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Ronch [10]
4 years ago
13

It’s bath time while filling the tub dad used the last capful of bubble bath.after doing quick search for more to buy he notices

that he can get 36oz for $3.79 or 24 oz for $2.89. Which of these options is the better deal ?
Mathematics
1 answer:
WINSTONCH [101]4 years ago
6 0

If you divide the price by the number of ounces, you will get how much you are paying for a single ounce.

\frac{3.79}{36} ≈ 0.11 dollars per ounce

\frac{2.89}{24} ≈ 0.12 dollars per ounce

As you can see, you are paying about 0.01 dollars more for every ounce of bubble bath in the second option. So you are better off buying 36 ounces for 3.79 dollars.

You can also do this buy dividing ounces by dollars.

\frac{36}{3.79} ≈ 9.5 ounces per dollar

\frac{24}{2.89} ≈ 8.3 ounces per dollar

Again, you can see that you get about 1 more ounce for every dollar you pay in the first option. So 36 ounces for 3.79 dollars is the way to go.

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In how many ways can a subcommittee of 6 students be chosen from a committee which consists of 10 senior members and 12 junior m
givi [52]

Answer:

The number of ways is 13860 ways

Step-by-step explanation:

Given

Senior Members = 10

Junior Members = 12

Required

Number of ways of selecting 6 students students

The question lay emphasis on the keyword selection; this implies combination

From the question, we understand that

<em>4 students are to be selected from senior members while 2 from junior members;</em>

The number of ways is calculated as thus;

<em>Ways = Ways of Selecting Senior Members * Ways of Selecting Junior Members</em>

<em />Ways = ^{10}C_4 * ^{12}C2

Ways = \frac{10!}{(10-4)!4!)} * \frac{12!}{(12-2)!2!)}

Ways = \frac{10!}{(6)!4!)} * \frac{12!}{(10)!2!)}

Ways = \frac{10 * 9 * 8 * 7 *6!}{(6! * 4*3*2*1)} * \frac{12*11*10!}{(10!*2*1)}

Ways = \frac{10 * 9 * 8 * 7}{4*3*2*1} * \frac{12*11}{2*1}

Ways = \frac{5040}{24} * \frac{132}{2}

Ways = 210 * 66

Ways = 13860

<em>Hence, the number of ways is 13860 ways</em>

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Sunny_sXe [5.5K]

The difference is  $\frac{3 x+81}{x(x-9)(x+9)}$

Explanation:

The expression is $\left(\frac{9}{x^{2}-9 x}\right)-\left(\frac{6}{x^{2}-81}\right)$

Removing the parenthesis, we have,

$\left\frac{9}{x^{2}-9 x}\right-\left\frac{6}{x^{2}-81}\right$

Factoring the terms $x^{2}-9 x$ and $x^{2}-81$, we get,

$\frac{9}{x(x-9)}-\frac{6}{(x+9)(x-9)}$

Taking LCM, we get,

$\frac{9(x+9)-6x}{x(x-9)(x+9)}}$

Simplifying the numerator, we get,

$\frac{9x+81-6x}{x(x-9)(x+9)}}$

Subtracting the numerator, we have,

$\frac{3 x+81}{x(x-9)(x+9)}$

Hence, the difference is $\frac{3 x+81}{x(x-9)(x+9)}$

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