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Lynna [10]
4 years ago
15

A motorcycle manufacturer offers 3 different models, each available in 6 different colors. how many different combinations of mo

del and color are available? a- 9 b- 6 c-18 d- 12 e- 24
Mathematics
1 answer:
NARA [144]4 years ago
5 0
The is a choice of 6 colors for each of the 3 models. Therefore the number of combinations of model and color is: 3 x 6 = 18.
The answer is c - 18.
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H e l p. m e. p l e a s e
Nuetrik [128]

Answer:

Step-by-step explanation:

Your revenue should be the price demand function * the price:

(x) 350 - 2x =

R(X) = 350x - 2x^{2}

your profit function should be the

Revenue - cost

350x - 2x^{2} - 14204 -4x

- 2x^{2}  + 346x - 14204

b.e.p  is where R=0

- 2x^{2}  + 346x - 14204 =0

Factored: f(x) = -2(x - 106)(x - 67)

bep = 67

6 0
3 years ago
Which could be the first step in solving the equation represented by the model below? HELP PLZ
kotegsom [21]

Answer:

take away four from both -4

5 0
4 years ago
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Please help! What is 1 1/2+1/6?<br><br> -giving out 20 points
Mice21 [21]

1  \frac{2}{3}

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7 0
3 years ago
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Tim has an after-school delivery service that he provides for several small retailers in town. He uses his bicycle and charges $
nlexa [21]

Answer:

<em>$4.84</em>

<em></em>

Step-by-step explanation:

Given that

For delivery within 1\frac{1}2 mi, charges = $1.25

For delivery within 1\frac{1}2 mi - 1\frac{3}4 mi, charges = $1.70

For delivery within 1\frac{3}4 mi - 2 mi, charges = $2.15

and

so on.

i.e.

For delivery within 2 mi - 2\frac{1}4 mi, charges = $2.60

For delivery within 2\frac{1}4 mi - 2\frac{1}2 mi, charges = $3.05

For delivery within 2\frac{1}2 mi - 2\frac{3}4 mi, charges = $3.50

For delivery within 2\frac{3}4 mi - 3 mi, charges = $3.95

For delivery within 3 mi - 3\frac{1}4 mi, charges = $4.40

So, every 0.25 mi or \frac{1}4 mi increase in distance, there is an increase of $0.45 in the charges.

It is given that there is increase of 10% in the rates.

i.e. for every 0.25 mi increase in distance, there is an increase of 0.45*1.10 = $0.495 in the charges.

The distance of 3\frac{1}8 miles lies within the range 3 mi - 3\frac{1}4 mi.

So actual charges after increase of 10%

\Rightarrow \$4.40 \times \dfrac{110}{100}\\\Rightarrow \$4.40 \times 1.10\\\Rightarrow \bold{\$4.84}

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4 years ago
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