The human resources manager at a company records the length in hours, of one shift at work, X. He creates the probabity

Mrs.Helio has 2 3/4 acres of farmland. she will plant corn on 1/4 of this land, potatoes on 1/12 of the land, wheat on 5/8 of th

Kazeer [188]

2 3/4 acres is the same as 11/4 acres
To find how much of the land is beans, find how much is already taken up, and the rest is beans.
Farmland-Corn-Potatoes-Wheat=Beans
1-(1/4)-(1/12)-(5/8)=1/24

Corn: (1/4)x(11/4)=11/16 acres
Potatoes: (1/12)x(11/4)=11/48 acres
Wheat: (5/8)x(11/4)=55/32 acres or 1 23/32 acres
<span>Beans: (1/24)x(11/4)=11/96 acres</span>

4 0

2 years ago

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a

Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q = -t/100 + c

$Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c} = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}$

when t = 0, Q = 200 L × 1 g/L = 200 g

$Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}$

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

$2 = 200e^{(-t/100)}\\\frac{2}{200} = e^{(-t/100)}$

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0

2 years ago

The following observations are on stopping distance (ft) of a particular truck at 20 mph under specified experi- mental conditio

zzz [600]

Answer:

see explaination

Step-by-step explanation:

Data : 32.1 , 30.6 , 31.4 , 30.4 , 31.0 , 31.9

Mean X-bar = 31.23

SD = 0.689

a)

Null Hypothesis : Xbar = mu

Alternate Hypothesis : Xbar > mu

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 30 ) /(0.689/sqrt(6))

= 4.372

P-value = ~0

Since P-vale < 0.01 , we will reject null hypothesis.

The data suggest that true average stopping ditance exceeds the maximum.

b)

i) SD = 0.65

mu = 31

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 31) /(0.65/sqrt(6))

= 0.867

P-value = 0.3859 Answer

ii) SD = 0.65

mu = 32

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 32) /(0.65/sqrt(6))

= -2.9

P-value = 0.0037 Answer

c)

i) SD = 0.8

mu = 31

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 31) /(0.8/sqrt(6))

= 0.704

P-value = 0.4814 Answer

ii) SD = 0.8

mu = 32

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 32) /(0.8/sqrt(6))

= -2.357

P-value = 0.0184 Answer

The probabilities obtained in part c are comparatively higher than that of part b.

d)

i) For alpha =0.01

z = (Xbar - mu) / (SD/sqrt(n))

=> -2.32 = (31.23 - 31) /(0.65/sqrt(n))

=> (0.65/sqrt(n)) = (31.23 - 31)/-2.32

=> sqrt(n) = (0.65*(-2.32)) / (31.23 - 31)

=> n = 43 Answer

ii) For beta =0.10

z = (Xbar - mu) / (SD/sqrt(n))

=> -1.28 = (31.23 - 31) /(0.65/sqrt(n))

=> (0.65/sqrt(n)) = (31.23 - 31)/-1.28

=> sqrt(n) = (0.65*(-1.28)) / (31.23 - 31)

=> n = 13 Answer

4 0

2 years ago

If the ratio of the sides of two squares is 3:1, what is the ratio of their perimeters?

Anvisha [2.4K]

Answer:

Ratio of the perimeters =3:1

Step-by-step explanation:

We have given that : Ratio of the sides of two squares is 3:1

To find : Ratio of their perimeters

Solution : Let the length of the sides are 3:1 = 3x : x

Formula of perimeter of square = 4(side)

Using the formula ,

Perimeter of 1 square = 4×3x= 12x

Perimeter of 2 square = 4×x= 4x

Ratio of the perimeter of 1 square and 2 square = 12x : 4x

= 3 : 1

8 0

2 years ago

Ralph bought a computer monitor with an area of 384 square inches. The length of the monitor is six times the quantity of five l

Alenkasestr [34]

Draw the picture and label the
width = w
The length of the monitor is six times the quantity of five less than half its width:
length = 6(w/2-5)
length = 3w-30
Area = (length)*(width)
384=(3w-30)*(w)
384=3w^2-30w
Answer:
the dimensions of the monitor in terms of its width is:
384=3w^2-30w

7 0

2 years ago