P((A = 0) \bigsqcup (A = 1)) = 1 - P(A = 0) - P (A = 1)

P (A \geq 1) = 1 - P ((A \geq 1)^c) = 1 - P(A = 0)

We have now to calculate P(A = 0) and P(A = 1).

For the event A = 0, we have to pick 13 cards and obtain no ace at all. Since there are 4 aces on the deck, we need to pick 13 cards from a specific group of 48. The total of favourable cases is equivalent to the ammount of subsets of 13 elements of a set of 48, in other words it is 48 \choose 13. The total of cases is 52 \choose 13. We obtain

P(A = 0) = {48 \choose 13}/{52 \choose 13} = \frac{48! * 39!}{52!*35!} \simeq 0.303  

For the event A = 1, we pick an Ace first, then we pick 12 cards that are no aces. Since we can pick from 4 aces, that would multiply the favourable cases by 4, so we conclude

P(A=1) = 4*{48 \choose 12}/{52 \choose 13} = \frac{4*13*48! * 39!}{52!*36!} \simeq 0.438      

Hence,  

1 - P(A = 1)-P(A=0) /1-P(A=1) = 1 - 0.438 - 0.303/1-0.303 = 0.371

We conclude that the probability of having two aces provided we have one is 0.371

b) For this problem, since we are guaranteed to obtain the ace of spades, we can concentrate on the other 12 cards instead. Those 12 cards have to contain at least one ace (other that the ace of spades).

We can interpret this problem as if we would have removed the ace of spades from the deck and we are dealt 12 cards instead of 13. We need at least one of the 3 remaining aces. We will use the random variable B defined by the amount of aces we have other that the ace of spades. We have to calculate the probability of B being greater or equal than 1. In order to calculate that we can compute the probability of the <em>complementary set</em> and substract that number from 1.

P(B \geq 1) = 1-P(B=0)

In order to calculate P(B=0), we consider the number of favourable cases in which we dont have aces. That number is equal to the amount of subsets of 12 elements from a set with 48 (the deck without aces). Then, the amount of favourable cases is 48 \choose 12. Without the ace of spades, we have 51 cards on the deck, therefore

P(B = 0) = {48 \choose 12} / {51 \choose 12} = \frac{48!*39!}{51!*36!} = 0.438

We can conclude

P(B \geq 1) = 1- 0.438 = 0.561

The probability to obtain at least 2 aces if we have the ace of spades is 0.561

4 0
3 years ago
A triangular prism is 32 inches long and has a triangular face with a base of 18 inches and a
worty [1.4K]

Answer:

1752

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
A graph of average resting heart rates is shown below. The average resting heart rate for adults is 72 beats per
Fed [463]

Answer:

Option (1)

Step-by-step explanation:

Characteristics of the graph attached,

1). Heart rates of the teenagers (0 to 20 years) varies from 112 beats per minute to 72 beats per minute.

2). Heart rates of the adults (20 years to 50 years) remains constant (72 beats per minute)

Option 1). Since average resting heart rate of the teenagers decreases from 0 to 10 years, so at the 20 years of age heart rate will be less than a teenager of 20 years.

Therefore, statement is not supported by the graph.

Option (2). A 20-year-old has the same average resting heart as at 30-year-old. [True]

Therefore, this option is supported by the graph.

Option (3). A 40-year-old may have the same average resting heart rate for 10 years.

[True]

Therefore, this statement is supported by the graph.

Option (4). The average resting heart rate for teenagers steadily decreases.

[True]. There is a linear decrease in the teenagers heart rate shown in the graph.

Therefore, the given statement is true and supported by the graph.

Option (1) will be the answer.

3 0
3 years ago
I need to know the answer asap
Doss [256]

subtract lowest time from highest time for each kind

129-29 = 100

320-210 = 110

 the ranges are about the same  but the times are higher for the dark chocolate

7 0
3 years ago
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