All categories

3 years ago

For every 5 red beads there are 3 white beads. What is the ratio of red to white?

Mathematics

1 answer:

ivanzaharov [21]3 years ago

3 0

5:3 5/3 5to3. It's because there are 5 red beds and 3 white beds... So 5 red beds to 3 white beds

You might be interested in

7. The startup cost for a boutique is $15,000, and each floral arrangement costs $17 for the boutique to make. If each arrangeme

Snowcat [4.5K]

Answer:

1875 arrangements

Step-by-step explanation:

Break-Even is the point when costs are equal to profit.

The cost is 15,000

We need to cover this up with the profit we get from sales.

Each arrangement is 17 (cost) and is sold for 25, so the profit from each arrangement is:

25 - 17 = 8

So, with each arrangement sale, we make profit of $8. How many of these we need to sell in order to break even (in order to make 15,000)??

We simply divide this amount (15,000) by the profit we make from each arrangement ($8), so that would be:

Number of Arrangements Needed to Break-Even = 15,000/8 = 1875

After 1875 arrangements, the boutique breaks even.

5 0

3 years ago

If you get a link, DO NOT CLICK IT OR PASTE IT! its probably an ip finder... if someone gives you a link report the person.

strojnjashka [21]

Answer:

Sweet thx

Step-by-step explanation:

5 0

2 years ago

Please help, thank you :)

BARSIC [14]

Answer:

Step-by-step explanation:

You already have 40 points and every movie is two points. 100-40=60. 60 divided by 2 is 30, so you still need to watch 30 more movies.

8 0

2 years ago

Suppose that three numbers are selected one by one, at random, and without replacement, from the set of integers {1, 2, 3, . . .

fiasKO [112]

A - the third number falls between the first two

B - the first number is smaller than the second

$P(A|B)=\dfrac{P(A\cap B)}{P(B)}\\\\ |\Omega|=n(n-1)(n-2)\\\\ |B|=\dfrac{n!}{2}\\\\ |A\cap B|=\dfrac{\dfrac{n!}{2}}{3}=\dfrac{n!}{6}\\\\ P(A\cap B)=\dfrac{\dfrac{n!}{6}}{n(n-1)(n-2)}=\dfrac{n!}{6n(n-1)(n-2)}=\dfrac{(n-3)!}{6}\\ P(B)=\dfrac{\dfrac{n!}{2}}{n(n-1)(n-2)}=\dfrac{n!}{2n(n-1)(n-2)}=\dfrac{(n-3)!}{2}\\\\ P(A|B)=\dfrac{\dfrac{(n-3)!}{6}}{\dfrac{(n-3)!}{2}}\\\\ P(A|B)=\dfrac{(n-3)!}{6}\cdot\dfrac{2}{(n-3)!}=\dfrac{1}{6}=\dfrac{1}{3}$

3 0

3 years ago

Figure BCDE was transformed using the composition rx-axis ◦ T2,2. If point B on the pre-image lies at (3, 4), what are the coord

kaheart [24]

Thank you for posting your question here at brainly. If point B on the pre-image lies at (3, 4), the are the coordinates of B'' on the final is (5, -6). Below is the solutions, I hope the answer helps.

rx-axis ◦ T2,2 (3,4)
<span>= rx-axis ( T2,2 (3,4) ) </span>
<span>= rx-axis ( 5,6 ) </span>
= (5, -6)

4 0

3 years ago

Read 2 more answers