Answer:
(a) average calls = 5
(b) probability that there is exactly one call in 6 consecutive monts = 0.038
Step-by-step explanation:
Let event of a customer requiring help in a particular month = H
and event of a customer not requiring help in a particular month = ~H
Given
p= 0.01, therefore
Number of households, n = 500.
Binomial distribution:
x = number of households requiring help in a particular month
P(x,n,p) = C(x,n)*p^x*(1-p)^(n-x)
where
C(x,n) = n!/(x!(n-x)!) is the the number of combinations of x objects out of n
(a) Average number of households requiring help = np = 500*0.01 = 5
(b)
Probability that there are no calls requiring help in a particular month
P(0), q= C(0,n)*p^0(1-p)^(n-0)
= 1*1*0.99^500
= 0.006570483
Applying binomial probability over six months,
q = 0.006570483
n = 6
x = 1
P(x,n,q)
= C(x,n)*q^x*(1-q)^(n-x)
= 6!/(1!*5!) * 0.006570483^1 * (1-0.006570483)^5
= 0.038145
Therefore the probability that in 6 consecutive months there is exactly one month that no customer has called for help = 0.038
please helpppppppppppp
Answer:
3
Step-by-step explanation:
4.5x 2 = 9
[tex]\sqrt{9}[/tex]= 3
Hellllllllllp! Someone help
Answer:
4. (2, 3)
5. (0, 1)
7. (-1, 2)
Step-by-step explanation:
I hope this helps! Have a nice dayy! :)
You're very close. Choices D and E are two of the three answers. The third answer is choice G (-1,2)
In short, the 3 answers are Choices D, E and GIf you plug the coordinates of the points into the inequality, you should get a true statement.
For instance, let's try the coordinates of choice G
[tex]-2x + 3y\ge 3\\\\-2(-1) + 3(2)\ge 3\\\\2 + 6\ge 3\\\\8\ge 3\\\\[/tex]
Which is true since 8 is indeed greater than 3. That verifies point G is a solution point. A similar story happens with points D and E as well.
----------
If you tried something like choice A, then,
[tex]-2x + 3y\ge 3\\\\-2*(2) + 3(-3)\ge 3\\\\-4 - 9\ge 3\\\\-13\ge 3\\\\[/tex]
Which is false because -13 is not greater than 3, and -13 is not equal to 3 either. So choice A is a non-answer. You should find that choices B, C, and F are also non-answers.
----------
The graph is below. Notice how points D, E and G are either in the blue shaded region or on the boundary. The boundary line is solid (due to the "or equal to" as part of the inequality sign), so points on the solid boundary line are part of the solution set. The graph is a quick way to visually confirm the answers. I used GeoGebra to make the graph.
So that's why the 3 answers are D, E and G.
A college admissions officer takes a simple random sample of 90 entering freshman and computes their mean mathematics sat score to be 436. assume the population standard deviation is σ = 101. Based on a 99% confidence interval for the mean mathematics SAT score, is it likely that the mean mathematics SAT score for entering freshmen class is greater than 460?
Answer:
460 is part of the confidence interval, which means that we cannot say that there is significant evidence that the mean mathematics SAT score for entering freshmen class is greater than 460
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1 - 0.99}{2} = 0.005[/tex]
Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.005 = 0.995[/tex], so Z = 2.575.
Now, find the margin of error M as such
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 2.575\frac{101}{\sqrt{90}} = 27.4[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 436 - 27.4 = 408.6.
The upper end of the interval is the sample mean added to M. So it is 436 + 27.4 = 463.4.
460 is part of the confidence interval, which means that we cannot say that there is significant evidence that the mean mathematics SAT score for entering freshmen class is greater than 460
How would I solve this?
Answer:
20°
Step-by-step explanation:
perpendicular from the center on a chord of a circle always bisects the chord.
AR=BR
∴m arcAC=m arc BC=20°
Find the simple interest on a loan of $34,500 at 6.9% interest for 11 months
Give your answer to the nearest cent
Answer:
$2182.13
Step-by-step explanation:
Simple interest (I) is calculated as
I = [tex]\frac{PRT}{100}[/tex] ( P is principal, R is rate of interest, T is time in years )
Here P = $34,500 , R = 6.9 and T = [tex]\frac{11}{12}[/tex] , then
I = [tex]\frac{34500(6.9)(\frac{11}{12}) }{100}[/tex]
= 345 × 6.9 × [tex]\frac{11}{12}[/tex]
= $2182.13
The actual length is 1.53 km
scale 10 cm to 1 km. find the length on drawing.
Answer: 15.3 cm
Step-by-step explanation:
The actual length of the line in question is 1.53km.
The scale is 10 cm to 1 km.
You can therefore use direct proportion to solve this.
Assume the length on the drawing is x:
10cm : 1km
x cm : 1.53 km
Cross multiply:
x = 15.3 cm
There are 38 Legs in a group of goat and hens. How many goats and hens are there?
1) 13 Goats , 3 hens
2) 11 goats ,3hens
3) 7 goats,3 hens
4) 8 goats,3 hens
Answer:
4)8 goats, 3 hens.
Step-by-step explanation:
8*4=32
3*2=6
32+6=38
The Molokai Nut Company (MNC) makes four different products from macadamia nuts grown in the Hawaiian Islands: chocolate-coated whole nuts (Whole), chocolate-coated nut clusters (Cluster), chocolate-coated nut crunch bars (Crunch), and plain roasted nuts (Roasted). The company is barely able to keep up with the increasing demand for these products. However, increasing raw material prices and foreign competition are forcing MNC to watch its margins to ensure it is operating in the most efficient manner possible. To meet marketing demands for the coming week, MNC needs to produce at least 1,000 pounds of the Whole product, between 400 and 500 pounds of the Cluster product, no more than 150 pounds of the Crunch product, and no more than 200 pounds of the Roasted product. Each pound of the Whole, Cluster, Crunch, and Roasted product contains, respectively, 60%, 40%, 20%, and 100% macadamia nuts with the remaining weight made up of chocolate coating. The company has 1,100 pounds of nuts and 800 pounds of chocolate available for use in the next week. The various products are made using four different machines that hull the nuts, roast the nuts, coat the nuts in chocolate (if needed), and package the products. The following table summarizes the time required by each product on each machine. Each machine has 60 hours of time available in the coming week.
Machine Minutes Required per Pound
Whole Cluster Crunch Roasted
Hulling 1 1 1 1
Roasting 2 1.5 1 1.75
Coating 1 0.7 0.2 0
Packaging 2.5 1.6 1.25 1
The selling price and variable cost associated with each pound of product is summarized in the following table:
Per Pound Revenue and Costs
Whole Cluster Crunch Roasted
Selling Price $5.00 $4.00 $3.20 $4.50
Variable Cost $3.15 $2.60 $2.16 $3.10
Required:
a. Formulate an LP model for this problem.
b. Create a spreadsheet model for this problem, and solve it using Solver.
c. What is the optimal solution?
Answer:
Profit = Selling price - Variable cost
Formulas:
F2 =SUMPRODUCT(B2:E2,$B$16:$E$16) copy to F2:F12, F14
Optimal solution: The company should produce the following quantities (in pounds) of the four varieties of nuts.
Whole = 1000
Cluster = 500
Crunch = 80
Roasted = 200
Total profit = $ 2913.20
Step-by-step explanation:
4. The electrical resistance of a wire varies inversely as the square of its radius. If the resistance is 0.80 ohm when the radius is 0.4cm. Find the resistance whom the radius is 0.7cm.
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Answer:
0.261 ohm
Step-by-step explanation:
If the radius increases by a factor of 0.7/0.4= 7/4, the square of this factor is (7/4)^2 = 49/16. The inverse of this square is 16/49, which is the factor by which the resistance changed.
The resistance of the larger wire is ...
(16/49)(0.80 ohm) ≈ 0.261 ohm
A set of numbers is shown below:
{0, 0.6, 2, 4, 6}
Which of the following shows all the numbers from the set that make the inequality 2x + 3 ≥ 7 true?
{4, 6}
{0, 0.6, 2}
{0, 0.6}
{2, 4, 6}
Answer:
{2,4,6}
Step-by-step explanation:
Hope it helps you
In a class of 40 students, 2/5
of the total number of students like to study English and 3/5
of the
total students like to study Mathematics.
a) How many students like to study English?
b)How many students like to study Mathematics?
find the area of the shape
Answer:
The area is 91 cm²
Step-by-step explanation:
The shape is a kite.
area of a kite = ½(p*q)
Where, p and q are the diagonals of the kite.
p = 13 cm
q = 7 + 7 = 14 cm
The area of the kite = ½(13 * 14)
= ½(182)
Area = 91 cm²
Find an equation for the line parallel to 3x-5y=2 with y-intercept (0,1/5). Write the answer in slope-intercept form.
Show that the set of nonsingluar 2 by 2 matrices is not a vector space. Show also that the set of singular 2 by 2 matrices is not a vector space.
Answer:
a) 2 nonsingular 2 by 2 matrices are not closed when added together hence it is not a vector space( i.e. their sum = singular and not nonsingular )
b) 2 singular 2 by 2 matrices is not closed under addition, hence they are not a vector space. ( i.e. their sum = nonsingular )
Step-by-step explanation:
a) Prove that nonsingular 2 by 2 matrices is not a vector space
2 nonsingular matrices are not closed when added together hence it is not a vector space ( i.e. their sum = singular and not nonsingular )
vector A = [tex]\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right][/tex] + vector B = [tex]\left[\begin{array}{ccc}0&1\\1&0\\\end{array}\right][/tex] = [tex]\left[\begin{array}{ccc}1&1\\1&1\\\end{array}\right][/tex] ( singular vector )
b) Prove that singular 2 by 2 matrices is not a vector space
2 singular 2 by 2 matrices is not closed under addition, hence they are not a vector space. ( i.e. their sum = nonsingular )
Vector C = [tex]\left[\begin{array}{ccc}1&0\\0&0\\\end{array}\right][/tex] + vector D = [tex]\left[\begin{array}{ccc}0&0\\0&1\\\end{array}\right][/tex] = [tex]\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right][/tex] ( nonsingular vector )
find the simultaneous equation for:
4x+3y=7
2x+5y=7
Answer:
x = 1 and y = 1
Step-by-step explanation:
You can eliminate x first to get the value of y which is 1 and then replace it in one of the equations.
In which number is the value of the 9 ten
times the value of the number nine in the number 920
Answer:
9 hundreds that is the answer
An expression is shown below:
3(m + 5 + 9m)
Part A: Write two expressions that are equivalent to the given expression. (3 points)
Part B: Show that one of your expressions in Part A is equivalent to the given expression using algebraic properties. Explain which properties you used. (4 points)
Part C: Show that your other expression from Part A is equivalent to the given expression by substituting a number for m. (3 points)
The answers are as follows part A = 30m+15 part B =3m+15+27m
and partC = 30m+15
What is an expression?Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition,substraction, multiplication and division.
Part A:- Two expressions that are equivalent to the given expressions are:-
3m + 15 + 27m
30m + 15
Part B: Show that one of your expressions in Part A is equivalent to the given expression using algebraic properties.
3 ( m + 5 + 9m )
Open the bracket by multiplying 3 by what is in the bracket
3m + 15 + 27m
Part C: Show that your other expression from Part A is equivalent to the given expression by substituting a number for m.
3 ( m + 5 + 9m )
Open the bracket by multiplying 3 by what is in the bracket
3m + 15 + 27m
Collect like terms together
3m + 27m + 15
= 30m + 15
To know more about Expression follow
https://brainly.com/question/723406
#SPJ2
The half life of Co-60 is 5.20 years. How many milligrams of a 1.00mg sample remains after 6.55 years
Answer:
For cobalt-60, which has a half-life of 5.27 years, 50% remains after 5.27 years (one half-life), 25% remains after 10.54 years (two half-lives), 12.5% remains after 15.81 years (three half-lives), and so on.
Step-by-step explanation:
You can figue out the rest.
solution - 12. The digit in the tens place of a two-digit number is three times that in the units place. If the digits are reversed, the new number will be 36 less than the original number. Find the original number. Check your solution.
Answer:
62
Step-by-step explanation:
t=digit in the tens place
u=digit in the units place
t=3*u
original number=t*10+u
number with reversed digits=u*10+t
u*10+t=t*10+u-36
u*10+(3*u)=(3*u)*10+u-36
10u+3u-30u-u=-36
-18u=-36
u=2
t=3*u=6
Original number = 62
Check the solution:
6=3*2 ok
26=62-36 ok
(Right angle) Trigonometry
please help!
Answer:
A = 41.4°
Step-by-step explanation:
Reference angle = A
Length of Adjacent side = 6
Length of Hypotenuse = 8
Apply the trigonometric function, CAH.
Cos A = Adj/Hyp
Substitute
[tex] Cos(A) = \frac{6}{8} [/tex]
[tex] A = Cos^{-1}(\frac{6}{8}) [/tex]
A = 41.4° (approximated to the nearest tenth)
Rolling a fair Eight-sided die produces a uniformly distributed set of numbers between 1 and 8 with a mean of 4.5 and a standard deviation of 2.291. Assume that n eight-sided dice are rolled many times and the mean of the n outcomes is computed each time.
Required:
a. Find the mean and the standard deviation of the resulting distribution of sample means for n=36.
b. The mean of the resulting distribution of the sample means is:________
Answer:
a. The mean is 4.5 and the standard deviation is 0.3818.
b. 4.5
Step-by-step explanation:
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Mean of 4.5 and a standard deviation of 2.291.
This means that [tex]\mu = 4.5, \sigma = 2.291[/tex]
a. Find the mean and the standard deviation of the resulting distribution of sample means for n=36.
By the Central Limit Theorem, the mean is 4.5 and the standard deviation is [tex]s = \frac{2.291}{\sqrt{36}} = 0.3818[/tex]
The mean is 4.5 and the standard deviation is 0.3818.
b. The mean of the resulting distribution of the sample means is:________
By the Central Limit Theorem, 4.5.
Last question! Please show work. Really need to get this done in 1 hour
Answer:
x = 1, y = 2 , z = 3
Step-by-step explanation:
[tex]6x\:+\:2y\:-4z\:=\:-2 \ \ \ \ \ \ \ \ \ -----( 1 ) \:\\\\-3x-4y\:+2z\:=\:-5 \ \ \ \ \ \ ------( 2 ) \:\\\\4x-\:6y\:+3z\:=\:1 \ \ \ \ \ \ \ \ \ \ \ \ ------- ( 3 ) \\\\( 2 ) \times 2=> -6x -8y + 4z = -10 \ \ \ \ \ \ \ -----( 4)[/tex]
Now add ( 1 ) and (4)
[tex]6x +2y -4z = - 2\\\\-6x-8y - 4z = -10\\\\=>0x -6y + 0 = - 12\\\\-6y = -12[/tex]
y = 2
Now multiply ( 1 ) by 2 and (3) by 3
[tex](1) \times 2 => 12x + 4y -8z = -4\\\\(3) \times 3 => 12x -18y +9z = 3\\\\Subtract \ the \ equation : ( 1) - ( 3) => 0x +22y -17z = -7[/tex] ------ ( 5 )
Substitute y = 2 in ( 5 ) :
[tex]22(2) - 17z = - 7\\\\ 44 - 17z = - 7\\\\ -17z = - 7 - 44\\\\ -17z = -51\\\\[/tex]
z = 3
Substitute z = 3 and y = 2 in ( 1 ) :
[tex]6x + 2y - 4z = - 2\\\\6x + 2( 2) -4( 3) = -2\\\\6x + 4 - 12 = -2\\\\6x - 8 = - 2\\\\6x = - 2 + 8 \\\\6x = 6\\[/tex]
x = 1
How much is six dimes, 8 nickels, and three one-dollar bills? *
[tex]\huge\textsf{Hey there!}[/tex]
[tex]\large\textsf{Guide}\downarrow[/tex]
[tex]\large\textsf{Penny: 1 cent} \\\large\textsf{Nickel: 5 cents}\\\large\textsf{Dime: 10 cents}\\\large\textsf{Quarter: 25 cents}\\\large\textsf{Half dollar: 50 cents}\\\large\textsf{1 bill = 1 dollar}[/tex]
[tex]\large\textsf{6 dimes:}\downarrow[/tex]
[tex]\large\textsf{= 10 + 10 + 10 + 10 + 10 + 10}[/tex]
[tex]\large\textsf{= 20 + 20 + 20}[/tex]
[tex]\large\textsf{= 40 + 20}[/tex]
[tex]\large\textsf{= \bf 60}[/tex]
[tex]\large\textsf{6 dimes = \bf 60 cents}[/tex]
[tex]\large\textsf{8 nickels}[/tex]
[tex]\large\textsf{= 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5}[/tex]
[tex]\large\textsf{= 10 + 10 + 10 + 10}[/tex]
[tex]\mathsf{= 20 + 20}[/tex]
[tex]\large\textsf{= \bf 40}[/tex]
[tex]\large\textsf{8 nickels = \bf 40 cents}[/tex]
[tex]\large\textsf{3 one dollar bills}[/tex]
[tex]\large\textsf{= 1 + 1 + 1}[/tex]
[tex]\large\textsf{= 2 + 1 }[/tex]
[tex]\large\textsf{= \bf 3}[/tex]
[tex]\large\textsf{3 one dollar bills = \bf 3 dollars}[/tex]
[tex]\large\textsf{NOW LETS SOLVE FOR YOUR EQUATION}[/tex]
[tex]\large\textsf{3.00 + 0.60 + 0.40}[/tex]
[tex]\large\textsf{= 3.60 + 0.40 }[/tex]
[tex]\large\textsf{= \bf \$4}[/tex]
[tex]\boxed{\boxed{\large\textsf{Therefore, your answer is: \huge \bf \$4}}}\huge\checkmark[/tex]
[tex]\large\textsf{Good luck on your assignment and enjoy your day!}[/tex]
~[tex]\frak{Amphitrite1040:)}[/tex]
Share 30 000 bags of 20kg rice in the ratio 2:3:4:6 among Villages A, B, C and D in that order. (2 marks) How many bags of 20 kg rice will Villages B and C get?
Answer:
2+3+4+6 = 15 parts
Village B gets 3 of 15 parts
3/15 * 30000 = 6000 bags
Village C gets 4 of 15 parts
4/15 * 6000 = 8000 bags
Answer:
each "share" needs "15" (2+3+4+6) "portions/bags"
2000 distributions
A-4000 bags/80000 kg
B-6000 bags/120000 kg
C-8000 bags/1680000 kg
D-12000 bags/240000 kg
Step-by-step explanation:
A certain medicine is given in an amount proportional to a patient's body weight. Suppose a patient weighing 150 pound requires 144 milligrams of medicine. what is the weight of a patient who requires 174.72 milligrams of medicine
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Answer:
182 lbs
Step-by-step explanation:
You can write the proportion for the required patient weight (w) as ...
weight/medicine = w/(174.72 mg) = (150 lb)/(144 mg)
w = (150 lb)(174.72/144) . . . . . multiply by 174.72 mg
w ≈ 182 lb
The patient's weight is 182 pounds.
(a+b)²=hihihihihihihihiihihihi
Answer:
(a+b)²=a²+b²+2ab
Step-by-step explanation:
The square of sum of two terms is equal to the squared plus squared plus times product of and . In mathematics, the plus whole squared algebraic identity is called in three ways. The square of sum of two terms identity.
Find the area for me pls
this figure can be divided into three parts one is rectangle other one is semicircle and the third one is one fourth of the circle .so let's find the area of each figure one by one. For the rectangle 12×8 =96
for semicircle that is on the top it has the radius 6 which is a half of 12
so area of the semicircle is
[tex] \frac{1}{2} \pi \: r {}^{2} \\ \frac{1}{2} \times 3.14 \times 6 {}^{2} \\ 3.14 \times 18 \\ 56.52[/tex]
no that's fine. 80 of the 1/4 of the other Circle
[tex] \frac{1}{4} \pi {}^{2} \\ \frac{1}{4} 3.14 \times 8 {}^{2} \\ 3.14 \times 16 \\ 50.24[/tex]
add all these areas
96+56.52+50.24 =202.76
Find the net change in the value of the function between the given inputs.
h(t) = t2 + 9; from −4 to 7
For this question I am sure the answer is 81% as you divide 45 and 55. However, it is stating my answer is incorrect even though I put 0.81% as well. Did I round wrong or is the answer wrong completely?
Answer:
it says round to the nearest 10th so it wouldn't be 81, it would be 81.8%
A tank contains 1000L of pure water. Brine that contains 0.04kg of salt per liter enters the tank at a rate of 5L/min. Also, brine that contains 0.06kg of salt per liter enters the tank at a rate of 10L/min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15L/min. Answer the following questions. 1. How much salt is in the tank after t minutes
Answer:
s(t) = 160/3 ( 1 - e^(-3t / 200) )
Step-by-step explanation:
volume of pure water in tank = 1000 L
Brine contains 0.04kg of salt/L
Inflow rate of Brine containing 0.04kg of salt/L = 5L/min
Brine containing 0.06 kg of salt/L
Inflow rate of Brine containing 0.06 kg of salt/L = 10L/min
Solution is thoroughly mixed and drains from tank at 15L/min
a) Determine the amount of salt is in the tank after t minutes
rate of salt entering = 0.2 + 0.6 = 0.8 kg/min
rate of salt leaving = s/1000 * 15
amount of salt at time (t) = s(t)
initial condition s( 0 ) = 0
ds/dt = 0.8 - 15s/1000 = 0.8 - 3s/200
200 ds/dt = ( 160 - 3s )
-200/3 In ( 160 - 3s ) = t + c
Given that ; t = 0 , s = 0
c = - 200/3 In ( 160 )
∴ -200/3 In ( 160 - 3s ) = t - 200/3 In ( 160 )
- 200/3 [ In ( 60 - 3s ) - In ( 160 ) ] = t
therefore:
In ( 160 - 3s / 160 ) = -3t/200
= ( 160 - 3s / 160 ) = e ^ (-3t/200 )
Hence amount of salt in tank after t minutes
s(t) = 160/3 ( 1 - e^(-3t / 200) )