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2 years ago

Change to 2/5 to hundredths

Mathematics

2 answers:

snow_tiger [21]2 years ago

3 0

Answer:

40/100

Step-by-step explanation:

100/5 = 20

2/5 x 20 = 40/100

Nuetrik [128]2 years ago

3 0

This is how you solve 2/5 = 4/10 = .40

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Thrice the difference between x and 4 is less than9 what's the inequality

3241004551 [841]

(x-4)•2 < 9
the difference between x and 4 means to subtract
and twice that means to double or times by 2
and all that’s LESS than 9
so the sign opens up towards the 9

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3 years ago

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adell [148]

What is the question??

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3 years ago

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What is the meaning of the unknown factor and quotient

Komok [63]

I don't know what you mean by unknown factor but a quotient is the answer of a division problem

7 0

3 years ago

What is the sum of the prime factors of 2014

goblinko [34]

The sum of prime factors of 2014 is 74

<h3>Solution:</h3>

Given that to find sum of prime factors of 2014

Let us first find the prime factors of 2014

A prime number is a whole number greater than 1 whose only factors are 1 and itself

"Prime Factorization" is finding which prime numbers multiply together to make the original number.

Prime factors of 2014:

The Prime Factorization is:

$2014 = 2 \times 19 \times 53$

Thus the prime factors of 2014 are 2, 19, 53

Let us now find the sum of prime factors of 2014

sum of prime factors of 2014 = 2 + 19 + 53 = 74

Thus the sum of prime factors of 2014 is 74

7 0

2 years ago

A donut store has 11 different types of donuts. You can only buy a bag of 3 of them, where each donut has to be of a different t

MakcuM [25]

Answer:

$165$ .

Step-by-step explanation:

Since repetition isn't allowed, there would be $11$ choices for the first donut, $(11 - 1) = 10$ choices for the second donut, and $(11 - 2) = 9$ choices for the third donut. If the order in which donuts are placed in the bag matters, there would be $11 \times 10 \times 9$ unique ways to choose a bag of these donuts.

In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type $3 \times 2 \times 1 = 6$ times.

For example, if a bag includes donut of type $x$ , $y$ , and $z$ , the count $11 \times 10 \times 9$ would include the following $3 \times 2 \times 1$ arrangements:

$xyz$ .
$xzy$ .
$yxz$ .
$yzx$ .
$zxy$ .
$zyx$ .

Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count $11 \times 10 \times 9$ by $3 \times 2 \times 1 = 6$ to find the actual number of donut combinations:

$\begin{aligned} \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of $3$ objects from a set of $11$ distinct objects:

$\begin{aligned}\begin{pmatrix}11 \\ 3\end{pmatrix} &= \frac{11 !}{(11 - 3)! \times 3 !} \\ &= \frac{11 !}{8 ! \times 3 !} \\ &= \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

5 0

1 year ago