Do you know the Combination Formula? If Not it is (n!)/r!(n - r)!
I think n would equal 6, and r would equal 2. So it would be 6!/2!(4!)
I believe you can do this yourself now that I've given you an equation.
Answer: The correct answer is option A; 78
Step-by-step explanation: The equation representing the line of best fit which is given as
y = 0.9x - 1
depicts the relationship between a player's practice results and actual game results. Whatever value is given as x which is the result from practice would be a good way of predicting a player's result when engaged in the real game.
For the question above, a player with a practice shooting percentage of 88 would have his game shooting estimated by simply inserting the value of his practice percentage into the equation showing the line of best fit shown as follows;
y = 0.9x - 1
Where x = 88
y = 0.9 (88) - 1
y = 79.2 - 1
y = 78.2
y ≈ 78
Therefore the approximate shooting percentage for a player with a practice shooting percentage of 88 would be 78 percent.
Answer:
y=-2x+5. (4,-3)
Step-by-step explanation:
1) What's being asked in other words is to write a function. So the initial point is (1,3). In x-axis, a translation movement can only be to the right or to the left. Similarly for y-axis the possible movements are shifting up or down. The question says, 3 units right for x-cordinate and 6 units down for y cordinate. So the final point of this translation would be (1+3,3-6) =(4,-3)
2) To write a linear function we need to find out the slope:
y=-2x+5
<u>Proof:</u>
x=1
y=-2(1)+5
y=3 Then (1,3)
x=4
y=-2(4)+5
y=-8+5
y=-3 Then (4,-3)
<span>(a) This is a binomial
experiment since there are only two possible results for each data point: a flight is either on time (p = 80% = 0.8) or late (q = 1 - p = 1 - 0.8 = 0.2).
(b) Using the formula:</span><span>
P(r out of n) = (nCr)(p^r)(q^(n-r)), where n = 10 flights, r = the number of flights that arrive on time:
P(7/10) = (10C7)(0.8)^7 (0.2)^(10 - 7) = 0.2013
Therefore, there is a 0.2013 chance that exactly 7 of 10 flights will arrive on time.
(c) Fewer
than 7 flights are on time means that we must add up the probabilities for P(0/10) up to P(6/10).
Following the same formula (this can be done using a summation on a calculator, or using Excel, to make things faster):
P(0/10) + P(1/10) + ... + P(6/10) = 0.1209
This means that there is a 0.1209 chance that less than 7 flights will be on time.
(d) The probability that at least 7 flights are on time is the exact opposite of part (c), where less than 7 flights are on time. So instead of calculating each formula from scratch, we can simply subtract the answer in part (c) from 1.
1 - 0.1209 = 0.8791.
So there is a 0.8791 chance that at least 7 flights arrive on time.
(e) For this, we must add up P(5/10) + P(6/10) + P(7/10), which gives us
0.0264 + 0.0881 + 0.2013 = 0.3158, so the probability that between 5 to 7 flights arrive on time is 0.3158.
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