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FromTheMoon [43]
2 years ago
5

Brandi solved 16 / 5 by using a related multiplication expression. What multiplication expression did she is?

Mathematics
1 answer:
klemol [59]2 years ago
8 0

The given question is wrong.

Question:

Brandi solved \frac{1}{6}\div 5 by using a related multiplication expression. What multiplication expression did she is?

Answer:

The related multiplication expression is  $\frac{1}{6}\times \frac{1}{5}.

Solution:

Given expression is \frac{1}{6}\div 5.

We can write 5 as a fraction with denominator 1.

$ \frac{1}{6}\div5=\frac{1}{6}\div \frac{5}{1}

Using the fraction rule  $\frac{a}{b}\div \frac{c}{d}=\frac{a}{b}\times \frac{d}{c},

$\frac{1}{6}\div \frac{5}{1}=\frac{1}{6}\times \frac{1}{5}

Now multiply the fractions using the rule $\frac{a}{b}\times \frac{d}{c}=\frac{a\times d}{b\times c},

$\frac{1}{6}\times \frac{1}{5}=\frac{1\times1}{6\times5}

Let us multiply the numbers in the numerator and denominator separately, we get

$\frac{1\times1}{6\times5}=\frac{1}{30}

Hence the related multiplication expression is  $\frac{1}{6}\times \frac{1}{5}.

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Answer:

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Step-by-step explanation:

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The weight of a bull clad is 388 kilograms. If it’s weight increases at a rate of 1 2/5 kilograms per day how long it will take
Alex73 [517]

Answer:

It will take  <u><em>80 days</em></u>  for the bull calf to reach a weight of 500 kilograms.

Step-by-step explanation:

Given:

The weight of a bull calf is 388 kilograms.

Now, to find the weight of bull calf of how long it will take to reach a weight of 500 kilograms, if it’s weight increases at a rate of 1 2/5 kilograms per day.

Required weight which to be increased = 500 - 388 = 112 kilograms.

Rate of weight increase = 1\frac{2}{5}=\frac{7}{5}

                                        = 1.4\ kilograms.

Thus, the time required = \frac{required\ weight}{rate\ of\ weight}

                                        = \frac{112}{1.4}

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<em>The time required   =    80 days</em>.

Therefore, it will take 80 days for the bull calf to reach a weight of 500 kilograms.

5 0
3 years ago
A probability experiment is conducted in which the sample space of the experiment is upper s equals startset 4 comma 5 comma 6 c
olchik [2.2K]
The correct question statement is:

A probability experiment is conducted in which the sample space of the experiment is S = {4,5,6,7,8,9,10,11,12,13,14,15}. Let event E={7,8,9,10,11,12,13,14,15}. Assume each outcome is equally likely. List the outcomes in E^{c}. Find P(E^{c}).

Solution:

Part 1:

E^{c} means compliment of the set E. A compliment of a set can be obtained by finding the difference of the set from the universal set. The universal set is the set which contains all the possible outcomes of the events which is S in this case.

So, compliment of E will be equal to S - E. S - E will result in all those elements of S which are not present in E. So, we can write:

E^{c}=S-E \\  \\ &#10;E^{c}=(4,5,6,7,8,9,10,11,12,13,14,15)-(7,8,9,10,11,12,13,14,15) \\  \\ &#10;E^{c}=(4,5,6)

Thus the set compliment of E will contain the elements {4,5,6}.So

E^{c} = {4,5,6}

Part 2)

P(E^{c}) means probability that if we select any number from the Sample Space S, it will belong the set E compliment.

P(E^{c}) = (Number of Elements in E^{c})/Number of elements in S

Number of elements in set S = n(S) = 12
Number of elements in set E^{c} = n(E^{c})=3

So, 

P(E^{c})= \frac{n(E^{c}) }{n(S)} \\  \\ &#10;P(E^{c})= \frac{3}{12} \\  \\ &#10;P(E^{c})= \frac{1}{4}
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3 years ago
Equation equivalent to 2(x+3)
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Answer:

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Step-by-step explanation:

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a(b+c)

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ab+ac

plz brainliest :)

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Answer:

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