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2 years ago

Rob peeled 125 oranges in 5 hours. At that rate, how many oranges would he peel in 12 minutes?

Mathematics

2 answers:

marin [14]2 years ago

6 0

Answer:

$\boxed {\tt 5 \ oranges}$

Step-by-step explanation:

Let's set up a proportion.

$\frac{oranges}{minutes} =\frac{oranges}{minutes}$

He can peel 125 oranges in 5 hours. First, let's find the minutes in 5 hours.

There are 60 minutes in 1 hour, so multiply 60 and 5 (60*5=300). There are 300 minutes in 5 hours.

$\frac{125 \ oranges}{300 \ minutes}=\frac{oranges}{minutes}$

We don't know how many oranges he can peel in 12 minutes. So we say he can peel x oranges in 12 minutes.

$\frac{125 \ oranges}{300 \ minutes}=\frac{x \ oranges}{12 \ minutes}$

$\frac{125 \ }{300 \ }=\frac{x \ }{12 }$

Solve for x by isolating it. x is being divided by 12. The inverse of division is multiplication. Multiply both sides of the proportion by 12.

$12*\frac{125 \ }{300 \ }=\frac{x \ }{12 \ }*12$

$12*\frac{125 \ }{300 \ }=x$

$12*0.416666667=x$

$5=x$

Rob can peel 5 oranges in 12 minutes.

vlabodo [156]2 years ago

6 0

Answer:

5 oranges in 12 minutes

Step-by-step explanation:

$125 \div 5 = 25$

$25 \div 60 = 0.41$

$0.41 \times 12 = 5$

what i did here was I took the original amount (125), divided it by 5 which is 25 oranges per hour

1 hr = 60 mins

then I divided 25 oranges by 60 mins, then times it by 12 to get 5 around 5 oranges in 12 minutes.

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Which definite integral approximation formula is this: the integral from a to b of f(x)dx ≈ (b-a)/n * [<img src="https://tex.z-d

Stella [2.4K]

The answer is most likely A.

The integration interval [a, b] is split up into n subintervals of equal length (so each subinterval has width (b - a)/n, same as the coefficient of the sum of y terms) and approximated by the area of n rectangles with base (b - a)/n and height y.

n subintervals require n + 1 points, with

x₀ = a

x₁ = a + (b - a)/n

x₂ = a + 2(b - a)/n

and so on up to the last point x = b. The right endpoints are x₁, x₂, … etc. and the height of each rectangle are the corresponding y 's at these endpoints. Then you get the formula as given in the photo.

• "Average rate of change" isn't really relevant here. The AROC of a function G(x) continuous* over an interval [a, b] is equal to the slope of the secant line through x = a and x = b, i.e. the value of the difference quotient

(G(b) - G(a) ) / (b - a)

If G(x) happens to be the antiderivative of a function g(x), then this is the same as the average value of g(x) on the same interval,

$g_{\rm ave}=\dfrac{G(b)-G(a)}{b-a}=\dfrac1{b-a}\displaystyle\int_a^b g(x)\,\mathrm dx$

(* I'm actually not totally sure that continuity is necessary for the AROC to exist; I've asked this question before without getting a particularly satisfying answer.)

• "Trapezoidal rule" doesn't apply here. Split up [a, b] into n subintervals of equal width (b - a)/n. Over the first subinterval, the area of a trapezoid with "bases" y₀ and y₁ and "height" (b - a)/n is

(y₀ + y₁) (b - a)/n

but y₀ is clearly missing in the sum, and also the next term in the sum would be

(y₁ + y₂) (b - a)/n

the sum of these two areas would reduce to

(b - a)/n = (y₀ + 2 y₁ + y₂)

which would mean all the terms in-between would need to be doubled as well to get

$\displaystyle\int_a^b f(x)\,\mathrm dx\approx\frac{b-a}n\left(y_0+2y_1+2y_2+\cdots+2y_{n-1}+y_n\right)$