Call n, the first of those consecutive terms, then three consecutive terms are:n, n +1, and n +2.
So, the equation will be n + (n +1) + (n +2) = 467
=> n + n + 1 + n + 2 = 467
=> 3n + 3 = 467
All of them are valid forms of the equation for the sum of three consecutive integers is 467.
You can solve it now, if you want:
3n = 467 - 3
3n = 464
n = 464 / 3
n = 154.6
The solution means that there are not three consecutive numbers whose sume is 467.
You can verifiy that if you take 154 + 155 + 156 = 465
And 155 + 156 + 157 = 468
Answer:
The answer is 200.96 cm^2.
2*3.14*r=50.24
=>r=50.24/6.28
=>r=8 cm
Thus area = 3.14*r*r
= 3.14*8*8
=200.96 cm^2
Step-by-step explanation:
Answer:
Step-by-step explanation:
for 1st week=1 ¢
for 2nd week= 2¢
for 3rd week= 4¢
for 4th week=8¢
for 5th week=16¢
for 6th week=32¢
for 7th week=64¢
for 8th week=64+64=128¢
for 9th week= 128+128=256¢
for 10th week= 256+256=512¢
.: he saved a total of
1+2+4+8+16+32+64+128+256+512=1023¢
1023¢=1023/100=$10.23
<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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