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2 years ago

8

You have just timed a person doing a hair cut for the first time. It took 50 minutes. What unit improvement factor learning curv

e would you use if the person took 35 minutes on the second hair cut?
A. 35 percent
B. 50 percent
C. 70 percent
D. 75 percent
E. 80 percent

1 answer:

Sladkaya [172]2 years ago

3 0

Answer: C. 70 percent

Step-by-step explanation:

Given, Time for the first unit = 50 minutes

Time for the second unit = 35 minutes

The unit improvement factor learning curve = (The time for the second unit) ÷ (time for the first unit) x 100.

So, The unit improvement factor learning curve = 35÷ 50 × 100 = 70 percent.

Hence, the correct option is "C. 70 percent".

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Which graph represents the function of f(x) = the quantity of 9 x squared plus 9 x minus 18, all over 3 x plus 6

lora16 [44]

Answer:

The answer is the option C

graph of $3x$ minus $3$ , with discontinuity at negative $2$ , negative $9$

Step-by-step explanation:

we have

$f(x)=\frac{9x^{2}+9x-18}{3x+6}$

Simplify

$f(x)=9\frac{(x^{2}+x-2)}{3(x+2)}$

$f(x)=3\frac{(x^{2}+x-2)}{(x+2)}$

Step 1

Convert to a factored form the numerator

$x^{2}+x-2=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+x=2$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+x+0.25=2+0.25$

$x^{2}+x+0.25=2.25$

Rewrite as perfect squares

$(x+0.5)^{2}=2.25$

Square root both sides

$x+0.5=(+/-)1.5$

$x=-0.5(+/-)1.5$

$x=-0.5+1.5=1$

$x=-0.5-1.5=-2$

so

$x^{2}+x-2=(x-1)(x+2)$

Step 2

Simplify the function f(x)

$f(x)=3\frac{(x^{2}+x-2)}{(x+2)}=3\frac{(x-1)(x+2)}{(x+2)}$

The domain of the function f(x) is all real numbers except the number $x=-2$

Because the denominator can not be zero

$f(x)=3\frac{(x-1)(x+2)}{(x+2)}=3(x-1)=3x-3$

$f(x)=3x-3$ ------> with a discontinuity at $x=-2$

$f(-2)=3(-2)-3=-9$

The discontinuity is at point $(-2,-9)$

the answer in the attached figure

8 0

2 years ago

Read 2 more answers

Select all the pairs that could be reasonable approximations for the diameter and circumference of a cirlce 5 meters and 22 mete

bazaltina [42]

Answer:

Option a) circle 5 meters and 22 meters

Step-by-step explanation:

We are given the following information in the question:

A pair of diameter and the circumference is given. We have to find a correct approximations for the diameter and circumference.

a) circle 5 meters and 22 meters

$\text{Diameter} = 5\text{ meters}\\\text{Circumference} = \pi d = 3.14\times 5 = 15.7\text{ meters}$

b) 19 inches and 50 inches

$\text{Diameter} = 19\text{ inches}\\\text{Circumference} = \pi d = 3.14\times 19 = 59.66\text{ inches}$

c) 33 centimeters and 80 centimeters

$\text{Diameter} = 33\text{ centimeters}\\\text{Circumference} = \pi d = 3.14\times 33 =103.62\text{ centimeters}$

Thus, no pair gives a reasonable approximation. Only the circle with diameter 5 and circumference 22 meters have closest approximation.

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2 years ago

How much heat is required to raise the temperature of

Elina [12.6K]

Answer:

The amount of heat required to raise the temperature of liquid water is 9605 kilo joule .

Step-by-step explanation:

Given as :

The mass of liquid water = 50 g

The initial temperature = $T_1$ = 15°c

The final temperature = $T_2$ = 100°c

The latent heat of vaporization of water = 2260.0 J/g

Let The amount of heat required to raise temperature = Q Joule

Now, From method

Heat = mass × latent heat × change in temperature

Or, Q = m × s × ΔT

or, Q = m × s × ( $T_2$ - $T_1$ )

So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )

Or, Q = 50 g × 2260.0 J/g × 85°c

∴ Q = 9,605,000 joule

Or, Q = 9,605 × 10³ joule

Or, Q = 9605 kilo joule

Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer

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2 years ago

Match the following guess solutions ypyp for the method of undetermined coefficients with the second-order nonhomogeneous linear

Sunny_sXe [5.5K]

Answer:

Step-by-step explanation:

1 ) Given that

$(d^2y/dx^2) + 4y = x - x^2 + 20\\\\ (d^2y/dx^2) + 4y = - x^2 + x + 20$

For a non homogeneous part $- x^2 + x + 20$ , we assume the particular solution is

$y_p(x) = Ax^2 + Bx + C$

2 ) Given that

$d^2y/dx^2 + 6dy/dx + 8y = e^{2x}$

For a non homogeneous part $e^{2x}$ , we assume the particular solution is

$y_p(x) = Ae^{2x}$

3 ) Given that

y′′ + 4y′ + 20y = −3sin(2x)

For a non homogeneous part −3sin(2x) , we assume the particular solution is

$y_p(x) = Acos(2x)+Bsin(2x)$

4 ) Given that

y′′ − 2y′ − 15y = 3xcos(2x)

For a non homogeneous part 3xcos(2x) , we assume the particular solution is

$y_p(x) = (Ax+B)cos2x+(Cx+D)sin2x$

4 0

2 years ago

(4 tens 3 ones) x 10

adoni [48]

To solve this problem, we must first convert all of the numbers represented by words into their numeral equivalents so we can simplify. To do this, we must remember that the ones place is the first to the left of the decimal point, and the tens place is one place farther to the left. If we have 4 tens and 3 ones, this is the same as 43 (we put the digit 4 in the tens place and the digit 3 in the ones place). If we replace our words with our new numerical value, we get:

43 * 10

To solve this, we just multiply these two numbers together, which gives us:

430

Therefore, your answer is 430.

Hope this helps!

6 0

2 years ago