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NeTakaya
3 years ago
9

Which pair of number has the least common multiple of 12

Mathematics
1 answer:
liraira [26]3 years ago
5 0

Answer:

3 and 4.

Step-by-step explanation:

3 * 2 *2 = 12.

So 3 and 4 have an LCM of 12.

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Solve Proportions In the following exercises, solve.
Julli [10]

Answer:

Thus, a golden retriever should be given 19 teaspoon of medicine.

Step-by-step explanation:

  • Given that a golden retriever weighing 85 pounds has diarrhea. His medicine is prescribed as 1 teaspoon per 5 pounds.
  • To find how much medicine should he be given.
  • Assign a variable. Write a sentence that gives the information to find it.
  • Translate into a proportion be careful of the units.
  • Multiply both sides by the  Least Common Denominator and simplify. Check the answer.

Step 1 of 3

Write the data in mathematical form and simplify.

Let x=the number of teaspoon

If 1  teaspoon medicine is 5  pounds then 85  pounds is how many teaspoon of medicine.

Translate the statement into proportion.

\begin{aligned}\frac{\text { pounds }}{\text { teaspoon }} &=\frac{\text { pounds }}{\text { teaspoon }} \\\frac{5}{1} &=\frac{85}{x}\end{aligned}

Step 2 of 3

Multiply both sides by x and simplify

$x(5)=\frac{85}{\bcancel{x}}(\bcancel{x})

5x=85

Divide by 5,

x=\frac{85}{5}

<em>Simplify, x=19</em>

<em>Step 3 of 3</em>

<em>Check the answer.</em>

<em>Substitute x=19  in original proportion and simplify.</em>

<em />$$\begin{aligned}\frac{5}{1} &=\frac{85}{x} \\5 &=\frac{85}{19}\end{aligned}$\\ \\ Hence, $5=5$, True

5 0
2 years ago
A loan company knows that 5% of its loans will be delinquent. Of the company’s 400 loan accounts, what is the probability
Scrat [10]

Answer:

Step-by-step explanation:

Given that a loan company knows that 5% of its loans will be delinquent.

since each loan is independent of the other p , probability for any random loan to be delinquent is constant 0.05

X no of delinquent loan accounts is binomial with n =400 and p = 0.05

Since n is very large and also np = 20 and nq >5 we can approximate to normal

Mean = np = 20 :  Variance = npq = 19

Std dev = 4.36

X is N(1, 4.36)

With continuity correcton we calculate the prob

a) that exactly 25 accounts will be delinquent?

=P(24.5

b) that fewer than 30 accounts will be delinquent?

=P(X<29.5)

= 0.9854

c) more than 24 accounts will be delinquent

=P(X>24,5)

=0.1509

7 0
3 years ago
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