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4 years ago

Twenty people are going by van to a movie. Each van seats 8 people. How many vans are needed to take everyone?

1 answer:

7 0

Yes,you are correct.
The first van carries 8 people.
The second van carries 8 people.
The 3rd van carries the remaining 4 people.
8+8+4=20

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A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee

katrin [286]

Answer: A) 1260

Step-by-step explanation:

We know that the number of combinations of n things taking r at a time is given by :-

$^nC_r=\dfrac{n!}{(n-r)!r!}$

Given : Total multiple-choice questions = 9

Total open-ended problems=6

If an examine must answer 6 of the multiple-choice questions and 4 of the open-ended problems ,

No. of ways to answer 6 multiple-choice questions

= $^9C_6=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!3!}=84$

No. of ways to answer 4 open-ended problems

= $^6C_4=\dfrac{6!}{4!(6-4)!}=\dfrac{6\times5\times4!}{4!2!}=15$

Then by using the Fundamental principal of counting the number of ways can the questions and problems be chosen = No. of ways to answer 6 multiple-choice questions x No. of ways to answer 4 open-ended problems

= $84\times15=1260$

Hence, the correct answer is option A) 1260

5 0

3 years ago

Esmeralda received a bag of jellybeans that contained a mixture of regular flavored jellybeans and "surprise" jellybeans. The ba

Firlakuza [10]

Answer:

6 "surprise" flavors jellybeans

Step-by-step explanation:

From the information given:

The bag contained a mixture of regular flavored jellybeans and "surprise" jellybeans.

There are 40 jellybeans in the bag.

If 15% of the jellybean were "surprise" flavors.

Then, the number of expected "surprise" jellybeans will be:

= 15% of 40 jellybeans

= (15/100) × 40

= 6 "surprise" flavors jellybeans

If 6 surprise jellybean is contained in the bag;

Thus, the number of regular flavored jellybean will be

= 40 - 6

= 34 regular flavored jellybean

5 0

3 years ago

Gabriela asistió al buen fin y compró a crédito un refrigerador de $8,500.50, un televisor de $3,500.50, una computadora de $6,9

vazorg [7]

Answer:

23,999

Step-by-step explanation:

Un refrigerador de $8,500.50

Un televisor de $3,500.50

Una computadora de $6,900.00

Una lavadora de $4,999.00

8,500.50 + 3,500.50 + 6,999.00 + 4,999.00 = 23,999.00

7 0

3 years ago

A total of 200 video game players take a survey on their favorite game. Unknown Kingdom

zepelin [54]

Answer:

Could you explain it another way?

Step-by-step explanation:

4 0

3 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

Viefleur [7K]

Answer:

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.2087, 0.2507).

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:

Suppose a sample of 1537 tenth graders is drawn. Of the students sampled, 1184 read above the eighth grade level. So 1537 - 1184 = 353 read at or below this level. Then

$n = 1537, \pi = \frac{353}{1537} = 0.2297$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2297 - 1.96\sqrt{\frac{0.2297*0.7703}{1537}} = 0.2087$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2297 + 1.96\sqrt{\frac{0.2297*0.7703}{1537}} = 0.2507$

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.2087, 0.2507).

7 0

3 years ago