Answer:
1/25
Step-by-step explanation:
There are five students. They all have a 1/5 chance. You have to include the chance of all five students, so each student has a 1/25 chance.
Blair & Rosen (B&R) plc is a U.K. based brokerage firm that specializes In building investment portfolios designed to meet the specific needs of its clients who are mostly private investors willing to invest their r savings in stocks and shares. One client who contacted B&R recently has a maximum of $500,000 to invest. The company`s investment advisor has decided to recommend the portfolio consisting of two investment funds: An internet fund where the companies are all active in internet businesses of one kind or another and the blue-chip fund which is more conservative and traditional. The internet fund has a projected annual return over the next few years of 12 %, while the blue-chip fund has a projected annual return of 9%. The investment advisor has decided that at most, $350,000 of the client`s funds should be invested in the internet fund. B&R services include risk rating for each investment alternative. The internet fund which is more risky of the two alternatives has a risk rating of 6 for every thousand dollar invested while the blue-chip fund has a risk rating of 4 per thousand dollar invested. So, for example, if $10000 is invested in each of the two investments funds, B&R risk rating for the portfolio would be 6(10) + 4(10)= 100. Finally B&R has developed a questionnaire to measure each client`s risk tolerance. Based on the responses, each client is classified as conservative, moderate or aggressive investor. The questionnaire results have classified the current client as a moderate investor. B&R recommends that a client who`s a moderate investor limit his or her portfolio to a maximum risk rating of 240. You have been asked to help the B&R investment advisor. What is the recommended investment portfolio for this client? What is the annual return for the portfolio? A second client , also with $500,000 has been classified as aggressive. B&R recommends that the maximum portfolio risk rating for an aggressive investor is 320. What is the recommended investment portfolio for this aggressive investor
Answer:
Blair & Rosen (B&R) Plc.
Recommendation for moderate investor:
Internet fund = 96/240 * $500,000 = $200,000
Blue-chip fund = 144/240 * $500,000 = $300,000
Annual return for the portfolio:
Internet fund = $200,000 * 12% = $24,000
Blue-chip fund = $300,000 * 9% = $27,000
Total portfolio returns = $51,000
Annual returns of portfolio = $51,000/$500,000 * 100 = 10.2%
Recommendation for aggressive investor:
Internet fund = 192/320 * $500,000 = $300,000
Blue-chip fund = 128/320 * $500,000 = $200,000
Step-by-step explanation:
a) Data and Calculations:
Maximum investible savings = $500,000
Projected annual return of the internet fund = 12%
Projected annual return of the blue-chip fund = 9%
Maximum determined amount to invest in the internet fund = $350,000
Risk rating for the internet fund = 6/1,000
Risk rating for the blue-chip fund = 4/1,000
Maximum risk rating for a moderate investor = 240
Maximum risk rating for an aggressive investor = 320
Recommendation for moderate investor:
Internet fund = 96/240 * $500,000 = $200,000
Blue-chip fund = 144/240 * $500,000 = $300,000
Annual return for the portfolio:
Internet fund = $200,000 * 12% = $24,000
Blue-chip fund = $300,000 * 9% = $27,000
Total returns = $51,000
Annual returns of portfolio = $51,000/$500,000 * 100 = 10.2%
Recommendation for aggressive investor:
Internet fund = 192/320 * $500,000 = $300,000
Blue-chip fund = 128/320 * $500,000 = $200,000
Simplify the expression. 8x^-10 y^'6 -2x^2y^-8 Write your answer without negative exponents.
Answer:
[tex]8x^{-10}y^6 - 2x^2y^{-8} = \frac{8y^{14} - 2x^{12}}{x^{10}y^8}[/tex]
Step-by-step explanation:
Given
[tex]8x^{-10}y^6 - 2x^2y^{-8}[/tex]
Required
Simplify
Rewrite as:
[tex]8x^{-10}y^6 - 2x^2y^{-8} = \frac{8y^6}{x^{10}} - \frac{2x^2}{y^8}[/tex]
Take LCM
[tex]8x^{-10}y^6 - 2x^2y^{-8} = \frac{8y^6*y^8 - 2x^2 * x^{10}}{x^{10}y^8}[/tex]
Apply law of indices
[tex]8x^{-10}y^6 - 2x^2y^{-8} = \frac{8y^{14} - 2x^{12}}{x^{10}y^8}[/tex]
PLEASE HELP please I need this done now
The total cost of a truck rental, y, for x days, can be modeled by y = 35x + 25.
What is the rate of change for this function?
Answers
A- 35$
B-25$
C-60$
D-10$
Answer:
35
Step-by-step explanation:
y = 35x+23 is in the form
y = mx+b where m is the slope and b is the y intercept
The slope can also be called the rate of change
35 is the slope
Verify that the indicated family of functions is a solution of the given differential equation. Assume an appropriate interval I of definition for each solution.
d^2y/ dx^2 − 6 dy/dx + 9y = 0; y = c1e3x + c2xe3x When y = c1e3x + c2xe3x,
y'' - 6y' + 9y = 0
If y = C₁ exp(3x) + C₂ x exp(3x), then
y' = 3C₁ exp(3x) + C₂ (exp(3x) + 3x exp(3x))
y'' = 9C₁ exp(3x) + C₂ (6 exp(3x) + 9x exp(3x))
Substituting these into the DE gives
(9C₁ exp(3x) + C₂ (6 exp(3x) + 9x exp(3x)))
… … … - 6 (3C₁ exp(3x) + C₂ (exp(3x) + 3x exp(3x)))
… … … + 9 (C₁ exp(3x) + C₂ x exp(3x))
= 9C₁ exp(3x) + 6C₂ exp(3x) + 9C₂ x exp(3x))
… … … - 18C₁ exp(3x) - 6C₂ (exp(3x) - 18x exp(3x))
… … … + 9C₁ exp(3x) + 9C₂ x exp(3x)
= 0
so the provided solution does satisfy the DE.
Two workers finished a job in 12 days. How long would it take each worker to do the job by himself if one of the workers needs 10 more days to finish the job than the other worker
Two workers finished a job in 7.5 days.
How long would it take each worker to do the job by himself if one of the workers needs 8 more days to finish the job than the other worker?
let t = time required by one worker to complete the job alone
then
(t+8) = time required by the other worker (shirker)
let the completed job = 1
A typical shared work equation
7.5%2Ft + 7.5%2F%28%28t%2B8%29%29 = 1
multiply by t(t+8), cancel the denominators, and you have
7.5(t+8) + 7.5t = t(t+8)
7.5t + 60 + 7.5t = t^2 + 8t
15t + 60 = t^2 + 8t
form a quadratic equation on the right
0 = t^2 + 8t - 15t - 60
t^2 - 7t - 60 = 0
Factor easily to
(t-12) (t+5) = 0
the positive solution is all we want here
t = 12 days, the first guy working alone
then
the shirker would struggle thru the job in 20 days.
Answer:7 + 17 = 24÷2 (since there are 2 workers) =12. Also, ½(7) + ½17 = 3.5 + 8.5 = 12. So, we know that the faster worker will take 7 days and the slower worker will take 17 days. Hope this helps! jul15
Step-by-step explanation:
Please help!!! Find the domain of the function y = 2 cot(5∕8x).
A) All real numbers except odd integer multiples of 8π∕5
B) All real numbers except 0 and integer multiples of 8π∕5
C) All real numbers except 0 and integer multiples of 4π∕5
D) All real numbers except odd integer multiples of 4π∕5
Answer:
B) All real numbers except 0 and integer multiples of 8π∕5
Step-by-step explanation:
Cotangent function:
The cotangent function is given by:
[tex]y = \cot{ax} = \frac{\cos{ax}}{\sin{ax}}[/tex]
Domain:
All real values except those at which:
[tex]\sin{ax} = 0[/tex]
The sine is 0 for 0 and all integer multiples of [tex]\frac{1}{a}[/tex]
In this question:
[tex]a = \frac{5}{8}[/tex], so the values outside the domain are 0 and the integer multiples of [tex]\frac{8}{5}[/tex]. Then the correct answer is given by option b.
Identify the domain of the function shown in the graph.
A. -5
B. x> 0
C. 0
D. x is all real numbers.
The combined value of the ages of Mary, Kate and Tom is 26 years. What will be their age in total after 2 years?
Answer:
32
Step-by-step explanation:
they will each age two years, 3x2 is 6, add 6 to 26
Answer:
32
Step-by-step explanation:
they will each age two years, 3x2 is 6, add 6 to 26
What the distance between -6,2 -6,-15
Answer:
The answer is 17
Step-by-step explanation:
-15-2= -17
What is the percent increase from 250 to 900?
1. Write the percent change formula for an increase.
Percent Increase =
Amount of Increase
Original Amount
2. Substitute the known quantities for the amount of the increase and the original amount.
Percent Increase =
900 − 250
250
3. Subtract.
Percent Increase =
650
250
Answer:
260% is the correct answer
Step-by-step explanation:
i hope i helped
X = The set of months in a year?
there are 12 set of months in a year
The radius of a sphere is increasing at a rate of 3 mm/s. How fast is the volume increasing when the diameter is 60 mm
Answer:
The volume is increasing at a rate of 33929.3 cubic millimeters per second.
Step-by-step explanation:
Volume of a sphere:
The volume of a sphere of radius r is given by:
[tex]V = \frac{4\pi r^3}{3}[/tex]
In this question:
We have to derivate V and r implicitly in function of time, so:
[tex]\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}[/tex]
The radius of a sphere is increasing at a rate of 3 mm/s.
This means that [tex]\frac{dr}{dt} = 3[/tex]
How fast is the volume increasing when the diameter is 60 mm?
Radius is half the diameter, so [tex]r = 30[/tex]. We have to find [tex]\frac{dV}{dt}[/tex]. So
[tex]\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}[/tex]
[tex]\frac{dV}{dt} = 4\pi (30)^2(3) = 33929.3[/tex]
The volume is increasing at a rate of 33929.3 cubic millimeters per second.
A street light is mounted at the top of a 15-ft-tall pole. A man 6 feet tall walks away from the pole with a speed of 5 ft/s along a straight path. How fast (in ft/s) is the tip of his shadow moving when he is 45 feet from the pole
Answer:
25/3 ft/s
Step-by-step explanation:
Height of pole , h=15 ft
Height of man, h'=6 ft
Let BD=x
BE=y
DE=BE-BD=y-x
All right triangles are similar
When two triangles are similar then the ratio of their corresponding sides are equal.
Therefore,
[tex]\frac{AB}{CD}=\frac{BE}{DE}[/tex]
[tex]\frac{15}{6}=\frac{y}{y-x}[/tex]
[tex]\frac{5}{2}=\frac{y}{y-x}[/tex]
[tex]5y-5x=2y[/tex]
[tex]5y-2y=5x[/tex]
[tex]3y=5x[/tex]
[tex]y=\frac{5}{3}x[/tex]
Differentiate w.r.t t
[tex]\frac{dy}{dt}=\frac{5}{3}\frac{dx}{dt}[/tex]
We have dx/dt=5ft/s
Using the value
[tex]\frac{dy}{dt}=\frac{5}{3}(5)=\frac{25}{3}ft/s[/tex]
Hence, the tip of his shadow moving with a speed 25/3 ft/s when he is 45 feet from the pole.
Answer:
The tip pf the shadow is moving with speed 25/3 ft/s.
Step-by-step explanation:
height of pole = 15 ft
height of man = 6 ft
x = 45 ft
According to the diagram, dx/dt = 5 ft/s.
Now
[tex]\frac{y-x}{y}=\frac{6}{15}\\\\15 y - 15 x = 6 y \\\\y = \frac{5}{3} x\\\\\frac{dy}{dt} = \frac{5}{3}\frac{dx}{dt}\\\\\frac{dy}{dt}=\frac{5}{3}\times 5 =\frac{25}{3} ft/s[/tex]
Log6^(4x-5)=Log6^(2x+1)
Answer:
[tex]x = 3[/tex]
Step-by-step explanation:
Given
[tex]\log6^{(4x-5)} =\log6^{(2x+1)}[/tex]
Required
Solve for x
We have:
[tex]\log6^{(4x-5)} =\log6^{(2x+1)}[/tex]
Remove log6 from both sides
[tex](4x-5) = (2x+1)[/tex]
Collect like terms
[tex]4x - 2x = 5 + 1[/tex]
[tex]2x = 6[/tex]
Divide by 2
[tex]x = 3[/tex]
according to the fundemental theorem of algebra, how many roots exist for the polynomial function? f(x) = (x^3-3x+1)^2
Answer:
6
Step-by-step explanation:
First, we can expand the function to get its expanded form and to figure out what degree it is. For a polynomial function with one variable, the degree is the largest exponent value (once fully expanded/simplified) of the entire function that is connected to a variable. For example, x²+1 has a degree of 2, as 2 is the largest exponent value connected to a variable. Similarly, x³+2^5 has a degree of 2 as 5 is not an exponent value connected to a variable.
Expanding, we get
(x³-3x+1)² = (x³-3x+1)(x³-3x+1)
= x^6 - 3x^4 +x³ - 3x^4 +9x²-3x + x³-3x+1
= x^6 - 6x^4 + 2x³ +9x²-6x + 1
In this function, the largest exponential value connected to the variable, x, is 6. Therefore, this is to the 6th degree. The fundamental theorem of algebra states that a polynomial of degree n has n roots, and as this is of degree 6, this has 6 roots
Determine la razón de la siguiente progresión geométrica: 81,27,9,3,1,....
Answer:
BẠN BỊ ĐIÊN À
Step-by-step explanation:
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please help, it’s urgent !!!
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A
B
C
for more explanation please don't hesitate to just respond
Please help due tomorrow
Answer:
10x8=80 that would be the area for the picture 14x11=154 for the boards area
HELP ANYONE PLZZZ ?
1sr.
z(x)=x+1
If you input a 3 into z(x), what do you get for the output?
2nd.
n(x)=2/x
n(x) will give you an output for any number you use as an input except which of the following?
A. 0
B .3
C. 5
D. Trick question- you can get an output for every number you use as an input .
9514 1404 393
Answer:
4A. 0Step-by-step explanation:
1. Input 3 for x and do the arithmetic.
z(x) = x+1
z(3) = 3+1 = 4 . . . . . the output is 4
__
2. The expression for n(x) has x in the denominator. The expression will be undefined when the denominator is zero, so x=0 cannot be used.
Which of the following statements are correct? Select ALL that apply!
Select one or more:
O a. -1.430 = -1.43
O b. 2.36 < 2.362
O c.-1.142 < -1.241
O d.-2.33 > -2.29
O e. 2.575 < 2.59
O f. -2.25 -2.46
Given: F = {(0, 1), (2, 4), (4, 6), (6, 8)} and G = {(2, 5), (4, 7), (5, 8), (6, 9), (7, 5)}
(F + G) (2) =
4
5
9
9514 1404 393
Answer:
9
Step-by-step explanation:
The ordered pair (2, 4) in the relation for function F tells you F(2) = 4.
The ordered pair (2, 5) in the relation for function G tells you G(2) = 5.
Then the sum is ...
(F+G)(2) = F(2) +G(2) = 4 +5
(F+G)(2) = 9
Matthew actually drew the 10 of hearts and the 3 of clubs. If he keeps those to one side and selects two more from the pack, what is the chance that he'll get a pair of 10s this time? As before, give your answer in its simplest form. 2nd Attempt: Probability of getting a pair of 10s
What is the endpoint of a line segment if the midpoint M( – 3, 4) and the other endpoint is E(7, – 2)?
Answers
(– 13, 10)
(10, – 13)
(– 1, 2)
(2, – 1)
9514 1404 393
Answer:
(-13, 10)
Step-by-step explanation:
If M is the midpoint of segment DE, then ...
D = 2M -E
D = 2(-3, 4) -(7, -2) = (2(-3)-7, 2(4)+2) = (-13, 10)
The other end point is (-13, 10).
Translate the triangle. Then enter the new coordinates. A(-3, 4) A'([?], [?]) B'([ ], [ ] C([],[]) B(0, 1) C(-4,1)
or
Answer:
The new coordinates are [tex]A'(x,y) = (3, 0)[/tex], [tex]B'(x,y) = (6, -3)[/tex] and [tex]C'(x,y) = (2, -3)[/tex].
Step-by-step explanation:
Vectorially speaking, the translation of a point can be defined by the following expression:
[tex]V'(x,y) = V(x,y) + T(x,y)[/tex] (1)
Where:
[tex]V(x,y)[/tex] - Original point.
[tex]V'(x,y)[/tex] - Translated point.
[tex]T(x,y)[/tex] - Translation vector.
If we know that [tex]A(x,y) = (-3,4)[/tex], [tex]B(x,y) = (0,1)[/tex], [tex]C(x,y) = (-4,1)[/tex] and [tex]T(x,y) = (6, -4)[/tex], then the resulting points are:
[tex]A'(x,y) = (-3, 4) + (6, -4)[/tex]
[tex]A'(x,y) = (3, 0)[/tex]
[tex]B'(x,y) = (0,1) + (6, -4)[/tex]
[tex]B'(x,y) = (6, -3)[/tex]
[tex]C'(x,y) = (-4, 1) + (6, -4)[/tex]
[tex]C'(x,y) = (2, -3)[/tex]
The new coordinates are [tex]A'(x,y) = (3, 0)[/tex], [tex]B'(x,y) = (6, -3)[/tex] and [tex]C'(x,y) = (2, -3)[/tex].
A cable that weighs 6 lb/ft is used to lift 600 lb of coal up a mine shaft 500 ft deep. Find the work done. Show how to approximate the required work by a Riemann sum. (Let x be the distance in feet below the top of the shaft. Enter xi* as xi.)
Answer:
A cable that weighs 6 lb/ft is used to lift 600 lb of coal up a mine shaft 500 ft deep. Find the work done. Show how to approximate the required work by a Riemann sum.
Step-by-step explanation:
Simplify to the extent possible
(logx16)(log2x)
Answer:
[tex]{ \tt{ = ( log_{x}16)( log_{2}x) }}[/tex]
Change base x to base 2:
[tex]{ \tt{ = (\frac{ log_{2}16}{ log_{2}x } )( log_{2}x)}} \\ \\ { \tt{ = log_{2}(16) }} \\ = { \tt{ log_{2}(2) }} {}^{4} \\ = { \tt{4 log_{2}(2) }} \\ = { \tt{4}}[/tex]
How many tens are in 6 hundreds
Answer:
60
Step-by-step explanation:
10 x 6 = 60
Hope this helped! :)
HELP ME WITH THIS MATHS QUESTION
PICTURE IS ATTACHED
Answer:
In picture.
Step-by-step explanation:
To do this answer, you need to count the boxes up to the mirror line. This will give us the exact place to draw the triangle.
The picture below is the answer.
Evaluate −3w − 6p for w=2 and p = −7
-3w-6p when w=2 and p=-7
-3(2)-6(-7)
= -6 + 42
= 36
Answer:
48
Step-by-step explanation:
-3w-6p when w=2 and p--7
you want to plug in the numbers to their letters
-3(2)-6(-7)
then you want to times the numbers.
-6-42
=48
Please help me out really need it
Answer:
[tex]{ \tt{hypotenuse = { \boxed{5}}}} \\ { \tt{opposite = { \boxed{3}}}} \\ { \tt{adjacent = { \boxed{4}}}} \\ \\ { \tt{ \sin \angle R = \frac{{ \boxed{3}}}{{ \boxed{5}}} }} \\ \\ { \tt{ \cos \angle R = \frac{{ \boxed{4}}}{{ \boxed{5}}} }} \\ \\ { \tt{ \tan \angle R = \frac{ \boxed{3}}{{ \boxed{4}}} }}[/tex]
