Answer:
8
Step-by-step explanation:
5V-6W
V=4. W=2.
5(4)-6(2)
20-6(2)
20-12=
8
Answer:
Step-by-step explanation:
<ACB = <ECD
These 2 angles are vertically opposite and are equal.
<B = <D
They are both right angles are therefore equal.
The answer is the AA postulate.
A
Note
ASA is a congruence postualate. If S is between two angles that can be shown to be corresponding and equal, then you will have 2 congruent triangles.
SSS if three sides of 1 triangle = 3 sides of a second triangle, then the 2 triangles are congruent. If the the three sides of one triangle are in a ratio with 3 sides of the other triangle, then the triangles could be similar, but that is not the case here.
SAS this is the terminology for congruence as well. We don't know enough to use it for similarity. Some sort of ratio would have to be mentioned to do that.
You are intended to use AA as your answer.
Answer:
I believe its 8
Step-by-step explanation:
9 + 10 + 13 + 2 +6 = 40 / 5 = 8
<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.
</span>
G(x)/f(x) will be simplified to (x+3)(x-3)/2-x^1/2,
which will give you [0,4) ∪(4, ∞).
Choice B
