16(x^3+1)^2-22(c^3+1)-3=0 as a quadratic equation

a certain flight arrives on time percent of the time. suppose flights are randomly selected. use the normal approximation to the

iVinArrow [24]

The probability that exactly 128 flights are on time is 88.27% and the probability that atleast 1 flight is on time is 0.022.

Given that the chances of reaching the flight on time be 88%.

We are required to find the probability that exactly 128 flights are on time, atlleast 1 flight on time.

Probability is basically the calculation of likeliness of happening an event among all the events possible. It lies between 0 and 1.

Probability=Number of items/Total items.

It cannot be negative.

Probability that the flight will be on time if one flight is selected=0.88

Probability that exactly 128 flights are on time=

128=1145*P/100

P=12800/145

P=88.27%

b) Atleast 1 flight is on time

=( $128C_{0}(0.88)^{0} (0.12)^{1}$ + $128C_{1} (0.88)^{1} (0.12)^{11}$ )/128

=(0.12+128*0.88*0.245)/128

=(0.12+2.575)/128

=0.022

Hence the probability that exactly 128 flights are on time is 88.27% and the probability that atleast 1 flight is on time is 0.022.

Learn more about probability at brainly.com/question/24756209

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Other options are incomplete and not found anywhere else.

8 0

2 years ago

Off Topic: when you delete a place on Life360, does the other person in your circle get the notification that you have remove it

Komok [63]

I honestly don’t think so

8 0

3 years ago

Suppose 40% of DC area adults have traveled outside of the United States. Nardole wants to know if his customers are atypical in

Sonbull [250]

Answer:

We conclude that % of DC area adults who have traveled outside of the United States is different from 40%.

Step-by-step explanation:

We are given that 40% of DC area adults have traveled outside of the United States. Nardole wants to know if his customers are typical in this respect. He surveys 40 customers and finds 60% have traveled outside of the U.S.

We have to test is this result a statistically significant difference.

Let p = % of DC area adults who have traveled outside of the United States

SO, Null Hypothesis, $H_0$ : p = 40% {means that 40% of DC area adults have traveled outside of the United States}

Alternate Hypothesis, $H_a$ : p $\neq$ 40% {means that % of DC area adults who have traveled outside of the United States is different from 40%}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = % of customers who have traveled outside of the United States

in a survey of 40 customers = 60%

n = sample of customers = 40

So, test statistics = $\frac{0.60-0.40}{\sqrt{\frac{0.60(1-0.60)}{40} } }$

= 2.582

Since in the question we are not given with the significance level so we assume it to be 5%. So, at 0.05 level of significance, the z table gives critical value of 1.96 for two-tailed test. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject null hypothesis as it will fall in the rejection region.

Therefore, we conclude that % of DC area adults who have traveled outside of the United States is different from 40%.

7 0

3 years ago

Savannah bought barn wood for $60 for 5 pieces. She spent $40 on materials to paint

Kisachek [45]

Her profit per each wood is $5 so her total profit is $25

because
60 divided by 5 is 12, so each wood is 12 dollars
she spent 40 dollars on materials so 40 divided by 5 is 8, so she spent $20 on each piece of wood to paint it because 12(price of wood)+8(cost of materials per wood)= 20
and shes sells them for 25
so 25-20=5
5x5=25
so the answer is $25 profited

3 0

3 years ago

Which one of the sentences would be correct

spin [16.1K]

Answer:

Nobody was in the hall

Nobody will be in the hall

7 0

3 years ago