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galben [10]
4 years ago
9

A basketball coach will select the members of a five-player team from among 9 players, including John and Peter. If the five pla

yers are chosen at random, what is the probability that the coach chooses a team that includes both John and Peter?
Mathematics
1 answer:
vivado [14]4 years ago
6 0

Answer:

Step-by-step explanation:

Given that a basketball coach will select the members of a five-player team from among 9 players, including John and Peter.

Out of nine players five are chosen at random.

The team consists of John and Peter.

Hence we can sort 9 players as I group, John and Peter and II group 7 players.

Now the selection is 2 from I group and remaining 3 from II group.

Hence no of ways of selecting a team that includes both John and Peter=2C2*7C3=35

Total no of ways = 9C5=126

= \frac{2C2*7C3}{9C5}=\frac{35}{126} =\frac{5}{18}

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If we take a look at the data presented in the chart we can see the damage that the head gasket on truck suffers according to the ambient temperature. The correct answer is letter C) because there is a negative slope that is less than one and the y-intercept begins 14 units up the y-axis.


<span>I hope this helps, Regards.</span>

6 0
4 years ago
Read 2 more answers
I'd really appreciate it if anyone could help! :) I'll give Brainliest!
Dmitrij [34]

B and C are correct!

8 0
3 years ago
Hi, I have this question to answer and I am just really confused I don't know where to begin we haven't had a question like this
Step2247 [10]
   i think it is D. 43.7 and 59.1 seconds
6 0
3 years ago
If f(x)=2x-5 and g(x)=4x, find f(g(3)).<br> i’m sloww, 15 points
VARVARA [1.3K]

Answer:

  • 19

Step-by-step explanation:

<u>Given</u>

  • f(x)=2x-5
  • g(x)=4x,
  • f(g(3))= ?

<u>Solution:</u>

  • g(3) = 4*3 = 12
  • f(g(3)) = f(12) = 2*12 - 5 = 24 - 5= 19
8 0
3 years ago
The tree diagram below illustrates the possible outcomes of three tosses of a balanced coin. What is the probability that “heads
wariber [46]
<h3>Answer:   3/8</h3>

3/8 = 0.375 = 37.5%

====================================================

Explanation:

Start at the very left hand side of the diagram. Also, start at the top as well. So you'll start at "heads". Trace along the top most path and you'll get "heads, heads, heads". This says we got 3 heads in a row. But we want a case where heads only comes up exactly twice. So we rule out this scenario.

Go back to the first "heads". Now trace along the top, but don't land on that third "heads". Instead, go to the "tails" that is underneath in that last column. So you'll have "heads, heads, tails". This time we got exactly two heads out of three tosses. Off to the side somewhere, make a tally of how many times this scenario happens. Another time it happens is when we get "heads, tails, heads". Another time is when you get "tails, heads, heads".

There are only three situations where we get exactly two heads out of three tosses. This is equivalent to the event "getting exactly one tail", so that might be an easier scenario to track down.

If you traced out all of the possible paths, you would find that there are 8 different paths. Note how 2^3 = 2*2*2 = 8 is a numerical way to compute this, due to the fact that each coin toss has 2 outcomes and we have three tosses total.

Since we have three outcomes where we got exactly two heads, out of 8 outcomes total, this leads to the answer being 3/8

Side note: 3/8 = 0.375 = 37.5%

----------------------

We could also solve this problem by listing out all the possible scenarios. I'll use H for heads and T for tails. Something like HHT means we got heads twice in a row, then tails later. The order matters.

Listed below are the 8 possible outcomes. The items in bold represent situations where we get exactly two heads, which is the same as getting exactly one tail.

  1. HHH
  2. HHT
  3. HTH
  4. HTT
  5. THH
  6. THT
  7. TTH
  8. TTT

The order of each of those eight items is the same as the order if you were to trace out the possible paths of the table given (starting at the top and moving down). We can see that there are 3 cases where we get exactly two 'H's out of 8 possible outcomes total. So that's another way to arrive at 3/8

4 0
3 years ago
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