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rosijanka [135]
2 years ago
5

Can the sides of a triangle have lengths 2, 13, and 14?

Mathematics
1 answer:
ohaa [14]2 years ago
7 0

Answer:

Yes

Step-by-step explanation:

The larger side is smaller than the other sides combined

(2 + 13) > 14

Hope this helps

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jok3333 [9.3K]

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4 0
3 years ago
3x+4(x-8)-14 evaluate for x=11
djverab [1.8K]

Answer:

For X = 11, the expression is equal to 31

Step-by-step explanation:

Hello

For X = 11 you only have to replaced the value into the expression, so:

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Best regards

4 0
3 years ago
Read 2 more answers
According to an​ airline, flights on a certain route are on time 80 ​% of the time. suppose 10 flights are randomly selected and
natulia [17]
<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>
6 0
3 years ago
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