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3 years ago

9

How do you multiply fractions by improper fractions

1 answer:

katovenus [111]3 years ago

4 0

Well you have to Multiply the numerators together and then <span>Multiply the denominators together after that you simplify </span><span>Convert the improper fraction (if it is improper) to a mixed fraction answer. To do this divide the denominator into the numerator finding the quotient and remainder.</span>the fraction and then

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What an equivalent expression for 48 + 18x

FromTheMoon [43]

Answer:

6(8+3x)

Step-by-step explanation:

48+18x

We can factor out an 6 from each term in the expression

6(8+3x)

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4 years ago

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Factor <img src="https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B3%7D" id="TexFormula1" title="-\frac{1}{3}" alt="-\frac{1}{3}" align=

ivanzaharov [21]

10/3 I did the math

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2 years ago

Someone please help me?

Anna35 [415]

Answer:

43

Step-by-step explanation:

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2 years ago

(Anderson, 1.14) Assume that P(A) = 0.4 and P(B) = 0.7. Making no further assumptions on A and B, show that P(A ∩ B) satisfies 0

Goshia [24]

Answer with Step-by-step explanation:

We are given that

P(A)=0.4 and P(B)=0.7

We know that

$P(A)+P(B)+P(A\cap B)=P(A\cup B)$

We know that

Maximum value of $P(A\cup B)$ =1 and minimum value of $P(A\cup B)$ =0

$0\leq P(A\cup B )\leq 1$

$0\leq P(A)+P(B)-P(A\cap B)\leq 1$

$0\leq 0.4+0.7-P(A\cap B)\leq 1$

$0\leq 1.1-P(A\cap B)\leq 1$

$0\leq 1.1-P(A\cap B)$

$P(A\cap B)\leq 1.1$

It is not possible that $P(A\cap B)$ is equal to 1.1

$1.1-P(A\cap B)\leq 1$

$-P(A\cap B)\leq 1-1.1=-0.1$

Multiply by (-1) on both sides

$P(A\cap B)\geq 0.1$

Again, $P(A\cup B)\geq P(B)$

$0.4+0.7-P(A\cap B)\geq 0.7$

$1.1-P(A\cap B)\geq 0.7$

$-P(A\cap B)\geq -1.1+0.7=-0.4$

Multiply by (-1) on both sides

$P(A\cap B)\leq 0.4$

Hence, $0.1\leq P(A\cap B)\leq 0.4$

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3 years ago

nignag [31]

132/4 = 33, x = 33 boxes

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3 years ago

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