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2 years ago

A new section of freeway will be

Mathematics

1 answer:

Ann [662]2 years ago

6 0

<h3>Answer: 52.5 miles</h3>

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Explanation:

Multiply 1/6 by 63 to get (1/6)*63 = 10.5

This is the same as dividing by 6, so, 63/6 = 10.5

We can see that 10.5 miles have already been built.

So the amount missing is 63-10.5 = 52.5 miles

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I'm begging you please please help please please ASAP please please help please please ASAP please please help please please ASA

Tresset [83]

Answer:

Step-by-step explanation:

Hello,

This is related to ratios and proportions. We have a 12 foot tree that casts a 6 foot shadow, so now we are trying to figure out how tall is a tree that casts a 4 foot shadow.

We can set an easier method to get this.

12 = 6 (this means the 12 foot tree casts a 6 foot shadow)

x = 4 (this means how tall is the tree to cast a 4 foot shadow, the unknown height of the tree is denoted as x)

Then, we can cross multiply.

48 = 6x

Solve for x, which is the unknown height we are trying to figure out.

x = 8

8 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