Ask question

Ask question

All categories

3 years ago

10

If the plane was full,had a crew of 20 people,and 140 passengers with window seats,then there were 406 passengers who did not ge

t window seats.How many people can fit the plane?

1 answer:

iris [78.8K]3 years ago

3 0

Answer:

466 people can fit in the plane.

Step-by-step explanation:

20 + 140 + 406 = 466

You might be interested in

Jim earns $1,600 per month after taxes. He is working on his budget and has the first three categories finished. Housing $576 Fo

Afina-wow [57]

B.

He has used the highest recommended percentages to calculate the amounts for the three categories.

5 0

3 years ago

Read 2 more answers

The gift shop at a science museum sells lollipops that are made to look like the planets. Each lollipop is shaped like a sphere

adelina 88 [10]

Answer:

7234.56

Step-by-step explanation:

Using 3.14 as pi the answer is

3 0

2 years ago

What is the y intercept

Lostsunrise [7]

Answer:

1

Step-by-step explanation:

b= y intercept

6 0

2 years ago

The price of 8 -minute phone call is $2.80 What is the price of 18 -minute phone call?

notsponge [240]

Answer:

6.3

Step-by-step explanation:

what you do is divide 8 minutes by 2.80 then multiply that answer by the answer from what you divided.

7 0

3 years ago

A soft-drink manufacturer purchases aluminum cans from an outside vendor. A random sample of 70 cans is selected from a large sh

NNADVOKAT [17]

Answer:

The 95% confidence interval for the proportion of cans in the shipment that meet the specification is (0.741, 0.917).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

70 cans, 58 meet the specification for puncture resistance. This means that $n = 50, \pi = \frac{58}{70} = 0.829$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.829 - 1.96\sqrt{\frac{0.829*0.171}{70}} = 0.741$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.829 + 1.96\sqrt{\frac{0.829*0.171}{70}} = 0.917$

The 95% confidence interval for the proportion of cans in the shipment that meet the specification is (0.741, 0.917).

5 0

3 years ago