They don’t need to be graphed pleas help answer them all 4

According to an airline, flights on a certain route are on time 90 90% of the time. Suppose 20 20 flights are randomly selecte

Brilliant_brown [7]

Answer:

(a) Yes, the above experiment is a binomial distribution.

(b) Probability that exactly 18 flights are on time is 28.5%.

(c) Probability that at least 18 flights are on time is 67.7%.

(d) Probability that fewer than 18 flights are on time is 32.3%.

(e) Probability that between 17 and 19 flights, inclusive, are on time is 74.5%.

Step-by-step explanation:

We are given that according to an airline, flights on a certain route are on time 90% of the time.

Suppose 20 20 flights are randomly selected and the number of on time flights is recorded.

The above situation can be represented through binomial distribution;

$P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,.......$

where, n = number trials (samples) taken = 20 flights

r = number of success

p = probability of success which in our question is probability that

flights on a certain route are on time, i.e; p = 0.90

Let X = Number of flights on a certain route that are on time

So, X ~ Binom(n = 20, p = 0.90)

(a) Yes, the above experiment is a binomial distribution as the probability of success is the probability in a single trial.

And also, each flight is independent of another.

(b) Probability that exactly 18 flights are on time is given by = P(X = 18)

P(X = 18) = $\binom{20}{18} \times 0.90^{18} \times (1-0.90)^{20-18}$

= $190 \times 0.90^{18} \times 0.10^{2}$

= 0.285

Therefore, probability that exactly 18 flights are on time is 28.5%.

(c) Probability that at least 18 flights are on time is given by = P(X $\geq$ 18)

P(X $\geq$ 18) = P(X = 18) + P(X = 19) + P(X = 20)

= $\binom{20}{18} \times 0.90^{18} \times (1-0.90)^{20-18}+\binom{20}{19} \times 0.90^{19} \times (1-0.90)^{20-19}+\binom{20}{20} \times 0.90^{20} \times (1-0.90)^{20-20}$

= $190 \times 0.90^{18} \times 0.10^{2}+20 \times 0.90^{19} \times 0.10^{1}+1 \times 0.90^{20} \times 0.10^{0}$

= 0.677

Therefore, probability that at least 18 flights are on time is 67.7%.

(d) Probability that fewer than 18 flights are on time is given by = P(X<18)

P(X < 18) = 1 - P(X $\geq$ 18)

= 1 - 0.677 = 0.323

Therefore, probability that fewer than 18 flights are on time is 32.3%.

(e) Probability that between 17 and 19 flights, inclusive, are on time is given by = P(17 $\leq$ X $\leq$ 19)

P(17 $\leq$ X $\leq$ 19) = P(X = 17) + P(X = 18) + P(X = 19)

= $\binom{20}{17} \times 0.90^{17} \times (1-0.90)^{20-17}+\binom{20}{18} \times 0.90^{18} \times (1-0.90)^{20-18}+\binom{20}{19} \times 0.90^{19} \times (1-0.90)^{20-19}$

= $1140 \times 0.90^{17} \times 0.10^{3}+190 \times 0.90^{18} \times 0.10^{2}+20 \times 0.90^{19} \times 0.10^{1}$

= 0.745

Therefore, probability that between 17 and 19 flights, inclusive, are on time is 74.5%.

7 0

3 years ago

you throw a ball upward. its height h in feet after t seconds can be modeled by the function h=-16t^2+30t+6.after how many secon

professor190 [17]

When the ball hit the ground, the height will be equal to "0".
h=0
Then:

-16t²+30t+6=0

we have to solve this square equation:

t=[-30⁺₋√(900+384)] / (-32)
t=(-30⁺₋√1284)/(-32)
we have two solutions:
t₁=(-30-√1284)/(-32)=2.06
t₂=(-30+√1284)/(-32)=-0.18 This solution is not valid.

Answer: 2.06 seconds.
t

4 0

3 years ago

A 1,200-kg automobile is traveling at a velocity of 100 m/s. Is its energy potential or kinetic? How much energy does it possess

Rama09 [41]

Answer:

$K. E. = 6 \times {10}^{6} \: Joule$

Step-by-step explanation:

Since the automobile is in motion, therefore it has kinetic energy (K. E.)

Here, m = 1200 kg, v = 100 m/s

$\because \: K. E. = \frac{1}{2} m {v}^{2} \\ \\ \therefore \: K. E. = \frac{1}{2} \times 1200 \times {(100)}^{2} \\ \\ \therefore \: K. E. = 600 \times 10000\\ \\ \therefore \: K. E. = 6000000 \: Joule \\ \\\therefore \: K. E. = 6 \times {10}^{6} \: Joule$

3 0

4 years ago

J+9-3<8 Please show work?

jeyben [28]

9 - 3 = 6
J + 6 < 8 subtract 6 from both sides
J + 6 - 6 < 8 - 6
J < 2

5 0

3 years ago

The incubation period for the eggs of a House Wren has a normal probability distribution with a mean time of 336 hours and a sta

geniusboy [140]

Answer:

2.28%

Step-by-step explanation:

If X is the random variable that measures the time of incubation needed before hatching, we want

P(X>343)

This is the area under the normal curve whose mean is 336 hours and its standard deviation 3.5 hours to the right of the point 343. (See picture attached)

This can be done easily with the help of a computer and we find

P(X>343) = 0.0228 or 2.28%

6 0

3 years ago