49) according to the graph shown, if this economy were

49) according to the graph shown, if this economy were to open to trade, consumers would

Solution

D)

Because the prices in that economy were higher than world price, In open economy what happens is that, The price of goods will eventually be at world price. This is not just the███, █████████ ██████ ██████████████ ███████████████ ████ █████████ ████ █████████ █████ ███████████████ █████ ████████████ ███████████ █████ █████████ ███ ████████ ███████████████ ██████████ █████ ████████████ ██ ██████████████ ███ ███████████████ ███████ █████████████ ██████████████ ██████ ███████████████ ███████████████████████ █████████████

Consider the following. (Give your answers correct to t

Consider the following. (Give your answers correct to two decimal places.)

Solution

a) 1-a=0.83

a=0.17

confidence coefficient is Za/2

=Z0.17/2

=Z0.085

look in the cumulative standard normal table for value 1-0.085 = 0.9150

closest value is 0.9147 for z=1.37

██████████ = ███████

███) ███████=██████████

███=█████

███████ = ████████████

████████ ████ ███████ ████████████████████████ ████████████████ ██████████ ██████████ █████ ███████████ ████████████ = ████████████

███████████ = ██████

Growth of Bacteria THe number N of bacteria present in

Growth of Bacteria: THe number N of bacteria present in a culture at time t (in hours) obeys the law of uninhibited growth N(t) = 1000e^(0.01*t)

a) Determine the number of bateria at t + 0 hours

b) What is the growth rate of the bateria?

c) What is the population after 4 hours?

d) When will the number of bateria reach 1700?

e) When will the number of bateria double?

Solution

<a> At t+0 hours, the number of bacteria = 1000e^(0.01t)

<b> Growth rate = 0.01 = 1%

<c> After t = 4 hourss,

N = 1000e^(0.04) = 1040.81

<d> Let at t, bacteria be 1700.

Then , 1700 = 1000e^(0.01t)

=> 1.7 = e^(0.01t) ██ ██████ █████ ███ ███ ███████████ █████████████&#██████;

=&██████; █████(█████) = ██████████

=&██; ███ = ███████████████ ██████████

&█████;██&███; ████ ███ ██, ██████ ██████████████████ ███████████ ██████████,

=&██████; ██████ = ███████████^(█████████)

=&█████; ███ = █^(██████████) █ ███████ ███████████ ███ ██████ █████████ █████████,

████(██) = █████████

█ = █████████ ████████

A sample of 23 of 171 funded projects revealed that 1

A sample of 23 of 171 funded projects revealed that 11 were valued at $18,000 each and 12 were valued at $20,940 each. From the sample data, estimate the total value of the funding for all the projects. (Enter your answers to two decimal places.)

Solution

p1=11/23=0.4782 x1=18000

p2=12/23=0.5217 x2=20940

x=(p1x1+p2████) = ██████████████

██ ██████████ ████████ █████ ██████████████ = ████████████████*██████

=$███████████████████

maximum force Newtons minimum force NewtonsSolutionF m

maximum force =  Newtons
minimum force =  Newtons

Solution

F = mg*mu / ( sin theta + mu cos theta)

value of g = 9.8 m / s^2

find df/dt

df/dt = (- mu *mg(cos t – mu sin t) / ( sin t + cos mu t )^2 = 0

mu = .3

plugging the value we get

df/dt =  (- .3 *mg(cos t – .3 sin t) / ( ████████ ███ + █████ ██ ██ )^█ = █

(█ ████ *██████(██████ ██ &#█████████; ████ ███████ ███) = ███

███

█████████ = █████████ ███

██ = █████^██████ (████████) = ██████████

█████ █ = ██████████

███ = ██(██████) ███ (██████████████+████*████████) = ██(████) █ ██████████████

██ ██=██ ██████ █= ████████

█ ████████████ ██████████ █████ ███████████████████

A computer system uses passwords that are exactly seven

A computer system uses passwords that are exactly seven characters and each character is one of the 26 letters (a-z) or 10 integers (0-9). You maintain a password for this computer system. Let A denote the subset of passwords that begin with a vowel (either a, e, i, o, or u) and let B denote the subset of passwords that end with an even number (either 0, 2, 4, 6, or 8).

a) Suppose a hacker selects a password at random. What is the probability that your password is selected?

(b) Suppose a hacker knows your password is in event A and selects a password at random from this subset. What is the probability that your password is selected?

(c) Suppose a hacker knows your password is in A and B and selects a password at random from this subset. What is the probability that your password is selected?

Solution

(1/5) (1/36)^5 (█████) = ████████████████████████████ = ███████████████████████████

oo ATampT LTE 936 PM imathas.rationalreasoning.net Pre

oo AT&T; LTE 9:36 PM imathas.rationalreasoning.net Preview 0 g(z) i. Function g is exponential 11. g(x)- |8(0.8)% 6.4 5.12 Preview -3 h(x)3.5845.68.75 i. Function h is exponential ii. h(x)= 18(5.6)% 13.671875 Preview License Points possible: 1 Unlimited attempts Score on last attempt: 0.9. Score in gradebook: 0.9 Post this question to forum Submit

Solution

Let h(x)=a*b^x.

ATQ,

a*b=8.75 ……(1)

a/b=5.6 ……(2)

From equation 1 with equation 2, we get

(a*b) * (a/b) = 8████████ * █████

=&███; ██^█ = ████

=&████; █ = ███

███████████████ ███████ █████ █ ███ ███████████████ ██,

██ = ███████████

=&█████; █ = ███████

███,

█(███) = ██*(█████)^███

Please explain your reasoning behind your answers so I

Please explain your reasoning behind your answers so I am able to understand! I will give 5 stars to the person who gives both accurate answers AND provides thorough explanations!(:


Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.5. (Round your answers to four decimal places.)



Solution

z = x – m / (sd / sqrt(n) )

a) P(x>51) = P(z > (51 – 50) / (1.5 / sqrt(10) ) = P(z > 2.11) = 1 – P(z < 2.11) = 1 – ██████████████ = █████████

███) █(█&██████;██████) = ███(██ &██████; (████ &#██████████; █████) ███ (███████ ███ ██████████(████) ) = ███(█ &█████; ████████) = █ &#██████; ███(█ &███; ████████) = █ &#███████████; ████████████████ = ███████████

Using a financial statements model like the one shown b

Using a financial statements model like the one shown below, record the appropriate amounts for the following two events: (In the Cash Flow column, use the initials OA to designate operating activity, IA for investing activity, FA for financing activity, and NC for net change in cash. Amounts to be deducted should be indicated with minus sign. Leave no cells blank – be certain to enter \”0\” wherever required. Omit the \”$\” sign in your response.) January 1, 2010, issue of the note payable. December 31, 2010, payment on the note payable.

Solution

(1) assets( +cash ) = liability ( + notes payable) = FA statement of cash flow

(2) assets(- cash) = liability (- notes payable) = FA   \”<████>

                                                                           = $██ ██████ ███████

In the year 2003, a total of 6.6 million passengers too

In the year 2003, a total of 6.6 million passengers took a cruise vacation. The number of cruise passengers has been growing by 8% each year since 2003. Assume that this growth rate continues.

A. Complete the formula that computes N , the number of cruise passengers (in millions) t years after 2003.

B. How many cruise passengers (in millions) were there in the year 2013?

Solution