Answer:
Range:
UCL = 4.73
LCL = 18.08
MEAN :
UCL = 27.115
LCL = 31.219
Step-by-step explanation:
Given the data:
The mean and range of each sample :
Sample __ Thread wear __ xbar __ R
1 ___31 __ 42 ___ 28 ____ 33.67 _14
2___ 26 _ 18 ____35____ 26.33 _17
3___25 __30 ___ 34____29.67 _ 9
4 __ 17 __ 25 ___ 21 _____ 21 ___ 8
5 __ 38 _ 29 ___ 35 _____ 34 __ 9
6 __ 41 __42 ___36 _____39.67_ 6
7 __ 21 __ 17 ___29 _____22.33 _12
8 __ 32 __26___28 ____ 28.67 _ 6
9 __ 41 __ 34 __ 33 ______ 36 __8
10__29___17___30 _____25.33_ 13
11 __26 __ 31 __ 40 _____32.33_ 14
12__23 __ 19 __ 45 _____12.33 __6
13 __17 __ 24 __ 32_____24.33__15
14 __43__ 35___17_____ 31.67 _ 26
15__18 ___25__ 29_____ 24 ___ 11
16__30___42___31 ____34.33__ 12
17__28___36 __ 32____ 32 ____8
18__40 __ 29 __ 31 ____33.33 __ 11
19__18 ___29__ 28____ 25 ____11
20_ 22 __ 34 __ 26 ___ 27.33 __12
Size per sample, sample size, n = 3
Number of samples, k = 20
We calculate the sample mean and range average :
Sample mean, x-- = Σxbar/n = 29.167
Range average, Rbar = ΣR/n = 11.4
The mean control limit :
x-- ± A2Rbar
From the x chart ;
A2 for n = 20 is A2 = 0.180
29.167 ± 0.180(11.40)
LCL = 29.167 - 0.180(11.40) = 27.115
UCL = 29.167 + 0.180(11.40) = 31.219
The Range control limit :
Rbar(1 ± 3(d3/d2))
From the R-chart :
d2 at n = 20 ; d2 = 3.735
d3 at n = 20 ; d3 = 0.729
LCL = 11.40(1 - 3(0.729/3.735)) = 4.725
UCL = 11.40(1 + 3(0.729/3.735)) = 18.075
When studying radioactive material, a nuclear engineer found that over 365 days,
1,000,000 radioactive atoms decayed to 970,258 radioactive atoms, so 29,742 atoms
decayed during 365 days.
a. Find the mean number of radioactive atoms that decayed in a day.
b. Find the probability that on a given day, 50 radioactive atoms decayed.
a. The mean number of radioactive atoms that decay per day is
(Round to three decimal places as needed.)
Answer:
a) The mean number of radioactive atoms that decay per day is 81.485.
b) 0% probability that on a given day, 50 radioactive atoms decayed.
Step-by-step explanation:
Poisson distribution:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\lambda[/tex] is the mean in the given interval.
a. Find the mean number of radioactive atoms that decayed in a day.
29,742 atoms decayed during 365 days, which means that:
[tex]\lambda = \frac{29742}{365} = 81.485[/tex]
The mean number of radioactive atoms that decay per day is 81.485.
b. Find the probability that on a given day, 50 radioactive atoms decayed.
This is P(X = 50). So
[tex]P(X = 50) = \frac{e^{-81.485}*(81.485)^{50}}{(50)!} = 0[/tex]
0% probability that on a given day, 50 radioactive atoms decayed.
the expression when b=3 and y= -3
5b-y
Answer:
18
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to RightStep-by-step explanation:
Step 1: Define
Identify
b = 3
y = -3
5b - y
Step 2: Evaluate
Substitute in variables: 5(3) - -3Multiply: 15 - - 3Subtract: 18Solve the system of linear equations below.
6x + 3y = 33
4x + y = 15
A.
x = 2, y = 7
B.
x = -13, y = 7
C.
x = - 2/3, y = 12 2/3
D.
x = 5, y = 1
Answer:
A. x=2 y=7
Step-by-step explanation:
-12x -3 = 3y
6x + 3y = 33
sooo you add them up...
so its
-6x = -12
x=2
and then you plug in the x value into one of the equations
6x + 3y = 33
6(2) + 3y = 33
12 + 3y = 33
3y = 33 - 12
3y = 21
21/3=7
y=7
What does y equal in the solution of the system of equations below? 5y-3x-4z=22 2z-2x=-6 2z+3x=-6
9514 1404 393
Answer:
y = 2
Step-by-step explanation:
Subtracting the second equation from the third gives ...
(2z +3x) -(2z -2x) = (-6) -(-6)
5x = 0
x = 0
Using this in the third equation, we have ...
2z +0 = -6
z = -3
And substituting these values into the first equation, we have ...
5y -3(0) -4(-3) = 22
5y = 10 . . . . . subtract 12
y = 2
__
The solution to the system is (x, y, z) = (0, 2, -3).
Nikola thinks that the model that reflects the growth of smartphones shipped from manufacturers to stores around the world may be logistic rather than exponential. Do you agree with Nikola
Answer:
When most people have a smartphone, that is, the variable starts getting closer to its capacity, the demand will start to have a slight decrease, until it stabilizes, so yes, Nikola is correct.
Step-by-step explanation:
Exponential model:
The variable keeps growing consistently, at a fixed rate.
Logistic model:
The variable starts growing, but as it approaches a limit, for example, the carry capacity of an environment, the growth rate starts to decrease, until the variable stabilizes at a fixed value.
Growth of smartphones shipped from manufacturers to stores around the world.
When most people have a smartphone, that is, the variable starts getting closer to its capacity, the demand will start to have a slight decrease, until it stabilizes, so yes, Nikola is correct.
If the domain of a function that is reflected over the x-axis is (1, 5), (2, 1), (-1, -7), what is the range?
A. (1, -5), (2, -1), (-1, 7)
B. (5, 1), (1, 2), (-7, -1)
C. (-5, -1), (-1, -2), (7, 1)
D. (-1, 5), (-2, 1), (1, -7)
Answer:
A. (1, -5), (2, -1), (-1, 7)
Step-by-step explanation:
Reflecting a function over the x-axis:
When a function is reflected over the x-axis, the x-value stays the same, while y changes the signal, so the transformation rule is:
[tex](x,y) \rightarrow (x,-y)[/tex]
To find the range:
We apply the transformation to the points in the domain. Thus:
[tex](1,5) \rightarrow (1,-5)[/tex]
[tex](2,1) \rightarrow (2,-1)[/tex]
[tex](-1,-7) \rightarrow (-1,-(-7)) = (-1, 7)[/tex]
Thus the correct answer is given by option a.
Answer:
It is letter A and please give me brainliest
Step-by-step explanation:
Find the measure of ∠C in the image below. 60+55+m∠C=180
Answer:
angle C= 65 degree
Step-by-step explanation:
60+55+x= 180
115+x= 180
x= 180-115
x= 65
angle C= 65 degree
Please mark me as brainliest.
The population of a colony of 300 bacteria grows exponentially. After 2 hours, the population reaches 500. How much time will it take for the population to reach 9,600
Answer:
t = 13.56915448 hrs.
Step-by-step explanation:
500 = 300 [tex]e^{2k}[/tex]
5/3 = [tex]e^{2k}[/tex]
ln(5/3) = 2k ln(e)
k = ln(5/3)/2
k= 0.255412812
~~~~~~~~~~~~
9600 = 300 [tex]e^{0.255412812 t}[/tex]
32= [tex]e^{0.255412812 t}[/tex]
ln(32) = [tex]0.255412812 t[/tex] ln(e)
t = ln(32)/0.255412812
t = 13.56915448 hrs.
A basketball team is to play two games in a tournament. The probability of winning the first game is .10.1 the first game is won, the probability of winning the second game is 15. If the first game is lost, the probability of winning the second game is 25. What is the probability the first game was won if the second game is lost? Express the answer with FOUR decimal points.
Answer:
[tex]P(\frac{A}{B'})[/tex]=0.111
Step-by-step explanation:
Given:
The probability of winning the first game is 10.1
The first game is won
The probability of winning the second game is 15
If the first is lost, the probability of winning the second game is 25
Solution:
[tex]P(B)=P(A)P(\frac{B}{A})+P(A')P(\frac{B}{A'})\\ =0.1(0.15)+(0.3)*0.25)\\P(B)=0.24 ------(1)\\P(\frac{A}{B})=\frac{P(\frac{B}{A})P(A) }{P(B)}\\ =\frac{0.15(0.1)}{0.24}\\ =0.0625 ------(2)\\P(B')=1-P(B)=0.76 ------(3)\\P(A)=P(B)P(\frac{A}{B})+P(B')P(\frac{A}{B'})\\0.1=0.24(0.0625)+0.76(p(\frac{A}{B'} ))\\P(\frac{A}{B'})=0.111[/tex]
Answer:
[tex]P(W_1/W_2')=0.1110[/tex]
Step-by-step explanation:
Probability of winning the first game be considering the given factors be, [tex]W_1=0.1[/tex]
Probability of winning the second game be considering the given factors be, [tex]W_2[/tex]= probability of winning the second game when the first game is won + probability of winning the second game when the first game is lost:
[tex]P(W_2)=P(W_1).P(W_2/W_1)+P(W_1').P(W_2/W_1')[/tex]
[tex]P(W_2)=0.1\times 0.15+0.9\times 0.25[/tex]
[tex]P(W_2)=0.24[/tex]
Hence the probability of losing the second game:
[tex]P(W_2')=1-P(W_2)[/tex]
[tex]P(W_2')=0.76[/tex]
Probability of winning the first game when the second game is won:
[tex]P(W_1/W_2)=\frac{P(W_2/W_1).P(W_1)}{P(W_2)}[/tex]
[tex]P(W_1/W_2)=\frac{0.15\times 0.1}{0.24}[/tex]
[tex]P(W_1/W_2)=0.0625[/tex]
Probability of winning the first game be considering the given factors, [tex]W_1[/tex]= probability of winning the first game when the second game is won + probability of winning the first game when the second game is lost:
[tex]P(W_1)=P(W_2).P(W_1/W_2)+P(W_2').P(W_1/W_2')[/tex]
[tex]0.1=0.24\times0.0625+0.76\times P(W_1/W_2')[/tex]
[tex]P(W_1/W_2')=0.1110[/tex]
Express these system specifications using the propositions p “The user enters
a valid password,” q “Access is granted,” and r “The user has paid the
subscription fee” and logical connectives (including negations).
a) “The user has paid the subscription fee, but does not enter a valid
password.”
b) “Access is granted whenever the user has paid the subscription fee and
enters a valid password.”
c) “Access is denied if the user has not paid the subscription fee.”
d) “If the user has not entered a valid password but has paid the subscription
fee, then access is granted.”
Answer:
a) r ⋀~p
b)(r⋀p)⟶q
c) ~r ⟶ ~q
d) (~p ⋀r) ⟶q
Step-by-step explanation:
To solve this question we will make use of logic symbols in truth table.
We are told that;
p means "The user enters
a valid password,”
q means “Access is granted,”
r means “The user has paid the
subscription fee”
A) The user has paid the subscription fee, but does not enter a valid
password.”
Fist part of the statement is correct and so it will be "r". Second part of the statement is a negation and will be denoted by ~p. Since both statements are joined together in conjunction, we will use the conjuction symbol in between them which is "⋀" Thus, we have; r ⋀~p
B) Still using logic symbols, we have;
(r⋀p)⟶q
⟶ means q is true when r and p are true.
C) correct symbol is ~r ⟶ ~q
Since both statements are negation of the question. And also, if ~r is true then ~q is also true.
D) Similar to answer A to C above, applying similar conditions, we have (~p ⋀r) ⟶q
Construct a frequency distribution and a relative frequency histogram for the accompanying data set using five classes. Which class has the greatest relative frequency and which has the least relative frequency?Complete the table below. Use the minimum data entry as the lower limit of the first class.Class Frequency, f Relative frequencyx-x x xx-x x xx-x x xx-x x xx-x x x sumf= X?(Type integers or decimals. Round to the nearest thousandth as needed.)DATA:Triglyceride levels of 26 patients (in milligrams per deciliter of blood)138 199 240 143 294 175 240 216 223180 138 266 161 175 402 172 459 147391 152 199 294 188 320 421 161
Answer:
[tex]\begin{array}{cc}{Class}& {Frequency} & 138 - 202 & 14 & 203 - 267 & 5 & 268 - 332 & 3 & 333 - 397 & 1 & 398 - 462 & 3 \ \end{array}[/tex]
The class with the greatest is 138- 202 and the class with the least relative frequency is 333 - 397
Step-by-step explanation:
Solving (a): The frequency distribution
Given that:
[tex]Lowest = 138[/tex] --- i.e. the lowest class value
[tex]Class = 5[/tex] --- Number of classes
From the given dataset is:
[tex]Highest = 459[/tex]
So, the range is:
[tex]Range = Highest - Lowest[/tex]
[tex]Range = 459 - 138[/tex]
[tex]Range = 321[/tex]
Divide by the number of class (5) to get the class width
[tex]Width = 321 \div 5[/tex]
[tex]Width = 64.2[/tex]
Approximate
[tex]Width = 64[/tex]
So, we have a class width of 64 in each class;
The frequency table is as follows:
[tex]\begin{array}{cc}{Class}& {Frequency} & 138 - 202 & 14 & 203 - 267 & 5 & 268 - 332 & 3 & 333 - 397 & 1 & 398 - 462 & 3 \ \end{array}[/tex]
Solving (b) The relative frequency histogram
First, we calculate the relative frequency by dividing the frequency of each class by the total frequency
So, we have:
[tex]\begin{array}{ccc}{Class}& {Frequency} & {Relative\ Frequency} & 138 - 202 & 14 & 0.53 & 203 - 267 & 5 & 0.19 & 268 - 332 & 3 & 0.12 & 333 - 397 & 1 & 0.04 & 398 - 462 & 3 & 0.12 \ \end{array}[/tex]
See attachment for histogram
The class with the greatest is 138- 202 and the class with the least relative frequency is 333 - 397
The sum of two numbers is 8 and their difference is half of the sum. Find the
two numbers.
Answer:
6 and 2
Step-by-step explanation:
6+2= 8, so they can combine to equal 8. First box checked. Then, they can also be subtracted to get 4, half of 8. 8/2= 4 and 6-2=4. Second box checked.
Hope this helped :D
Step-by-step explanation:
Let's take the numbers as m and n and classify the info
m+n=8
m-n=4
So solving the sums Simultaneously you can get
2n=4
n=2
Substituting this value gives
2+m=8
m=6
Inorder to check if your answers are valid add them to the equations
6+2=8✅
6-2=4✅
Therefore the answers are valid
Solve the given system by the substitution method
5x + y 19
7x-2y = 13
Answer:
x = 3 , y = 4
Step-by-step explanation:
5x + y = 19 --------- ( 1 )
=> y = 19 - 5x
7x - 2y = 13 ------------ ( 2 )
Substitute y in ( 2 ) :
7x - 2( 19 - 5x ) = 13
7x - 38 + 10x = 13
17x = 13 + 38
17x = 51
x = 3
Substitute x in ( 1 ) :
5x + y = 19
5( 3 ) + y = 19
15 + y = 19
y = 19 - 15
y = 4
2. Find the area of a trapezium shaped field with a base of 45m, top is 35m and with a height of 55m applying the formula for trapezium = 0.5x b+axh
Given:
Base=
Top (a) =
Height =
3. Find the area of a Parallelogram shaped field where the base measures 19m and with a h of 37m.
Applying the formula for parallelogram=bxh
Given:
Base=
Height=
pahelp po thanks
Answer:
2. 2200 m²
3. 703 m²
Step-by-step explanation:
2. Given,
Base (b) = 45m
Top (a) = 35m
Height (h) = 55m
Area = (a+b)*h/2
= (45+35)*55/2
= 85*55/2 = 2200 m²
3. Given,
Base (b) = 19m
Height (h) = 37m
Area = b*h
= 19*37
= 703 m²
(The * sign represents the multiplication sign)
Answered by GAUTHMATH
Answer:
2. area = 2200 m²
3. area = 703 m²
Step-by-step explanation:
2. Find the area of a trapezium shaped field with a base of 45m, top is 35m and with a height of 55m applying the
formula for trapezium = 0.5 * (b+a) * h
Given:
Base= 45 m
Top (a) = 35 m
Height = 55 m
area = 0.5 * (b+a) * h
area = 0.5 * (45 m + 35 m) * 55 m
area = 2200 m²
3. Find the area of a Parallelogram shaped field where the base measures 19m and with a h of 37m.
Applying the formula for parallelogram = b * h
Given:
Base= 19 m
Height= 37 m
area = b * h
area = 19 m * 37 m
area = 703 m²
-10 degrees Celsius is what Fahrenheit
Answer:
Step-by-step explanation:
i think if its -10 degrees i think the fahrenheit would be 50 degrees
a) Read section 1.5 in the Yakir textbook. If you were a teacher and had 30 students in your class and wanted to know the class average on the first quiz, would you use a parameter or a statistic
Answer:
Parameter
Step-by-step explanation:
Required
Parameter of Statistic
From the question, we understand that the teacher is to calculate the class average.
To calculate the class average, the teacher will use the mean function/formula, which is calculated as:
[tex]Mean = \frac{\sum x}{n}[/tex]
Generally, mean is an example of a parameter.
So, we can conclude that the teacher will use parameer
Find the range of the data.
Scores: 81, 79, 80, 88, 72, 96, 86, 73, 79, 88
Answer:
24
Step-by-step explanation:
To find the range, you must subtract the lowest value from the highest value in the data set. If you organize the set from least to greatest, 72 is the lowest, and 96 is the highest.
So, 96 - 72 = 24, which is the range.
Match the number of significant figures to the value or problem.
1
?
0.008
4
?
54
3
?
1002. 43.2
2
?
1.068
Answer:
answer is 1 2 3 and 4 respectively of given match the following
The thickness X of aluminum sheets is distributed according to the probability density function f(x) = 450 (x2 - x) if 6 < x < 12 0 otherwise 5-1 Derive the cumulative distribution function F(x) for 6 < x < 12. The answer is a function of x and is NOT 1! Show the antiderivative in your solution. 5-2 What is E(X) = {the mean of all sheet thicknesses)? Show the antiderivative in your solution.
Solution :
Given :
[tex]f(x) = \left\{\begin{matrix}\frac{1}{450}(x^2-x) & \text{if } 6 < x < 12 \\ 0 & \text{otherwise}\end{matrix}\right.[/tex]
1. Cumulative distribution function
[tex]$P(X \leq x) = \int_{- \infty}^x f(x) \ dx$[/tex]
[tex]$=\int_{- \infty}^6 f(x) dx + \int_{6}^x f(x) dx $[/tex]
[tex]$=0+\int_6^x \frac{1}{450}(x^2-x) \ dx$[/tex]
[tex]$=\frac{1}{450} \int_6^x (x^2-x) \ dx$[/tex]
[tex]$=\frac{1}{450}\left[\frac{x^3}{3}-\frac{x^2}{2}\right]_6^x$[/tex]
[tex]$=\frac{1}{450}\left[ \left( \frac{x^3}{3} - \frac{x^2}{2}\left) - \left( \frac{6^3}{3} - \frac{6^2}{2} \right) \right] $[/tex]
[tex]$=\frac{1}{450}\left[\frac{x^3}{3} - \frac{x^2}{2} - 54 \right]$[/tex]
2. Mean [tex]$E(x) = \int_{- \infty}^{\infty} \ x \ f(x) \ dx$[/tex]
[tex]$=\int_{6}^{12}x . \left( \frac{1}{450} \ (x^2-x)\right)\ dx$[/tex]
[tex]$=\frac{1}{450} \int_6^{12} \ (x^3 - x^2) \ dx$[/tex]
[tex]$=\frac{1}{450} \left[\frac{x^4}{4} - \frac{x^3}{3} \right]_6^{12} \ dx$[/tex]
[tex]$=\frac{1}{450} \left[ \left(\frac{(12)^4}{4} - \frac{(12)^3}{3} \right) - \left(\frac{(6)^4}{4} - \frac{(6)^3}{3} \right) $[/tex]
[tex]$=\frac{1}{450} [4608 - 252]$[/tex]
= 17.2857
A radioactive material is known to decay at a yearly rate proportional to the amount at each moment. There were 1000 grams of the material 10 years ago. There are 980 grams right now. What will be the amount of the material right after 20 years
Answer:
x = 960.4
Step-by-step explanation:
980 = 1000[tex]e^{kt}[/tex]
.98 = [tex]e^{10 k}[/tex]
ln(.98) = 10k ln(e)
k = ln(.98)/10
k=-0.00202
~~~~~~~~~~~~~~
x = 1000[tex]e^{20 *-.00202}[/tex]
x = 960.4
The amount of the material right after 20 years will be x = 960.4.
What is an exponential expression?Powers can simply be expressed in concise form using exponential expressions. The exponent shows how many times the base has been multiplied. Since 2 is the "base" and 5 is the "exponent," it can be represented as 2x2x2x2=25 for the number 32. This phrase should be understood as "two to the fifth power."
Given that radioactive material is known to decay at a yearly rate proportional to the amount at each moment. There were 1000 grams of the material 10 years ago. There are 980 grams right now.
The amount of the material will be calculated as,
980 = 1000
[tex]0.98 = e^{10k}[/tex]
ln(.98) = 10k ln(e)
k = ln(.98)/10
k=-0.00202
The value after 20 years will be,
[tex]x = 1000e^{20\times 0.00202}[/tex]
x = 960.4
Therefore, the amount of the material right after 20 years will be x = 960.4.
To know more about an exponential expression follow
https://brainly.com/question/2456547
#SPJ5
A rectangle's length is three times as long as it is wide. Which expression represents the change in area if the width of the rectangle is increased by 1?
1. 3x^2
2. 3x
3. 3x^2+3x
4. the area increases by 3
Step-by-step explanation:
Let's say the rectangle's width is equal to y. We know that the length is three times the width, so the length = 3 * y. We also know that the area for a rectangle is equal to length * width, so the area, z, is equal to
(3*y) * y = z
3 * y² = z
Now, let's increase the width of the rectangle by 1. We can replace y with y+1 (as y+1 is 1 greater than y), and 3 * y with 3 * (y+1) to get
3*(y+1) * (y+1) = new area
(3y+3)*(y+1) = new area
3y²+3y +3 y + 3 = new area
3y² + 6y + 3 = new area
The difference in area is equal to the new area subtracted by the old area, or
3y²+6y+3 - 3y² = 6y +3. The variable for x is not given, so if x = (2y+1), the answer would be the second choice. However, solely using the information given, it is impossible to determine a solution outside of saying that it is not option 4, as 6y + 3 ≠ 3
Need help please I don’t get it
the e-function stuff can be confusing sometimes, but notices that g(x) / the blue line, is just somewhat lower, rest is the same.
how much lower? look at the y-intercepts
f(0)= "about 5"
g(0)= "about -3"
with this y-intercept only option c can work
1). A population of 20 rabbits is released into a wildlife region. The population is growing at a rate of 60% per year.
A) the General Equation from the Video was: P(x) = (blank)
What is the population of rabbits after 5 years?
B) the Evaluated equation I used to get the following answer is (blank)
and After five years there will be(blank)
rabbits.
And What is the population of rabbits after 8 years?
c) the Evaluated equation I used to get the following answer is(blank)
and After eight years there will be(blank)
rabbits.
Answer:
(a) A = 20(1.6)^t
(b) 210 rabbits
Step-by-step explanation:
Initial number of rabbits = 20
rate of growth, R = 60 % annually
(A) The general equation is
[tex]A = P \left ( 1+\frac{R}{100} \right )^t\\\\A = 20\left ( 1+\frac{60}{100} \right )^t\\\\A = 20 (1.6)^t[/tex]
(B) Let the time, t = 5 years
So, the population after 5 years is
[tex]A = 20 (1.6)^5\\\\A = 209.7 = 210 rabbits[/tex]
Find the volume
h=9cm
8cm
8cm
Answer: (8x8x9)/3=192
51
What is the inverse of the function f(x) = 2x + 1?
Oh(x) =
1
2x-
o h«x)= kx +
- 3x-2
Oh(x) =
Oh(x) =
Mark this and return
Save and Exit
Next
Submit
Type here to search
81
O
10:49 AM
^ D 0x
mamman
Answer:
let inverse f(x) be m:
[tex]m = \frac{1}{2x + 1} \\ 2x + 1 = \frac{1}{m} \\ 2x = \frac{1 - m}{m} \\ x = \frac{1 - m}{2m} [/tex]
substitute x in place of m:
[tex]{ \bf{ {f}^{ - 1}(x) = \frac{1 - x}{2x } }}[/tex]
help please i don’t understand it at this moment
Answer:
it's H. 1/2 in.=1,000 ft
F. 1 in.= 100ft
[tex]{hope 8 helps}}[/tex]
The time spent waiting in the line is approximately normally distributed. The mean waiting time is 7 minutes and the standard deviation of the waiting time is 1 minute. Find the probability that a person will wait for more than 6 minutes. Round your answer to four decimal places.
Answer:
0.15866
Step-by-step explanation:
6-7/1
=-1
p(x>-1)=1-p(x<1)
=0.15866
Seth and Ted can paint a room in 5 hours if they work together. If Ted were to work by himself, it would take him 2 hours longer than it would take Seth working by himself. How long would it take Seth to paint the room by himself if Ted calls in sick?
Answer:
9 hours
Step-by-step explanation:
Let
x = number of hours it would take Seth to work by himself
He would paint 1/x in 1 hour
x + 2 = number of hours it would take Ted to work by himself
He would paint 1/(x + 2) in 1 hour
Seth and Ted = 5 hours
They would paint 1/5 in 1 hour
The equation is this:
1/x + 1/(x + 2) = 1/5
(x + 2)+x/x(x+2) = 1/5
x+2+x / x(x+2) = 1/5
2x + 2 / x(x+2) = 1/5
2x + 2 = x(x + 2)1/5
2x + 2 = (x² + 2x)1/5
5(2x + 2) = x² + 2x
10x + 10 = x² + 2x
x² + 2x - 10x - 10 = 0
x² - 8x - 10 = 0
x = -b ± √b² - 4ac/2a
= -(-8) ± √(-8)² - 4(1)(-10) / 2(1)
= 8 ± √64 - (-40) / 2
= 8 ± √64 + 40) / 2
= 8 ± √104 / 2
= 8 ± 2√26 / 2
= 8/2 ± 2√26/2
= 4 ± √26
= 4 ± 5.0990195135927
= 4 + 5.0990195135927 or 4 - 5.0990195135927
= 9.0990195135927 or -1.Answer:
Step-by-step explanation:
Let
x = number of hours it would take Seth to work by himself
He would paint 1/x in 1 hour
x + 2 = number of hours it would take Ted to work by himself
He would paint 1/(x + 2) in 1 hour
Seth and Ted = 5 hours
They would paint 1/5 in 1 hour
The equation is this:
1/x + 1/(x + 2) = 1/5
(x + 2)+x / x(x+2) = 1/5
x+2+x / x(x+2) = 1/5
2x + 2 / x(x+2) = 1/5
Cross product
2x + 2 = x(x + 2)1/5
2x + 2 = (x² + 2x)1/5
Cross product
5(2x + 2) = x² + 2x
10x + 10 = x² + 2x
x² + 2x - 10x - 10 = 0
x² - 8x - 10 = 0
x = -b ± √b² - 4ac/2a
= -(-8) ± √(-8)² - 4(1)(-10) / 2(1)
= 8 ± √64 - (-40) / 2
= 8 ± √64 + 40) / 2
= 8 ± √104 / 2
= 8 ± 2√26 / 2
= 8/2 ± 2√26/2
= 4 ± √26
= 4 ± 5.0990195135927
= 4 + 5.0990195135927 or 4 - 5.0990195135927
= 9.0990195135927 or -1.0990195135927
Approximately,
x = 9 hours or -1 hour
It can't take Seth negative hours to work
Therefore,
x = number of hours it would take Seth to work by himself = 9 hours
What is net cash flow
The volume of a pyramid is 240 cubic centimeters. The pyramid has a rectangular base with sides 6cm by 4cm. Find the altitude and lateral surface area of the pyramid if the pyramid has equal lateral edges
Answer:
altitude = 30 cm
lateral surface area = 301 cm² (approximately)
Step-by-step explanation:
let the altitude be x,
240=6*4*x/3
or, x=30 cm
Lateral surface area,
=l×√(w/2)²+h²]+w×√[(l/2)²+h²]
=6×√[(4/2)²+30²]+4×√[(6/2)²+30²]
≈300.99806
≈ 301 cm²
Answered by GAUTHMATH
