Give an example of a real-world situation that could be represented by an algebraic expression

1 answer:

5 0

Gas is $2.53⁹ per gallon today, at that filling station
down the street from my house.
I have $10 in my pocket.

a). How much gas can I put in the car if I go there ?

b). My car gets 25 miles per gallon. How many 'miles' will my $10 buy ?

c). In Calgary (Alberta, Canada) today, gas is selling for
89¢ (Canadian) per liter. What is the percentage more or less,
compared to the price of gas down the street from my house ?

If I posted this problem on Brainly, I'd probably give it
something like 5 points for part-a, another 5 for part-b,
and 10 or 15 for part-c because in addition to math,
that part takes some research.

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"The Fountains of Bellagio" is a choreographed water display set to light and music that takes place in front of the Bellagio Ho

Pepsi [2]

Answer:

a) 4/15

b) 4/15

Step-by-step explanation:

maximum wait is 1 min,2 min,3 min,4 min,5 min,6 min,7 min,8 min,9 min,10 min,11 min,12 min,13 min,14 min,15 min =total 15 values

a) less than 4 min means he has to wait for 1 min, 2 min, 3 min, 4 min=total 4 values

Probability of wait for less than 4 minutes=4/15

b) more than 11 minutes wait means he has to wait for either 12 min, 13 min, 14 min, 15 min=4 values

Probability of wait for more than 11 minutes=4/15

6 0

3 years ago

16 x <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B8%7D" id="TexFormula1" title="\frac{1}{8}" alt="\frac{1}{8}" align="absm

Dmitry_Shevchenko [17]

The answer is 2. Because 16/8 is 2.

4 0

3 years ago

Read 2 more answers

The estimated value of the integral from 0 to 2 of x cubed dx , using the trapezoidal rule with 4 trapezoids is

bulgar [2K]

The integral is approximated by the sum,

$\displaystyle\int_0^2f(x)\,\mathrm dx\approx\sum_{n=0}^4\frac12\times\frac{f(x_n)+f(x_{n+1})}2=\frac14\sum_{n=0}^3(f(x_n)+f(x_{n+1}))$

where $f(x)=x^3$ and $x_n=\dfrac12n$ , giving you

$\displaystyle\frac14\sum_{n=0}^3\bigg(\left(\frac n2\right)^3+\left(\frac{n+1}2\right)^3\bigg)$
$\displaystyle\frac1{32}\sum_{n=0}^3(n^3+(n+1)^3)$
$\displaystyle\frac1{32}\sum_{n=0}^3(2n^3+3n^2+3n+1)$

Faulhaber's formulas make short work of computing the sum. You have

$\displaystyle\sum_{n=0}^k1=k+1$
$\displaystyle\sum_{n=0}^kn=\frac{k(k+1)}2$
$\displaystyle\sum_{n=0}^kn^2=\frac{k(k+1)(2k+1)}6$
$\displaystyle\sum_{n=0}^kn^3=\frac{k^2(k+1)^2}4$

which gives

$\displaystyle\frac1{16}\sum_{n=0}^3n^3+\frac3{32}\sum_{n=0}^3n^2+\frac3{32}\sum_{n=0}^3n+\frac1{32}\sum_{n=0}^31$
$\displaystyle\frac{36}{16}+\frac{42}{32}+\frac{18}{32}+\frac4{32}$
$\implies\displaystyle\int_0^2x^3\,\mathrm dx\approx\frac{17}4=4.25$