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valentina_108 [34]
2 years ago
5

According to an​ airline, flights on a certain route are on time 80 ​% of the time. suppose 10 flights are randomly selected and

the number of​ on-time flights is recorded. ​(a) explain why this is a binomial experiment. ​(b) find and interpret the probability that exactly 7 flights are on time. ​(c) find and interpret the probability that fewer than 7 flights are on time. ​(d) find and interpret the probability that at least 7 flights are on time. ​(e) find and interpret the probability that between 5 and 7 ​flights, inclusive, are on time.
Mathematics
1 answer:
natulia [17]2 years ago
6 0
<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>
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marusya05 [52]

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Sin \angle B =0.60

Cos \angle B=0.80

Step-by-step explanation:

Given

I will answer this question using the attached triangle

Solving (a): Sine and Cosine A

In trigonometry:

Sin \theta =\frac{Opposite}{Hypotenuse} and

Cos \theta =\frac{Adjacent}{Hypotenuse}

So:

Sin \angle A =\frac{BC}{BA}

Substitute values for BC and BA

Sin \angle A =\frac{8cm}{10cm}

Sin \angle A =\frac{8}{10}

Sin \angle A =0.80

Cos \angle A=\frac{AC}{BA}

Substitute values for AC and BA

Cos \angle A=\frac{6cm}{10cm}

Cos \angle A=\frac{6}{10}

Cos \angle A=0.60

Solving (b): Sine and Cosine B

In trigonometry:

Sin \theta =\frac{Opposite}{Hypotenuse} and

Cos \theta =\frac{Adjacent}{Hypotenuse}

So:

Sin \angle B =\frac{AC}{BA}

Substitute values for AC and BA

Sin \angle B =\frac{6cm}{10cm}

Sin \angle B =\frac{6}{10}

Sin \angle B =0.60

Cos \angle B=\frac{BC}{BA}

Substitute values for BC and BA

Cos \angle B=\frac{8cm}{10cm}

Cos \angle B=\frac{8}{10}

Cos \angle B=0.80

Using a calculator:

A = 53^{\circ}

So:

Sin(53^{\circ}) =0.7986

Sin(53^{\circ}) =0.80 -- approximated

Cos(53^{\circ}) = 0.6018

Cos(53^{\circ}) = 0.60 -- approximated

B = 37^{\circ}

So:

Sin(37^{\circ}) = 0.6018

Sin(37^{\circ}) = 0.60 --- approximated

Cos(37^{\circ}) = 0.7986

Cos(37^{\circ}) = 0.80 --- approximated

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3 years ago
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