Answer:
8x +15
Step-by-step explanation:
f(x) = 10 + 2x and g(x) = 6x + 5
(f+ g)(x) = 10 + 2x + 6x + 5
Combine like terms
= 8x +15
Answer:
8x+15
Step-by-step explanation:
(f+g)(x) = f(x)+g(x)
= (10+2x) + (6x+5)
= 8x+15
hope it helped, please mark me brainliest.
A rectangular tank 4 feet long, 3 feet wide, and 5 feet deep is full of oil with weight density 50 lb ft 3 lbft3 . Calculate the work required to pump all of the oil out over the top of the tank.
The work required for the given task of pumping all of the oil out over the top of the tank is 7,500 ft·lb
The known parameters;
The length of the rectangular tank, l = 4 feet
The width of the tank, w = 2 feet
The depth of the tank, h = 5 feet
The weight density of the oil with which the tank is filled, ρ × g = 50 lb/ft³
The unknown parameter
The work required to pump all of the oil out over the top of the tank
Method;
Calculate the force required to lift each slice (layer) of the oil to the top multiplied by the distance, y, the slice moves and summing the result as an integration as follows;
The volume of each slice, [tex]\mathbf{V_i}[/tex] = l × w × dy
The force required to move each slice, [tex]\mathbf{F_i}[/tex] = ρ × g × l × w × dy
The work done, [tex]\mathbf{W_i}[/tex], in moving the slice a distance, y, is given as follows;
[tex]\mathbf{W_i}[/tex] = ρ × g × l × w × y × dy
Therefore, the total work done, W, in pumping all the water located from y = 0, to y = 5, to the top of the tank, is given as follows;
[tex]\mathbf{W = \int\limits^5_0 {(\rho \times g \times l \times w \times y) } \, dy}[/tex]
Therefore;
W = (ρ × g × l × w × y²)/2
Plugging in the values, gives;
W = (50 lb/ft³ × 4 ft. × 3 ft. × (5 ft.)²)/2 = 7,500 ft·lb
The work required to pump all of the oil out over the top of the tank, W = 7,500 ft·lb.
Learn more about the use of integration to calculate the amount of work required for a given task here;
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It is estimated that 75% of all young adults between the ages of 18-35 do not have a landline in their homes and only use a cell phone at home.
(a) On average, how many young adults do not own a landline in a random sample of 100?
(b) What is the standard deviation of probability of young adults who do not own a landline in a simple random sample of 100?
(c) What is the proportion of young adults who do not own a landline?
(d) What is the probability that no one in a simple random sample of 100 young adults owns a landline?
(e) What is the probability that everyone in a simple random sample of 100 young adults owns a landline?
(f) What is the distribution of the number of young adults in a sample of 100 who do not own a landline?
(g) What is the probability that exactly half the young adults in a simple random sample of 100 do not own a landline?
Answer:
a) 75
b) 4.33
c) 0.75
d) [tex]3.2 \times 10^{-13}[/tex] probability that no one in a simple random sample of 100 young adults owns a landline
e) [tex]6.2 \times 10^{-61}[/tex] probability that everyone in a simple random sample of 100 young adults owns a landline.
f) Binomial, with [tex]n = 100, p = 0.75[/tex]
g) [tex]4.5 \times 10^{-8}[/tex] probability that exactly half the young adults in a simple random sample of 100 do not own a landline.
Step-by-step explanation:
For each young adult, there are only two possible outcomes. Either they do not own a landline, or they do. The probability of an young adult not having a landline is independent of any other adult, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
75% of all young adults between the ages of 18-35 do not have a landline in their homes and only use a cell phone at home.
This means that [tex]p = 0.75[/tex]
(a) On average, how many young adults do not own a landline in a random sample of 100?
Sample of 100, so [tex]n = 100[/tex]
[tex]E(X) = np = 100(0.75) = 75[/tex]
(b) What is the standard deviation of probability of young adults who do not own a landline in a simple random sample of 100?
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100(0.75)(0.25)} = 4.33[/tex]
(c) What is the proportion of young adults who do not own a landline?
The estimation, of 75% = 0.75.
(d) What is the probability that no one in a simple random sample of 100 young adults owns a landline?
This is P(X = 100), that is, all do not own. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 100) = C_{100,100}.(0.75)^{100}.(0.25)^{0} = 3.2 \times 10^{-13}[/tex]
[tex]3.2 \times 10^{-13}[/tex] probability that no one in a simple random sample of 100 young adults owns a landline.
(e) What is the probability that everyone in a simple random sample of 100 young adults owns a landline?
This is P(X = 0), that is, all own. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{100,0}.(0.75)^{0}.(0.25)^{100} = 6.2 \times 10^{-61}[/tex]
[tex]6.2 \times 10^{-61}[/tex] probability that everyone in a simple random sample of 100 young adults owns a landline.
(f) What is the distribution of the number of young adults in a sample of 100 who do not own a landline?
Binomial, with [tex]n = 100, p = 0.75[/tex]
(g) What is the probability that exactly half the young adults in a simple random sample of 100 do not own a landline?
This is P(X = 50). So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 50) = C_{100,50}.(0.75)^{50}.(0.25)^{50} = 4.5 \times 10^{-8}[/tex]
[tex]4.5 \times 10^{-8}[/tex] probability that exactly half the young adults in a simple random sample of 100 do not own a landline.
A student found the solution below for the given inequality.Which of the following explains whether the student is correct?The student is completely correct because the student correctly wrote and solved the compound inequality.The student is partially correct because only one part of the compound inequality is written correctly.The student is partially correct because the student should have written the statements using “or” instead of “and.”The student is completely incorrect because there is “ no solution “ to this inequality.
Answer:
The student is completely incorrect because there is no solution to this inequality.
Answer:
D on edge
Step-by-step explanation:
Question
(X-5y/y3)-1=
Answer:
[tex]x = y^3+5y[/tex]
Step-by-step explanation:
Complete question
[tex]\frac{x - 5y}{y^3} - 1=0\\[/tex]
Required
Solve for x
We have:
[tex]\frac{x - 5y}{y^3} - 1=0[/tex]
Collect like terms
[tex]\frac{x - 5y}{y^3} = 1[/tex]
Multiply through by [tex]y^3[/tex]
[tex]x - 5y = y^3[/tex]
Make x the subject
[tex]x = y^3+5y[/tex]
help pls! I need the answer quickly and pls explain. thank you!
Answer:
h = 6[tex]\sqrt{3}[/tex]
Step-by-step explanation:
The given is the special right triangle with angle measures : 90-60-30
and the side lengths for the given angles are represented by :
2a-a[tex]\sqrt{3}[/tex]-a
the side length that sees 60 degrees is represented by a[tex]\sqrt{3}[/tex] (h in this case)
the area of a triangle is calculated by multiplying height and base and that is divided by 2
a[tex]\sqrt{3}[/tex]*a/2 = 18[tex]\sqrt{3}[/tex] multiply both sides by 2
a^2[tex]\sqrt{3}[/tex] = 36[tex]\sqrt{3}[/tex] divide both sides by [tex]\sqrt{3}[/tex]
a^2 = 36 find the roots for both sides
a = 6
since h sees angle measure 60 and is represented by a[tex]\sqrt{3}[/tex]
h = 6[tex]\sqrt{3}[/tex]
[tex] \frac{3x - 2}{7} - \frac{5x - 8}{4} = \frac{1}{14} [/tex]
Answer:
[tex]x=2[/tex]
Step-by-step explanation:
[tex]\frac{3x-2}{7}-\frac{5x-8}{4}=\frac{1}{14}[/tex]
In order to factor an integer, we need to divide it by the ascending sequence of primes 2, 3, 5.
The number of times that each prime divides the original integer becomes its exponent in the final result.
In here, Prime number 2 to the power of 2 equals 4.
[tex]\frac{3x-2}{7}-\frac{5x-8}{2^{2} }=\frac{1}{14}[/tex]
First, We need to add fractions-
Rule:-
[tex]\frac{A}{B} +\frac{C}{D} =\frac{\frac{LCD}{B}+\frac{LCD}{D}C }{LCD}[/tex]
LCD = [tex]7 \cdot 2^{2}[/tex]
[tex]\frac{4(3x-2)+7(-(5x-8))}{7*2^{2} } =\frac{1}{14}[/tex]
[tex]x=2[/tex]
OAmalOHopeO
What is the difference between these two linear equations?
Y=3x - y=-3x
help it's due in 20 minutes
Answer:
Its the first one | -9 | < | 9 |
Step-by-step explanation:
its false because both sides are identical
Hope this helps!
Answer:
Choice AStep-by-step explanation:
|-9| = |9| = 9, not true-2² = -4 < 3, true-7≤-5, true|-1| = 1 ≥ 0, trueon the same graph draw line 2y-x=10 and y=3x
Answer:
Step-by-step explanation:
the ages of two students are in the ratio of 3:5,if the older is 40yrs. How old is the younger student
The ratio/proportion is young / old = young / old.
We know that one ratio is 3 / 5, so we need to complete the other.
3 / 5 = young / 40
5 goes into 40, 8 times, therefore we need to multiply the numerator by 8 also.
3 x 8 = 24
The younger student is 24 years old.
Hope this helps!
Answer:
24
Step-by-step explanation:
younger : older
3 :5
The older is 40
40/5 = 8
Multiply each by 8
younger : older
3 *8 :5 *8
24 : 40
The younger is 24
Differentiate y=2x+200/x with respect to x
Answer:
Hello,
[tex]\boxed{y'=2-\dfrac{200}{x^2} }\\[/tex]
Step-by-step explanation:
[tex](f(x)+g(x))'=f'(x)+g'(x)\\\\(2x)'=2*(x)'=2*1=2\\\\(\dfrac{200}{x} )'=200*(x^{-1})'=200*(-1)*x^{-1-1})=-\dfrac{200}{x^2} \\\\\\\boxed{y'=2-\dfrac{200}{x^2} }\\[/tex]
Answer:
[tex] \frac{dy}{dx } = 2 - \frac{200}{ {x}^{2} } [/tex]
Step-by-step explanation:
[tex]the \: equation \: can \: be \: rewriten \: as \\ y = 2x + 200 {x}^{ - 1} \\ \\ now \: differentiate \: the \: equation\ \\ \frac{dy}{dx} = 2 - 200 {x}^{ - 2} \\ \frac{dy}{dx} = 2 - \frac{200}{ {x}^{2} } [/tex]
Which figure always has exactly one line of symmetry?
A. rectangle
B. trapezoid
C. isosceles right triangle
D. circle
A shape of a trapezoid has exactly one line of symmetry. The correct option is B.
What is a trapezoid?An open, flat object with four straight sides and one pair of parallel sides is referred to as a trapezoid or trapezium.
A balanced and proportionate likeness between an object's two halves is referred to as symmetry in geometry. It implies that one half is the other's mirror image.
A trapezium's non-parallel sides are referred to as the legs, while its parallel sides are referred to as the bases. The legs of a trapezium can also be parallel. The parallel sides may be vertical, horizontal, or angled.
Therefore, the shape of a trapezoid has exactly one line of symmetry. The correct option is B.
To know more about Trapezoids follow
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A telescope contains both a parabolic mirror and a hyperbolic mirror. They share focus , which is 46feet above the vertex of the parabola. The hyperbola's second focus is 6 ft above the parabola's vertex. The vertex of the hyperbolic mirror is 3 ft below . Find the equation of the hyperbola if the center is at the origin of a coordinate system and the foci are on the y-axis. Complete the equation.
the center is at the origin of a coordinate system and the foci are on the y-axis, then the foci are symmetric about the origin.
The hyperbola focus F1 is 46 feet above the vertex of the parabola and the hyperbola focus F2 is 6 ft above the parabola's vertex. Then the distance F1F2 is 46-6=40 ft.
In terms of hyperbola, F1F2=2c, c=20.
The vertex of the hyperba is 2 ft below focus F1, then in terms of hyperbola c-a=2 and a=c-2=18 ft.
Use formula c^2=a^2+b^2c
2
=a
2
+b
2
to find b:
\begin{gathered} (20)^2=(18)^2+b^2,\\ b^2=400-324=76 \end{gathered}
(20)
2
=(18)
2
+b
2
,
b
2
=400−324=76
.
The branches of hyperbola go in y-direction, so the equation of hyperbola is
\dfrac{y^2}{b^2}- \dfrac{x^2}{a^2}=1
b
2
y
2
−
a
2
x
2
=1 .
Substitute a and b:
\dfrac{y^2}{76}- \dfrac{x^2}{324}=1
76
y
2
−
324
x
2
=1 .
What is the GCF of the expression 5x2y + 10xy2?
Answer:
[tex]5xy[/tex]
Step-by-step explanation:
[tex]\mathrm{Factor\:}:5x^2y[/tex]
[tex]5\cdot \:x\cdot \:x\cdot \:y[/tex]
[tex]\mathrm{Factor\:}:10xy^2[/tex]
[tex]2\cdot \:5\cdot \:x\cdot \:y\cdot \:y[/tex]
Common factor:-
[tex]5\cdot \:x\cdot \:y[/tex]
OAmalOHopeO
The sum of the 3rd and 7th terms of an A.P. is 38, and the 9th term is 37. Find the A.P.
Answer:
The AP is 1, 11/2, 10, 29/2, 19, ....
Step-by-step explanation:
Let the first term be a and d be the common difference of the arithmetic progression.
ATQ, a+2d+a+6d=38, 2a+8d=38 and a+8d=37. Solving this, we will get a=1 and d=9/2. The AP is 1, 11/2, 10, 29/2, 19, ....
Slope 0; through (-5, -1)
Answer:
y = -1
Step-by-step explanation:
The graph of y= -2x + 10 is:
O A. a line that shows only one solution to the equation.
O B. a point that shows the y-intercept.
O C. a line that shows the set of all solutions to the equation.
O D. a point that shows one solution to the equation.
SUBM
9514 1404 393
Answer:
C. a line that shows the set of all solutions to the equation.
Step-by-step explanation:
Any graph shows the set of all solutions to the equation being graphed.
The graph of a linear function is a straight line.
Plz help a beggar I don’t get it
Answer: 3
happy learning
Answer:
B.
Step-by-step explanation:
From the point (-1,0) the next point on the graph is up 3, right 1, making the slope a positive 3.
Consider an x distribution with standard deviation o = 34.
(a) If specifications for a research project require the standard error of the corresponding distribution to be 2, how
large does the sample size need to be?
B) If specifications for a research project require the standard error of the corresponding x distribution to be 1, how large does the sample size need to be?
Part (a)
The standard error (SE) formula is
[tex]\text{SE} = \frac{\sigma}{\sqrt{n}}\\\\[/tex]
where n is the sample size. We're given SE = 2 and sigma = 34, so,
[tex]\text{SE} = \frac{\sigma}{\sqrt{n}}\\\\2 = \frac{34}{\sqrt{n}}\\\\2\sqrt{n} = 34\\\\\sqrt{n} = \frac{34}{2}\\\\\sqrt{n} = 17\\\\n = 17^2\\\\n = 289\\\\[/tex]
So we need a sample size of n = 289 to have an SE value of 2.
Answer: 289========================================================
Part (b)
We'll use SE = 1 this time
[tex]\text{SE} = \frac{\sigma}{\sqrt{n}}\\\\1 = \frac{34}{\sqrt{n}}\\\\1*\sqrt{n} = 34\\\\\sqrt{n} = 34\\\\n = 34^2\\\\n = 1156\\\\[/tex]
Because we require greater precision (i.e. a smaller SE value), the sample size must be larger to account for this. In other words, as SE goes down, then n must go up, and vice versa.
Answer: 1156find the n^th root of z = -2i, n = 6
Answer:
2^(1/6) (cos(-pi/12)+i sin(-pi/12))
2^(1/6) (cos(3pi/12)+i sin(3pi/12))
2^(1/6) (cos(7pi/12)+i sin(7pi/12))
2^(1/6) (cos(11pi/12)+i sin(11pi/12))
2^(1/6) (cos(5pi/4)+i sin(5pi/4))
2^(1/6) (cos(19pi/12)+i sin(19pi/12))
Step-by-step explanation:
Let's convert to polar form.
-2i=2(cos(A)+i sin(A) )
There is no real part so cos(A) has to be zero and since we want -2 and we already have 2 then we need sin(A)=-1 so let's choose A=-pi/2.
So z=2(cos(-pi/2)+i sin(-pi/2)).
There are actually infinitely many ways we can write this polar form which we will need.
z=2(cos(-pi/2+2pi k)+i sin(-pi/2+2pi k))
where k is an integer
Now let's find the 6 6th roots or z.
2^(1/6) (cos(-pi/12+2pi k/6)+i sin(-pi/12+2pi k/6))
Reducing
2^(1/6) (cos(-pi/12+pi k/3)+i sin(-pi/12+pi k/3))
Plug in k=0,1,2,3,4,5 to find the 6 6th roots.
k=0:
2^(1/6) (cos(-pi/12+pi (0)/3)+i sin(-pi/12+pi (0)/3))
=2^(1/6) (cos(-pi/12)+i sin(-pi/12))
k=1:
2^(1/6) (cos(-pi/12+pi/3)+i sin(-pi/12+pi/3))
2^(1/6) (cos(3pi/12)+i sin(3pi/12))
k=2:
2^(1/6) (cos(-pi/12+2pi/3)+i sin(-pi/12+2pi/3))
2^(1/6) (cos(7pi/12)+i sin(7pi/12))
k=3:
2^(1/6) (cos(-pi/12+3pi/3)+i sin(-pi/12+3pi/3))
2^(1/6) (cos(11pi/12)+i sin(11pi/12))
k=4:
2^(1/6) (cos(-pi/12+4pi/3)+i sin(-pi/12+4pi/3))
2^(1/6) (cos(15pi/12)+i sin(15pi/12))
2^(1/6) (cos(5pi/4)+i sin(5pi/4))
k=5:
2^(1/6) (cos(-pi/12+5pi/3)+i sin(-pi/12+5pi/3))
2^(1/6) (cos(19pi/12)+i sin(19pi/12))
Team A scored 30 points less than four times the number of points that Team B scored. Team C scored 61 points more than half of the number of points that Team B scored. If Team A and Team C shared in the victory, having earned the same number of points, how many more points did each team have than Team B?
Answer:
team a and team c scored 74 points which is 48 points more than team b, scoring 26 points.
Step-by-step explanation:
The legend on a map states that 1 inch is 20 miles. If you measure 5 inches on the map, how many miles would the actual distance be? Actual distance = [ ? ] miles
Answer:
1 inch= 20 miles. 5*20=100 miles. The answer is 100 miles.
Step-by-step explanation:
On these types of questions just do that every time, then you don't need to ask, for example:
1 foot = 50 miles
If it measures 3 feet.
3*50=150 miles.
If you have any questions regarding my answer, tell me in the comments, and I will answer them.
Given the central angle, name the arc formed.
Major arc for ∠EQD
A. EQDˆ
B. GDFˆ
C. EGDˆ
D. EDˆ
9514 1404 393
Answer:
C. EGD
Step-by-step explanation:
A major arc is typically named using the end points and a point on the arc. Here, the end points are E and D, and points on the major arc include C, G, and F. The major arc ED could be named any of
arc ECDarc EGD . . . . choice Carc EFDOf course, the reverse of any of these names could also be used: DCE, DGE, DFE.
The height of a triangle is 5 yards greater than the base. The area of the triangle is 273 square yards. Find the length of the base and the height of the triangle.
Answer:
Base = 21 while Height = 16
Find the value of a.
A. 58
B. 130
C. 86
D. 65
Answer:
[tex]C. \ \ \ 86[/tex]°
Step-by-step explanation:
1. Approach
In order to solve this problem, one must first find a relationship between arc (a) and arc (c). This can be done using the congruent arcs cut congruent segments theorem. After doing so, one can then use the secants interior angle to find the precise measurement of arc (a).
2. Arc (a) and arc (c)
A secant is a line or line segment that intersects a circle in two places. The congruent segments cut congruent arcs theorem states that when two secants are congruent, meaning the part of the secant that is within the circle is congruent to another part of a secant that is within that same circle, the arcs surrounding the congruent secants are congruent. Applying this theorem to the given situation, one can state the following:
[tex]a = c[/tex]
3. Finding the degree measure of arc (a),
The secants interior angle theorem states that when two secants intersect inside of a circle, the measure of any of the angles formed is equal to half of the sum of the arcs surrounding the angles. One can apply this here by stating the following:
[tex]86=\frac{a+c}{2}[/tex]
Substitute,
[tex]86=\frac{a+c}{2}[/tex]
[tex]86=\frac{a+a}{2}[/tex]
Simplify,
[tex]86=\frac{a+a}{2}[/tex]
[tex]86=\frac{2a}{2}[/tex]
[tex]86=a[/tex]
Place the labels in the chart
If you can draw this out for me or describe were they are that will be very helpful:)
Answer:
Check the image
what is the volume of a cube with a length of a 10cm,
a width of 8cm and a height of 8cm
Answer:
640
Step-by-step explanation:
muntiply all the number length width height
The sum of three numbers is 124
The first number is 10 more than the third.
The second number is 4 times the third. What are the numbers?
Answer:
182/3,3 8/3, 152/3
Step-by-step explanation:
a+b+c=124
a trừ c= 10
4b=c
Answer:
a=29,b=79,c=19
Step-by-step explanation:
a=c+10
b=4c
=> a+b+c=c+10+4c+c=124
=> c=19
=> a= 29, b=79
(03.04) Use the graph below for this question: What is the average rate of change from x = −3 to x = 5? (1 point)
A.−1
B.0
C.1
D.8
Answer:
B. 0
Step-by-step explanation:
Rate of change from x = -3 to x = 5
Rate of change = [tex] \frac{f(b) - f(a)}{b - a} [/tex]
where, from the graph, we have:
a = -3, f(a) = -1,
b = 5, f(b) = -1,
Plug in the values
Rate of change = [tex] \frac{-1 -(-1)}{5 - (-3)} [/tex]
Rate of change = [tex] \frac{0}{8} [/tex]
Rate of change = 0
In an assembly-line production of industrial robots, gearbox assemblies can be installed in one minute each if holes have been properly drilled in the boxes and in ten minutes if the holes must be redrilled. Twenty-four gearboxes are in stock, 6 with improperly drilled holes. Five gearboxes must be selected from the 24 that are available for installation in the next five robots. (Round your answers to four decimal places.) (a) Find the probability that all 5 gearboxes will fit properly. (b) Find the mean, variance, and standard deviation of the time it takes to install these 5 gearboxes.
Answer:
The right answer is:
(a) 0.1456
(b) 18.125, 69.1202, 8.3139
Step-by-step explanation:
Given:
N = 24
n = 5
r = 7
The improperly drilled gearboxes "X".
then,
⇒ [tex]P(X) = \frac{\binom{7}{x} \binom {17}{5-x}}{\binom{24}{5}}[/tex]
(a)
P (all gearboxes fit properly) = [tex]P(x=0)[/tex]
= [tex]\frac{\binom{7}{0} \binom{17}{5}}{\binom{24}{5}}[/tex]
= [tex]0.1456[/tex]
(b)
According to the question,
[tex]X = 91+5[/tex]
Mean will be:
⇒ [tex]\mu = E(x)[/tex]
[tex]=E(91+5)[/tex]
[tex]=9E(1)+5[/tex]
[tex]=9.\frac{nr}{N}+5[/tex]
[tex]=9.\frac{5.7}{24} +5[/tex]
[tex]=18.125[/tex]
Variance will be:
⇒ [tex]\sigma^2=Var(X)[/tex]
[tex]=V(9Y+5)[/tex]
[tex]=81.V(Y)[/tex]
[tex]=81.n.\frac{r}{N}.\frac{N-r}{N}.\frac{N-n}{N-1}[/tex]
[tex]=81.5.\frac{7}{24}.\frac{24-7}{24}.\frac{24-5}{24-1}[/tex]
[tex]=69.1202[/tex]
Standard deviation will be:
⇒ [tex]\sigma = \sqrt{69.1202}[/tex]
[tex]=8.3139[/tex]