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2 years ago

10

Joan can make and bottle 8 smoothies an hour. if she works with a friend they can make and bottle 18 smoothies an hour. what is

the marginal product of labor? (a10 (b26 (c9 (d2the answer is (c9

2 answers:

alex41 [277]2 years ago

7 0

The marginal product of the labor which joan can make a bottle 8 smoothies in an hour and when she is wokring with her friend they can make a total of 18 bottle of smoothies an hour. The formula is 18 / 2 = 9.9 is the marginal product of labor.So the answer is C. 9

MAVERICK [17]2 years ago

4 0

Answer:

10

Step-by-step explanation:

Easiest possible explanation is because Joan can make 8 by herself with the help of a friend they're making 10 more (18/h)

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Which ordered pairs are in a proportional relationship with (0.2, 0.3)?

MAXImum [283]

Answer:

C and E

Step-by-step explanation:

We are given that

$x_1=0.2,y_1=0.3$

$\frac{y}{x}=\frac{0.3}{0.2}=\frac{3}{2}$

$k=\frac{y}{x}=\frac{3}{2}$

A. $x_2=1.2,y_2=2.3$

$\frac{y_2}{x_2}=\frac{2.3}{1.2}=\frac{23}{12}\neq=\frac{3}{2}$

Hence, it is not in proportional relationship with (0.2,0.3)

B.(2.7,4.3)

$x_3=2.7,y_3=4.3$

$\frac{y_3}{x_3}=\frac{4.3}{2.7}=\frac{43}{27}\neq\frac{3}{2}$

Hence, it is not in proportional relationship with (0.2,0.3).

C.(3.2,4.8)

$x_4=3.2,y_4=4.8$

$\frac{y_4}{x_4}=\frac{4.8}{3.2}=\frac{3}{2}$

Hence, the ordered pair (3.2,4.8) are in a proportional relationship with (0.2,0.3).

D.(3.5,5.3)

$\frac{y_5}{x_5}=\frac{5.3}{3.5}=\frac{53}{35}\neq \frac{3}{2}$

Hence, the ordered pair (3.5,5.3) are not in a proportional relationship with (0.2,0.3).

E.(5.2,7.8)

$\frac{y_6}{x_6}=\frac{7.8}{5.2}=\frac{3}{2}$

Hence, the ordered pair (5.2,7.8) are in a proportional relationship with (0.2,0.3).

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3 years ago

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2 years ago

Together Louisa and kill scored more than 30 points in the basketball game. If Kill scored 12 points how many points P did Louis

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Attached the solution with work shown.

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3 years ago

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AnsAnswerwer:

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Step-by-step explanation:

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2 years ago

A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

$t = \frac{ 5 + \sqrt{29} }{4} = 2.60$

or

$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

Since time cannot be negative, we discard the negative value and pick,

$t = 2.60s$

8 0

3 years ago