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dem82 [27]
3 years ago
11

Which property would be used to show that the product of two rational is always rational

Mathematics
1 answer:
KatRina [158]3 years ago
3 0

You multiply two rational numbers by multiplying numerators and denominators with each other. So, you have

\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{ac}{bd}

A rational number is a fraction where both numerator and denominator are integers. Since the product of two integers is an integer, the result fraction is also a rational number, because both ac and bd are integers.

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Derrick and Sam need 10 pounds of colored gravel for the bottom of their new fish tank. Derrick has 2 pounds of purple gravel an
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Answer:

4 yellow gravel . 2+4=6

10-6= 4

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3 years ago
You want to convert 1 kg to milligrams. You already know that 1 kg equals 1000 g explain how knowing that 1 gram equals 1000 mg
butalik [34]
1 kg=1000g*1000mg/1g=1000000mg =10⁶ mg
or you can use proportion

1000g - x mg
1 g     -1000mg
x=1000*1000=1000000mg 

1kg=10⁶ mg
6 0
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Read 2 more answers
Finish the value of d 2d - 5 = 17
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D is 11

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d = 11                                 answer
8 0
3 years ago
Which is correct? A.b.c.
Hatshy [7]
By geometry, there is a relationship between the tangent PA and the segments PB and PE, namely
PA^2=PB*PE
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6 0
3 years ago
A batch of 580 containers for frozen orange juice contains 8 that are defective. Two are selected, at random, without replacemen
serious [3.7K]

Answer:

a. 0.12109

b. 0.0001668

c .0.9726

d. 0.01038

e. 0.01211

f. 0.000001731

Step-by-step explanation:

Sample size = 580

Defective units = 8

Number of picks = 2

a) If the first container is defective, there 7 defective containers left in a population of 579. The probability of selecting a defective one is:

P=\frac{7}{579}=0.121

b) The probability that both are defective is given by:

P=\frac{8}{580}*\frac{7}{579}= 0.000167

c) The probability that both are acceptable is given by:

P=\frac{580-8}{580}*\frac{579-8}{579}= 0.9726

d) In this case, two defective units were removed from the batch, the probability that the third is also defective is:

P=\frac{6}{578}}= 0.0104

e) In this case, one acceptable and one defective unit were removed from the batch, the probability that the third is also defective is:

P=\frac{7}{578}= 0.01211

f) The probability that all three are defective is given by:

P=\frac{8}{580}*\frac{7}{579}*\frac{6}{578} = 0.000001731

7 0
3 years ago
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