All categories

3 years ago

Joe can clear a lot in 2.5 hours. His partner can do the same job in 7.5 hours. How long will it take them working together?

Mathematics

1 answer:

irina [24]3 years ago

7 0

Answer:

The time required by both persons to complete the work T = 1.458 hours

Step-by-step explanation:

Given data

Time required by Joe to clear a lot $T_1$ = 2.5 hours

Time required by his partner to clear a lot $T_2$ = 2.5 hours

Time required by both persons to complete the work by working together

$T = \frac{T_1 T_2}{T_1 + T_2}$

$T = \frac{(2.5)(3.5)}{2.5 + 3.5}$

T = 1.458 hours

Therefore the time required by both persons to complete the work T = 1.458 hours

You might be interested in

The number of cases of a new disease can be modeled by the quadratic regression equation y = -2x^2 + 40x + 8. What is the best p

Stels [109]

Answer:

158 cases

Step-by-step explanation:

Given tbe quadratic regression model :

y = -2x^2 + 40x + 8

y = number of cases of a new disease

x = number of years

The predicted number of cases of a new disease in 15 years can be calculated thus ;

Put x = 15 in the equation ;

y = -2(15)^2 + 40(15) + 8

y = - 2 * 225 + 600 + 8

y = - 450 + 600 + 8

y = 158

158 cases

5 0

2 years ago

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$