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4 years ago

35.83 x 0.0028 SHOW WORK! ! ! ! ! ! ! !

Mathematics

2 answers:

Travka [436]4 years ago

5 0

Answer:0.100324

Step-by-step explanation: Something tells me you are not understanding this subject. I already answered one of your questions similar to this one. Maybe pay attention in class.

ArbitrLikvidat [17]4 years ago

5 0

Answer:

1.0*10^(-1) (but 0.10 would also be appropriate)

Step-by-step explanation:

Rewrite 35.83 x 0.0028 as 3.583*10^1 and 0.0028 as 2.8*10^(-3).

Estimate the product: 36 times 3 is 108.

Multiply 3.583 and 2.8 together, obtaining 10.0324.

Noting that 10^1 * 10^(-3) = 10^(-2), write the intermediate answer as:

10.0234*10^(-2).

Move the decimal point one place to the left, and then, in compensation, rewrite this expression as 1.0234*10^(-1).

Finally, rewrite this result as 1.0*10^(-1). Since 0.0028 has just 2 signficant figures, 1.0 (2 significant figures) is appropriate. That's 1.0*10^(-1).

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3 years ago

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F(x)-6(x-1)+9 evaluate f(3)

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Answer:

fx-6x+15

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3 years ago

4. [5 pts] Describe how and why the formula for permutations differs from the formula for combinations.

ivann1987 [24]

Answer:

They are different because of the order in the permutation matters. In combination, the order doesn't matter. In other words in a permutation 123 and 132 are different but in a combination are the same group (they have the same digits 1,2, and 3).

Step-by-step explanation:

The formula of the permutation is $P(n,r)=\frac{n!}{(n-r)!}$ , when you are performing a permutation you pick r objects from a total of n, for the first pick you can choose from n, but for the second you have n-1, and this continues to your pick number r in which you will choose from n-r+1, and the total of permutation is the multiplicación of this number of choices for each pick, like this:

$n(n-1)(n-2)...(n-r+1)$

If $n!=n(n-1)(n-2)...(n-r+1)(n-r)(n-r-1)...(1)$ and $(n-r)!=(n-r)(n-r-1)(n-r-2)...(1)$

$\frac{n!}{(n-r)!}=\frac{n(n-1)(n-2)...(n-r+1)(n-r)(n-r-1)...(1)}{(n-r)(n-r-1)(n-r-2)...(1)}$

The factor equals above and under cancel each other.

$\frac{n!}{(n-r)!}=\frac{n(n-1)(n-2)...(n-r+1)(n-r)(n-r-1)...(1)}{(n-r)(n-r-1)(n-r-2)...(1)}\\\frac{n!}{(n-r)!}=n(n-1)(n-2)...(n-r+1)$

In combination, the order of the element isn't important, so from the total of permutation you have to eliminate the ones with the same objects with different order and counting just once each group, when choosing r objects the total of permutation for a single group of r objects is: $r(r-1)(r-2)...(1)=r!$ . If you divide the total of permutations of n taking r by r! you get the combinations (where the order is not important). The formula of the combination is $C(n,r)=\frac{n!}{r!(n-r)!}$ .

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4 years ago

A mixture contains forty ounces of glycol and water, and it is ten percent glycol. The mixture is to be strengthened to twenty-f

Archy [21]

I saw the answer for 10oz on here i tried that and it was marked wrong the correct answer is 4oz

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4 years ago

Carlos spent 1 1/4 hours doing his math homework. He spent 1/4 of this time practicing his multiplication facts.

olganol [36]

Carlos spent 15 minutes practicing his multiplication facts.

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First you must do 60 divided by 4 which is 15. 1/4 of an hour is 15 minutes.

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