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2 years ago

If JK = KL, find JL.

Mathematics

1 answer:

mafiozo [28]2 years ago

3 0

Answer:

umm your funny if u think im going to answer that im sorry but my brain is not working rn i prolly should go take a nap

Step-by-step explanation:

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Mkey [24]

Counting backwards next is 1

8 0

3 years ago

b) How many ways can you deal cards (from a deck of 52) to 4 people when each player gets 7 cards. 2 hidden and 5 visible. Assum

Mariana [72]

Answer:

${52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4$

Step-by-step explanation:

The dealing of the cards can be seen in the following way:

We first have to choose which 7 cards are going to be dealt to the 1st player. So we have to pick 7 cards out of the 52 available cards. This can be done in ${52 \choose 7}$ ways. Now that we have chosen which 7 cards are going to be dealt to the 1st player, we have to choose which 2 of them are going to be the hidden ones. So we have to pick 2 cards out of the 7 cards to be the hidden ones. This can be done in ${7 \choose 2}$ ways. At this point we now know which cards are being dealt to the 1st player, and which ones are hidden for him. Then we have to choose which 7 cards to deal to the 2nd player, out of the remaining ones. So we have to pick 7 out of 52-7=45. (since 7 have been already dealt to the 1st player). This can be done in ${45 \choose 7}$ . Then again, we have to pick which 2 are going to be the hidden cards for this 2nd player. So we have to pick 2 out of the 7. This can be done in ${7 \choose 2}$ ways. Then we continue with the 3rd player. We have to choose 7 cards out of the remaining ones, which at this point are 52-7-7=38. This can be done in ${38 \choose 7}$ . And again, we have to choose which ones are the hidden ones, which can be chosen in ${7 \choose 2}$ ways. Finally, for the last player, we choose 7 out of the remaining cards, which are 52-7-7-7=31. This can be done in ${31 \choose 7}$ ways. And choosing which ones are the hidden ones for this player can be done in ${7 \choose 2}$ ways. At the end, we should multiply all our available choices on each step, to get the total choices or total ways to deal the cards to our 4 players (since dealing the cards is a process of several steps with many choices on each step).

${52 \choose 7}{7 \choose 2}{45 \choose 7}{7 \choose 2}{38 \choose 7}{7 \choose 2}{31 \choose 7}{7 \choose 2}={52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4$

7 0

3 years ago

Simplify the expression. –42 ÷ (–6)

Vitek1552 [10]

The answer is 7
hope this helps!

4 0

3 years ago

Read 2 more answers

What is the formula for the area A of a trapezoid with parallel sides of length B and D, nonparallel sides of length A and C and

loris [4]

Answer:

$(B) \dfrac12H (B+D)$

Step-by-step explanation:

$\text{Area of a trapezoid }= \dfrac12 ($Sum of the parallel sides) \times $Height\\Parallel Sides = B and D\\Height =H\\Therefore:\\\text{Area of the trapezoid }= \dfrac12 (B+D) H$

The correct option is B.

4 0

3 years ago

The owner of a football team claims that the average attendance at games is over 523, and he is therefore justified in moving th

USPshnik [31]

Answer:

C) There is not sufficient evidence to support the claim that the mean attendance is greater than 523.

Step-by-step explanation:

Let μ be the the average attendance at games of the football team

The claim: the average attendance at games is over 523

Null and alternative hypotheses are:

$H_{0}$ : μ=523

$H_{a}$ : μ>523

The conclusion is failure to reject the null hypothesis.

This means that test statistic is lower than critical value. Therefore it is not significant, there is no significant evidence to accept the alternative hypothesis.

That is no significant evidence that the average attendance at games of the football team is greater than 523.

7 0

3 years ago