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2 years ago

A teacher reviews 4-1/2 papers per hour, how many papers will that teacher review in 6-1/3 hours

Mathematics

1 answer:

MrRa [10]2 years ago

5 0

Answer:

$28\frac{1}{2}\ papers$

Step-by-step explanation:

step 1

Convert mixed numbers to an improper fractions

$4\frac{1}{2}=\frac{4*2+1}{2}=\frac{9}{2}$

$6\frac{1}{3}=\frac{6*3+1}{3}=\frac{19}{3}$

step 2

we know that

A teacher reviews 9/2 papers per hour

using proportion

Find how many papers will the teacher review in 19/3 hours

Let

x-----> the number of papers

$(9/2)/1=x/(19/3)$

$x=(19/3)(9/2)$

$x=28.5\ papers$

Convert to mixed numbers

$28.5=28\frac{1}{2}\ papers$

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By selling a book for $16.50, a bookseller loses 12%. What is the cost price of the book?

Alika [10]

Answer: 18. 75

Step-by-step explanation:

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7 0

2 years ago

Determine the area (in units2) of the region between the two curves by integrating over the x-axis. y = x2 − 24 and y = 1

astra-53 [7]

Answer:

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

Step-by-step explanation:

This case represents a definite integral, in which lower and upper limits are needed, which corresponds to the points where both intersect each other. That is:

$x^{2} - 24 = 1$

Given that resulting expression is a second order polynomial of the form $x^{2} - a^{2}$ , there are two real and distinct solutions. Roots of the expression are:

$x_{1} = -5$ and $x_{2} = 5$ .

Now, it is also required to determine which part of the interval $(x_{1}, x_{2})$ is equal to a number greater than zero (positive). That is:

$x^{2} - 24 > 0$

$x^{2} > 24$

$x < -4.899$ and $x > 4.899$ .

Therefore, exists two sub-intervals: $[-5, -4.899]$ and $\left[4.899,5\right]$ . Besides, $x^{2} - 24 > y = 1$ in each sub-interval. The definite integral of the region between the two curves over the x-axis is:

$A = \int\limits^{-4.899}_{-5} [{1 - (x^{2}-24)]} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} [{1 - (x^{2}-24)]} \, dx$

$A = \int\limits^{-4.899}_{-5} {25-x^{2}} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} {25-x^{2}} \, dx$

$A = 25\cdot x \right \left|\limits_{-5}^{-4.899} -\frac{1}{3}\cdot x^{3}\left|\limits_{-5}^{-4.899} + x\left|\limits_{-4.899}^{4.899} + 25\cdot x \right \left|\limits_{4.899}^{5} -\frac{1}{3}\cdot x^{3}\left|\limits_{4.899}^{5}$