six+x is an example of_____ a variable. an expression a constant. a formula

there were a total of twenty-one baseball games during the three month season, and four were played at night. If the game are eq

kicyunya [14]

If there were a total of 21 games played over 3 months, you divide 21 by 3.

Your answer is 7 games were played each month.

The information about how many games were played at night is useless. It's just thrown in there to confuse you. :)

6 0

3 years ago

Consider quadrilateral BCEF inscribed in circle A. Diagonals EB and CF intersect at point D. Select all the statements that are

Elena L [17]

Answer:

$m\angle ECB+m\angle EFB=180$

$m\angle CDB\cong m\angle EDF$

$m\angle CDB+m\angle DCB+m\angle CBD=180 \degree$

Step-by-step explanation:

From the diagram, quadrilateral BCEF is a cyclic quadrilateral.

Opposite angles if a cyclic quadrilateral sum up to 180°

$m\angle ECB+m\angle EFB=180$

The diagonals intersect at D to form two pairs of vertical angles, and vertical angles are congruent.

$m\angle CDB\cong m\angle EDF$

Also sum of angles in triangle CBD is 180°.

$m\angle CDB+m\angle DCB+m\angle CBD=180 \degree$

6 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

Kira has $40 to spend and used $20 to buy a new pair of jeans. Which integer best represents the situation of spending $20?

pishuonlain [190]

-20 because 40 is a positive and spent is negative

5 0

2 years ago

Find the area of the triangle.

madreJ [45]

Start by finding the area of each smaller right triangle.
Right triangle area = (1/2)hb -> h = height & b = base
b = 4 & h = 3
(1/2)(4)(3) = 6
6*2 = 12 (there are two right triangles)
The area of the entire triangle is 12yd^2

8 0

3 years ago