Here, You need to calculate the slope of the graph by taking any two random points, as follows:
(x₁,y₁) = (0,0) & (x₂,y₂) = (1,55)
Now, we know,
y₂-y₁ = m(x₂-x₁)
55-0 = m(1-0)
m = 55/1
m = 55
Here m represents the slope of graph which is nothing but rate of change of distance.
In short, option A will be your answer.
Hope this helps!
M = (20 - 12)/(4 - 2)
= 8/2
= 4
slope = m = 4
If I’m correct negative one half
Answer:
A, B, C
Step-by-step explanation:
The top right corner is the positive side of the graph. If it goes below the X axis or to left of the Y axis, it is negative. (thus, the negative numbers)
<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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