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LenaWriter [7]
2 years ago
12

Jason wanted to place all 540 marbles in the 3 newly bought boxes. If he wants to

Mathematics
1 answer:
kotegsom [21]2 years ago
7 0

Answer:

The first box will have 90 marbles, the second 180 and the third 270.

Step-by-step explanation:

Division in a ratio of 1:2:3

1 + 2 + 3 = 6

So

The first box will have 1/6 of the marbles.

The second box will have 2/6 = 1/3 of the marbles

The third box will have 3/6 = 1/2 of the marbles.

First box:

One sixth, so:

(1/6)*540 = 540/6 = 90

Second box:

One third, so:

(1/3)*540 = 540/3 = 180

Third box:

One half, so:

(1/2)*540 = 540/2 = 270

The first box will have 90 marbles, the second 180 and the third 270.

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Which facts could be applied to simplify this expression? Check all that apply.
kykrilka [37]

The answer is B, C, and D. Like terms are terms with all the same variable, so 5x and -x are like terms.

C is correct. If we add -x to 5x, we get 4x. The other numbers remain unchanged because they have no like terms.

D is correct. Applying the rule of like terms, which is that like terms are numbers with the same variable, only add together numbers with the same variable.

Hope this helps!

3 0
2 years ago
Read 2 more answers
Melissa is planning a rectangular vegetable garden with a square patch for tomatoes. She wants the length of the garden to excee
aivan3 [116]

Answer: 8x + 14

Step-by-step explanation:

3x + 2 + 3x + 2 + x + 5 + x + 5 = 2(4x + 7) = 8x + 14

3 0
2 years ago
Please help me answer this ASAP! The two-way table shows the number of ninth and tenth graders who prefer going to sporting even
Elis [28]
<h2>Answer</h2>

0.43

<h2>Explanation</h2>

Remember that RelativeFrecuency=\frac{Frecuency}{SumOfAllFrecuencies}

Since the problem is telling us "Among tenth graders", we must focus on the 10th graders row only. From the row, we can infer that the frequency is the number of 10th graders who prefer going to sporting events, so Frecuency=6. Now, the sum of all frequencies will be the sum of all the 10th graders, so SumOfAllFrecuencies=6+8=14. Let's replace the values:

RelativeFrecuency=\frac{Frecuency}{SumOfAllFrecuencies}

RelativeFrecuency=\frac{6}{14}

RelativeFrecuency=0.4285

And rounded to the nearest hundredth:

RelativeFrecuency=0.43

4 0
2 years ago
Past studies indicate that about 60 percent of the trees in a forested region are classified as softwood. A botanist studying th
alukav5142 [94]

Answer:

1. If it is true that 60 percent of the trees in a forested region are classified as softwood, 0.015 is the probability of obtaining a population proportion greater than 0.6.

Step-by-step explanation:

Hello!

The historical information indicates that 60% of the forest trees are classified as softwood.

A botanist thinks that the proportion might be greater than 60%, so he tested his belief obtaining:

H₀: p = 0.60

H₁: p > 0.60

p-value: 0.015

You need to interpret this p-value. Little reminder:

The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis. It represents the % of size n samples from a population with proportion p=p₀, which will produce a measure that provides evidence as (or stronger) than the current sample that p is not equal to p₀.

The correct answer is:

1. If it is true that 60 percent of the trees in a forested region are classified as softwood, 0.015 is the probability of obtaining a population proportion greater than 0.6.

I hope this helps!

8 0
2 years ago
Roberto wants to display his 18 sports cards in an album. Some pages hold 2 cards and others hold 3 cards. How many different wa
WITCHER [35]
<span>65 = number of different arrangements of 2 and 3 card pages such that the total number of card slots equals 18. 416,154,290,872,320,000 = number of different ways of arranging 18 cards on the above 65 different arrangements of page sizes. ===== This is a rather badly worded question in that some assumptions aren't mentioned. The assumptions being: 1. The card's are not interchangeable. So number of possible permutations of the 18 cards is 18!. 2. That all of the pages must be filled. Since the least common multiple of 2 and 3 is 6, that means that 2 pages of 3 cards can only be interchanged with 3 pages of 2 cards. So with that said, we have the following configurations. 6x3 card pages. Only 1 possible configuration. 4x3 cards and 3x2 cards. These pages can be arranged in 7!/4!3! = 35 different ways. 2x3 cards and 6x2 cards. These pages can be arranged in 8!/2!6! = 28 ways 9x2 card pages. These can only be arranged in 1 way. So the total number of possible pages and the orders in which that they can be arranged is 1+35+28+1 = 65 possible combinations. Now for each of those 65 possible ways of placing 2 and 3 card pages such that the total number of card spaces is 18 has to be multiplied by the number of possible ways to arrange 18 cards which is 18! = 6402373705728000. So the total amount of arranging those cards is 6402373705728000 * 65 = 416,154,290,872,320,000</span>
6 0
2 years ago
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