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KonstantinChe [14]

2 years ago

7

12. Brad sells ice cream and soft drinks at outdoor festivals. He buys soft drinks for 50 cents per can and ice cream bars for $

75 per hundred. He marks up all items by 100 percent selling the drinks for $1.00 and ice cream bars for $1.50. One day he sold 100 cans of soft drinks and 90 ice cream bars. His expenses totaled $31.50. What was his net profit?

1 answer:

Damm [24]2 years ago

4 0

Well, he made $100 off of drinks, and $135 off of ice cream bars.

Then to find his profit take $235 - $31.50 = $203.50

There's your answer! $203.50 was his profit.

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Gavin likes biking. He never misses a chance to go for a ride when the weather is nice. This week his goal is to bike about 65 t

andreyandreev [35.5K]

Answer:

How far should he ride on each of the four days to reach his goal?

1st day: $8$ miles

2nd day: $12$ miles

3rd day: $18$ miles

4th day: $27$ miles

Step-by-step explanation: As the problem says, $x$ is the number of miles he rides on the first day. Let's start off with that.

1st day: $x$ miles

He want to ride 1.5 times as far as he rode the day before... no 1.5 more, but 1.5 <em>times</em> as far as he rode the day before; you would multiply 1.5 with the previous day's length.

2nd day: $1.5x$

Then you multiply $1.5$ to $1.5x$ to get the third day's.

3rd day: $2.25x$

4th day: $3.375x$

----------------------------

Phew! Gavin wants to ride a total of 65 miles over these four days, so if Gavin added all the miles of the four days, he should get 65...

1st+2nd+3rd+4th=65

$x+1.5x+2.25x+3.375x=65$

$8.125x=65$

$\frac{8.125x}{8.125}=\frac{65}{8.125}$

$x=8$

Yes! Now that we've got the hard part done... substitute 8 for ever single $x$ .

1st day: $8$ miles

2nd day: $1.5(8)=12$ miles

3rd day: $2.25(8)=18$ miles

4th day: $3.375(8)=27$ miles

-------------------------

Checking my answer:

Just add the miles!

$8+12+18+27=65$

$65=65$

✓

-----------------------

Hope that helps! :D

5 0

2 years ago

The word geometry has eight letters. three letters are chosen at random. what is the probability that two consonants and one vow

olga nikolaevna [1]

Answer:

0.536 is the required probability.

Step-by-step explanation:

We have been given the word the word "GEOMETRY"

we have to find the probability that two consonants and one vowel are chosen:

Number of consonants are: 5

Number of vowels are: 3

Hence, The required probability is: $\frac{^5C_2\cdot ^3C_1}{^8C_3}$

Using: $^nC_r=\frac{n!}{(r!)(n-r)!}$

$\frac{\frac{5!}{3!\cdot 2!}\cdot\frac{3!}{1!\cdot 2!}}{\frac{8!}{3!\cdot 5!}}$

Simplifying the above expression:

$\frac{\frac{5\cdot 4\cdot 3!}{3!\cdot 2}\cdot {\frac{3\cdot 2!}{2!}}}{\frac{8\cdot 7\cdot 6\cdot 5!}{5!\cdot 3\cdot 2}}$

Further simplification after cancelling out the common terms we get:

$\Rightarrow \frac{30}{56}=\frac{15}{28}=0.5357=0.536$

Hence, Option 1 is correct.

4 0

2 years ago

Read 2 more answers

What is the solution to the inequality 2 + 4/9x _> 4 + x

Vsevolod [243]

Multiply each side by 9:
18 + 4x > 36 + 9x
Subtract 18 from each side:
4x > 18 + 9x
Subtract 9x from each side:
-5x > 18
Divide each side by -5:
x < -3.6

Hope this helps!

8 0

2 years ago

If c (x) = StartFraction 5 Over x minus 2 EndFraction and d(x) = x + 3, what is the domain of (cd)(x)?

Volgvan

Answer:

The domain of the function (cd)(x) will be all real values of x except x = 2.

Step-by-step explanation:

The two functions are $c(x) = \frac{5}{x - 2}$ and d(x) = x + 3

So, (cd)(x) = $(\frac{5}{x - 2})(x + 3) = \frac{5(x + 3)}{x - 2}$

Then, for x = 2 the function (cd)(x) will be undefined as zero in the denominator will make the function (cd)(x) undefined.

Therefore, the domain of the function (cd)(x) will be all real values of x except x = 2. (Answer)

8 0

2 years ago

Read 2 more answers

R+5/mn=p solve for m

AnnZ [28]

Answer:

$\bold{m\ =\ \dfrac5{(p-r)n}}$

Step-by-step explanation:

$\bold{r+\dfrac5{mn}\ =\ p}\\\\ {}\quad-r\qquad-r\\\\{}\ \ \bold{\dfrac5{mn}\ =\ p-r}\\\\{}\ ^{_\times}(mn)\quad ^{_\times}(mn)\\\\{}\quad\bold{5\ =\ (p-r)^{_\times}(mn)}\\\\\div(p-r)\quad\div(p-r)\\\\{}\ \ \bold{\dfrac5{p-r}\ =\ mn}\\\\{}\quad \ \div n\quad\ \ \div n\\\\\bold{\dfrac5{(p-r)n}\ =\ m}$

If you mean (r+5)/mn then:

$\bold{\dfrac{r+5}{mn}\ =\ p}\\\\{}\ ^{_\times}(mn)\quad ^{_\times}(mn)\\\\{}\ \bold{r+5\ =\ pmn}\\\\\div(pn)\quad\div(pn)\\\\{}\ \ \bold{\dfrac{r+5}{pn}\ =\ m}$

4 0

2 years ago