Ask question

Ask question

All categories

2 years ago

14

1. A teacher had 8 packages of paper she wanted to split equally into 3 piles. How much should be in each pile? Between what two

whole numbers does your answer lie?

1 answer:

castortr0y [4]2 years ago

4 0

Answer:

If division of the packages in fractional numbers was possible, it should be 2.667 packages per pile.

The answer lies between 2 and 3 packages per pile.

Step-by-step explanation:

This can be solved as a simple division.

To split 8 packages in 3, this should correspond to 8/3≈2.667.

If division of the packages in fractional numbers was possible, it should be 2.667 packages per pile.

Other way to think about it is like mentally allocating a package in each pile. We can go up to 2 packages per pile (a total of 6 packages), and we can only add a package to 2 piles, letting the last pile to have only two packages.

Then, we have 2 piles with 3 packages and one pile with 2 packages.

You might be interested in

A rectangle has a perimeter of 30 yards. If one side is 5 yards long, how long is the other side?

vagabundo [1.1K]

Answer:

C. 10

Step-by-step explanation:

5+5=10

10+10=20

10+20=30

5 0

2 years ago

Read 2 more answers

Find c, the hypotenuse, rounded to four decimal places, in a right triangle with b = 16 and B = 55°

vredina [299]

C is 19.5
I believe not promising

8 0

2 years ago

Read 2 more answers

I need help with this someone very smart at math this algebra 2

Mrac [35]

Answer:

value if a =

$\frac{5}{4}$

Step-by-step explanation:

here's the solution :-

=》

$\frac{ 2(\sqrt{m}) {}^{3} }{ \sqrt[4]{m} }$

=》

$\frac{2(m {}^{ \frac{1}{2}} ) {}^{3} }{ {m}^{ \frac{1}{4} } }$

=》

$\frac{2 {m}^{ \frac{3}{2} }} { {m}^{ \frac{1}{4} } }$

=》

$2m {}^{ \frac{3}{2} - \frac{1}{4} }$

=》

$2m {}^{ \frac{6 - 1}{4} }$

=》

$2m {}^{ \frac{5}{4} }$

so, a = 5/4

6 0

2 years ago

The dwarves of the Grey Mountains wish to conduct a survey of their pick-axes in order to construct a 99% confidence interval ab

Dmitry_Shevchenko [17]

Answer:

The minimum sample size needed is 125.

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 $1 - \frac{\alpha}{2}$ .

The margin of error is:

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

For this problem, we have that:

$\pi = 0.25$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

What minimum sample size would be necessary in order ensure a margin of error of 10 percentage points (or less) if they use the prior estimate that 25 percent of the pick-axes are in need of repair?

This minimum sample size is n.

n is found when $M = 0.1$

So

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

$0.1 = 2.575\sqrt{\frac{0.25*0.75}{n}}$

$0.1\sqrt{n} = 2.575{0.25*0.75}$

$\sqrt{n} = \frac{2.575{0.25*0.75}}{0.1}$

$(\sqrt{n})^{2} = (\frac{2.575{0.25*0.75}}{0.1})^{2}$

$n = 124.32$

Rounding up

The minimum sample size needed is 125.

5 0

3 years ago

Jaelynn has 30 times as many stickers as her brother. Her brother has 8 stickers. How many stickers does Jaelynn have?

Dennis_Churaev [7]

30 x 8 = 240 so Jaelynn has 240 stickers :) hope I helped!

6 0

3 years ago

Read 2 more answers