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3 years ago

5

There are 144 pencils in a box. If each member of the class gets 8 pencils, there will be no pencils left over. How many members

of the class are there? Write an equation to represent the situation. Then solve the equation to answer the problem.
Pls help ASAP ILL GIVE BRAINLIST

1 answer:

kotykmax [81]3 years ago

3 0

Answer:18

Step-by-step explanation:

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Approximately 68% of films are rated R. If 720 films were recently rated, how many were rated R?

zloy xaker [14]

Answer:

Approximately 490 films were rated R.

Step-by-step explanation:

Total films recently rated is 720. About 68% of the films were rated R. Thus, 68% of 720 is calculated below.

$68\% \: of \: 720 \: films$

$= \frac{68}{100} \times 720$

$= 0.68 \times 720$

$= 489.6$

$= 490 \: films \: (approximated)$

Therefore, if 720 films were recently rated, 68% of films rated R is about 490 films.

3 0

3 years ago

Jennifer took a long multiple choice, end of year vocabulary test. The ratio of numbers of problems Jennifer got correct to the

noname [10]

Jennifer got 30% on her test!

7 0

3 years ago

Ethan lives 2 5/12 miles from school. Marquis lives 7/12 mile closer to school than ethan. How far does marquis live from school

goblinko [34]

Answer:

1 5/6 miles

Step-by-step explanation:

first, we take the amout ethan lives from school, and turn it into a mixed number

2 5/12 = 29/12

marquis lived 7/12 mile closer

so, we have to subtract that from how far away ethan lives

29/12 - 7/12 = 22/12

22/12 simplified is 1 10/12 or 1 5/6

7 0

2 years ago

The accompanying data came from a study of collusion in bidding within the construction industry. No. Bidders No. Contracts 2 6

viktelen [127]

Answer:

a)

The proportion of contracts involved at most five bidders is 0.667.

The proportion of contracts involved at least five bidders is 0.51.

b)

The proportion of contracts involved between five and 10 inclusive bidders is 0.5.

The proportion of contracts involved strictly between five and 10 bidders is 0.304.

Step-by-step explanation:

No. bidders No. contracts Relative frequency of contracts

2 6 6/102=0.0588

3 20 20/102=0.1961

4 24 24/102=0.2353

5 18 18/102=0.1765

6 13 13/102=0.1275

7 7 7/102=0.0686

8 5 5/102=0.049

9 6 6/102=0.0588

10 2 2/102=0.0196

11 1 1/102=0.0098

Total 102

a)

We have to find proportion of contracts involved at most five bidders.

Proportion of at most 5= Relative frequency 2+ Relative frequency 3+ Relative frequency 4+ Relative frequency 5

Proportion of at most 5=0.0588+ 0.1961+0.2353+0.1765

Proportion of at most 5=0.6667

The proportion of contracts involved at most five bidders is 0.667.

proportion of at least five bidders= proportion≥5= 1- proportion less than 5

Proportion less than 5=0.0588+ 0.1961+0.2353=0.4902

proportion of at least five bidders=1-0.4902=0.5098

The proportion of contracts involved at least five bidders is 0.51

b)

We have to find proportion of contracts involved between five and 10 inclusive bidders.

Proportion of contracts between five and 10 inclusive= Relative frequency 5+ Relative frequency 6+ Relative frequency 7+ Relative frequency 8+ Relative frequency 9+ Relative frequency 10

Proportion of between five and 10 inclusive=0.1765 +0.1275 +0.0686 +0.049 +0.0588 +0.0196

Proportion of between five and 10 inclusive=0.5

The proportion of contracts involved between five and 10 inclusive bidders is 0.5

We have to find proportion of contracts involved strictly between five and 10 bidders.

Proportion of contracts strictly between five and 10= Relative frequency 6+ Relative frequency 7+ Relative frequency 8+ Relative frequency 9

Proportion of strictly between five and 10=0.1275 +0.0686 +0.049 +0.0588

Proportion of strictly between five and 10=0.3039

The proportion of contracts involved strictly between five and 10 bidders is 0.304.

8 0

3 years ago

For a group of 100 people, compute(a) the expected number of days of the year that are birthdays of exactly 3 people.(b) the exp

Scilla [17]

Answer:

Step-by-step explanation:

Leaving leap years, a year contains 365 days.

For a group of 100 people, each person is independent of the other and probability of any day being his birthday has a chance of

$\frac{1}{365}$

a) Probability that exactly 3 people have same birthday = $\frac{1}{365^3}$

Each day is independent of the other

And hence no of days having exactly 3 persons birthday out of 100 persons is binomial with n = 365 and p = $\frac{1}{365^3}$

Expected value of days = np = $\frac{1}{365^2}$

b) Distinct birthdays is binomail with p =1/365 and n = 365

Hence

expected value = np =1

4 0

4 years ago