Find the product of all positive integer values of $c$ such that $3x^2+7x+c=0$ has two real roots. I will award a lot of points
Answer: 24
Step-by-step explanation:
Let's find one solution:
3x² + 7x + c = 0
a=3 b=7 c=c
First, let's find c so that it has REAL ROOTS.
⇒ Discriminant (b² - 4ac) ≥ 0
7² - 4(3)c ≥ 0
49 - 12c ≥ 0
-12c ≥ -49
[tex]c\leq\dfrac{-49}{-12}\quad \rightarrow c\leq \dfrac{49}{12}[/tex]
Since c must be a positive integer, 1 ≤ c ≤ 4
Example: c = 4
3x² + 7x + 4 = 0
(3x + 4)(x + 1) = 0
x = -4/3, x = -1 Real Roots!
You need to use Quadratic Formula to solve for c = {1, 2, 3}
Valid solutions for c are: {1, 2, 3, 4)
Their product is: 1 x 2 x 3 x 4 = 24
Answer:
$3x^2+7x+c=0$
comparing above equation with ax²+bx+c
a=3
b=7
c=1
using quadratic equation formula
[tex]x = \frac{ - b + - \sqrt{b {}^{2} - 4ac} }{ 2a} [/tex]
x=(-7+-√(7²-4×3×1))/(2×3)
x=(-7+-√13)/6
taking positive
x=(-7+√13)/6=
taking negative
x=(-7-√13)/6=
A pet store has 10 puppies, including 2 poodles, 3 terriers, and 5 retrievers. If Rebecka and Aaron, in that order, each select one puppy at random without replacement find the probability that both select a poodle.
The probability is
Answer:
2/10 for Rebecka and either 2/9 or 1/9 for Aaron depending on if Rebecka selects a poodle or not.
Step-by-step explanation:
do some math
a geometric series has second term 375 and fifth term 81 . find the sum to infinity of series .
Answer: [tex]\bold{S_{\infty}=\dfrac{3125}{2}=1562.5}[/tex]
Step-by-step explanation:
a₁, 375, a₃, a₄, 81
First, let's find the ratio (r). There are three multiple from 375 to 81.
[tex]375r^3=81\\\\r^3=\dfrac{81}{375}\\\\\\r^3=\dfrac{27}{125}\qquad \leftarrow simplied\\\\\\\sqrt[3]{r^3} =\sqrt[3]{\dfrac{27}{125}}\\ \\\\r=\dfrac{3}{5}[/tex]
Next, let's find a₁
[tex]a_1\bigg(\dfrac{3}{5}\bigg)=375\\\\\\a_1=375\bigg(\dfrac{5}{3}\bigg)\\\\\\a_1=125(5)\\\\\\a_1=625[/tex]
Lastly, Use the Infinite Geometric Sum Formula to find the sum:
[tex]S_{\infty}=\dfrac{a_1}{1-r}\\\\\\.\quad =\dfrac{625}{1-\frac{3}{5}}\\\\\\.\quad =\dfrac{625}{\frac{2}{5}}\\\\\\.\quad = \dfrac{625(5)}{2}\\\\\\.\quad = \large\boxed{\dfrac{3125}{2}}[/tex]
If the 2412 leaves are not a random sample, but the researchers treated the 2412 leaves as a random sample, this most likely made the data more:_____________.1. accurate, but not precise2. precise, but not accurate3. neither4. both accurate and precise
Answer:
2. Precise but not accurate
Step-by-step explanation:
In a high precision, low accuracy case study, the measurements are all close to each other (high agreement between the measurements) but not near/or close to the center of the distribution (how close a measurement is to the correct value for that measurement)
Solve by completing the square. x2−12x=−27 Select each correct answer. −9 −3 3 9 15
Answer:
x=9,3
Step-by-step explanation:
x²-12x=-27
x²-12x+(12/2)²=-27+(12/2)²
x²-12x+6²=-27+36
(x-6)²=9
x-6=[tex] \frac{ + }{ - } \sqrt{9} [/tex]
x-6=+3 and x-6=-3
x=9 and 3
The total area under the standard normal curve to the left of zequalsnegative 1 or to the right of zequals1 is
Answer:
0.3174
Step-by-step explanation:
Z-score:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the area under the normal curve to the left of Z. Subtracting 1 by the pvalue, we find the area under the normal curve to the right of Z.
Left of z = -1
z = -1 has a pvalue of 0.1587
So the area under the standard normal curve to the left of z = -1 is 0.1587
Right of z = 1
z = 1 has a pvalue of 0.8413
1 - 0.8413 = 0.1587
So the area under the standard normal curve to the right of z = 1 is 0.1587
Left of z = -1 or right of z = 1
0.1587 + 0.1587 = 0.3174
The area is 0.3174
I don't know what to do.
Answer:
104.93 in
Step-by-step explanation:
When we draw out a picture of our triangle, we should see that we need to use sin∅ to solve:
sin23° = 41/x
xsin23° = 41
x = 41/sin23°
x = 104.931
what is the simplest form of this expression 2(w-1) +(-2)(2w+1)
Answer:
-2w - 4
Step-by-step explanation:
What is the simplest form of this expression
2(w - 1) + (-2)(2w + 1) =
= 2w - 2 - 4w - 2
= -2w - 4
Answer: -2w-4
Step-by-step explanation:
subtract 4w of 2w
2w-2-4w-2
subtract 2 of -2
-2w-2-2
final answer
-2w-4
(0, 3) and (-2, -1)
Write an equation in slope intercept from of the line that passes through the given points.
Answer:
y = 2x + 3
Step-by-step explanation:
Slope Formula: [tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
Slope-Intercept Form: y = mx + b
Step 1: Find slope m
m = (-1 - 3)/(-2 - 0)
m = -4/-2
m = 2
y = 2x + b
Step 2: Rewrite equation
y = 2x + 3
*You are given y-intercept (0, 3), so simply add it to your equation.
the diagram shows a circle drawn inside a square the circle touches the edges of the square
Answer:
69.5309950592 cm²
Step-by-step explanation:
Area of Square:
Area = [tex]Length * Length[/tex]
Area = 18*18
Area = 324 square cm
Area of circle:
Diameter = 18 cm
Radius = 9 cm
Area = [tex]\pi r^2[/tex]
Area = (3.14)(9)²
Area = (3.14)(81)
Area = 254.469004941 square cm
Area of Shaded area:
=> Area of square - Area of circle
=> 324 - 254.469004941
=> 69.5309950592 cm²
If f(x)=2x−1, show that f(f(x))=4x−3. Find f(f(f(x))).
Answer: f(f(f(x)))=8x-7
Step-by-step explanation:
Since we were given f(x) and f(f(x)), We plug that into f(x) again to get f(f(f(x))).
2(4x-3)-1 [distribute]
8x-6-1 [combine like terms]
8x-7
Arrange the functions for which the result is a non-infinite value and the limit exists in ascending order of their limit values as x tends to infinity. Please see picture attached.
Answer:
see attached
Step-by-step explanation:
The limit as x gets large is the ratio of the highest-degree terms. In most cases, the limit can be found by evaluating that ratio. Where an absolute value is involved, the absolute value of the highest-degree term is used.
If the ratio gives x to a positive power, the limit does not exist. If the ratio gives x to a negative power, the limit is zero.
The arrangement of functions according to the given condition
[tex]m(x)=\frac{4x^{2}-6 }{1-4x^{2} }[/tex]
[tex]h(x)=\frac{x^{3} -x^{2} +4}{1-3x^{2} }[/tex]
[tex]k(x)=\frac{5x+1000}{x^{2} }[/tex]
[tex]i(x)=\frac{x-1}{|1-4x| }[/tex]
[tex]g(x)=\frac{|4x-1|}{x-4}[/tex]
[tex]l(x)=\frac{5x^{2} -4}{x^{2} +1}[/tex]
[tex]f(x)=\frac{x^{2} -1000}{x-5}[/tex]
[tex]j(x)=\frac{x^{2}-1 }{|7x-1|}[/tex]
What is limit?A limit is the value that a function approaches as the input approaches some value.
According to the given question
[tex]l(x)=\frac{5x^{2} -4}{x^{2} +1}[/tex]
⇒[tex]\lim_{nx\to \infty} \frac{5x^{2} -1}{x^{2} +1}[/tex]
⇒[tex]\lim_{x \to \infty} \frac{x^{2} }{x^{2} } \frac{5-\frac{1}{x^{2} } }{1+\frac{1}{x^{2} } }[/tex]
= 5 ([tex]\frac{1}{x^{2} } = 0[/tex] ,as x tends to infinity [tex]\frac{1}{x^{2} }[/tex] tends to 0)
[tex]i(x)=\frac{x-1}{|1-4x|}[/tex]
⇒[tex]\lim_{x \to \infty} \frac{x-1}{|1-4x|}[/tex] = [tex]\lim_{x \to \infty} \frac{x}{x} \frac{1-\frac{1}{x} }{|\frac{-1}{4}+\frac{1}{x} | }[/tex] =[tex]\frac{1}{\frac{1}{4} }[/tex] =[tex]\frac{1}{4}[/tex]
As x tends to infinity 1/x tends to 0, and |[tex]\frac{-1}{4}[/tex]| gives 1/4
[tex]f(x)= \frac{x^{2} -1000}{x--5}[/tex]
⇒[tex]\lim_{x \to \infty} \frac{x^{2} -1000}{x-5}[/tex]= [tex]\lim_{x \to \infty} \frac{x^{2} }{x} \frac{1-\frac{1000}{x^{2} } }{1-\frac{5}{x} }[/tex]= [tex]\lim_{x \to \infty} x\frac{1-\frac{1000}{x^{2} } }{1-\frac{5}{x} }[/tex] ⇒ limit doesn't exist.
[tex]m(x)=\frac{4x^{2}-6 }{1-4x^{2} }[/tex]
⇒[tex]\lim_{x\to \infty} \frac{4x^{2} -6}{1-4x^{2} }[/tex]=[tex]\lim_{x\to \infty} \frac{x^{2} }{x^{2} } \frac{4-\frac{6}{x^{2} } }{\frac{1}{x^{2} } -4}[/tex] [tex]= \lim_{n \to \infty} \frac{4}{-4}[/tex] = -1
As x tends to infinity [tex]\frac{1}{x^{2} }[/tex] tends to 0.
[tex]g(x)=\frac{|4x-1|}{x-4}[/tex]
⇒[tex]\lim_{x\to \infty} \frac{|4x-1|}{x-4}[/tex] = [tex]\lim_{x \to \infty} \frac{|x|}{x} \frac{4-\frac{1}{x} }{1 -\frac{4}{x} } }[/tex] = 4
as x tends to infinity 1/x tends to 0
and |x|=x ⇒[tex]\frac{|x|}{x}=1[/tex]
[tex]h(x)=\frac{x^{3}-x^{2} +4 }{1-3x^{3} }[/tex][tex]\lim_{x \to \infty} \frac{x^{3} -x^{2} +4}{1-3x^{3} }[/tex][tex]= \lim_{x \to \infty} \frac{x^{3} }{x^{3} } \frac{1-\frac{1}{x}+\frac{4}{x^{3} } }{\frac{1}{x^{3} -3} }[/tex] = [tex]\frac{1}{-3}[/tex] =[tex]-\frac{1}{3}[/tex]
[tex]k(x)=\frac{5x+1000}{x^{2} }[/tex]
[tex]\lim_{x \to \infty} \frac{5x+1000}{x^{2} }[/tex] = [tex]\lim_{x \to \infty} \frac{x}{x} \frac{5+\frac{1000}{x} }{x}[/tex] =0
As x tends to infinity 1/x tends to 0
[tex]j(x)= \frac{x^{2}-1 }{|7x-1|}[/tex]
[tex]\lim_{x \to \infty} \frac{x^{2}-1 }{|7x-1|}[/tex] = [tex]\lim_{x \to \infty} \frac{x}{|x|}\frac{x-\frac{1}{x} }{|7-\frac{1}{x}| }[/tex] = [tex]\lim_{x \to \infty} 7x[/tex] = limit doesn't exist.
Learn more about limit here:
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Evaluate. Write your answer as a fraction or whole number without exponents. 1/10^-3 =
Answer:
1000
Step-by-step explanation:
=> [tex]\frac{1}{10^{-3}}[/tex]
According to the law of exponents, [tex]\frac{1}{a^{-m}} = a^{m}[/tex]
So, it becomes
=> [tex]10^{3}[/tex]
=> 1000
Determine if the matrix below is invertible. Use as few calculations as possible. Justify your answer. 3 0 -4 2 0 6 -3 0 8
a. The matrix is invertible. The columns of the given matrix span R^3.
b. The matrix is not invertible. If the given matrix is A, the columns of A do not form a linearly independent set.
c. The matrix is invertible. The given matrix has 2 pivot positions.
d. The matrix is not invertible. If the given matrix is A, the equation Ax = 0 has only the trivial solution.
Answer:
b. The matrix is not invertible. If the given matrix is A, the columns of A do not form a linearly independent set.
Step-by-step explanation:
A square matrix is said to be invertible if the product of the matrix and its inverse result into an identity matrix.
3 0 -4
2 0 6
-3 0 8
Since the second column elements are all zero, the determinant of the matrix is zero ad this implies that the inverse of the matrix does not exist(i.e it is not invertible )
A square matrix is said to be invertible if it has an inverse.
The matrix is not invertible. If the given matrix is A, the columns of A do not form a linearly independent set.
The matrix is given as:
[tex]\left[\begin{array}{ccc}3&0&-4\\2&0&6\\-3&0&8\end{array}\right][/tex]
Calculate the determinant
The determinant of the matrix calculate as:
[tex]|A| = 3 \times(0 \times 8- 6 \times 0) - 0(2 \times 8 - 6 \times -3) -4(2 \times 0 - 0 \times -3)[/tex]
[tex]|A| = 3 \times(0) - 0(34) -4(0)[/tex]
[tex]|A| = 0 - 0 -0[/tex]
[tex]|A| = 0[/tex]
When a matrix has its determinant to be 0, then
It is not invertibleIt does not form a linear independent set.Hence, the correct option is (b)
Read more about matrix at:
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Please answer this correctly
Answer:
The second question
Step-by-step explanation:
The orca starts at -25 meters. She goes up 25 meters.
up 25 = +25
-25+25=0
Answer:
Option 2
Step-by-step explanation:
The orca swims at the elevation of -25 meters. The orca swims up 25 meters higher than before.
-25 + 25 = 0
The graph shows a gasoline tank being filled at a rate of 2,500 gallons of gas per
hour. How will the graph change if the rate slows?
The correct answer is The line will be less steep because the rate will be slower
Explanation:
The rate of the graph is defined by the number of gallons filled vs the time; this relation is shown through the horizontal axis (time) and the vertical axis (gallons). Additionally, there is a constant rate because each hour 2,500 gallons are filled, which creates a steep constant line.
However, if the rate decreases, fewer gallons would be filled every hour, and the line will be less steep, this is because the number of gallons will not increase as fast as with the original rate. For example, if the rate is 1,250 gallons per hour (half the original rate), after 8 hours the total of gallons would be 1000 gallons (half the amount of gallons); and this would make the line to be less steep or more horizontal.
Trucks in a delivery fleet travel a mean of 100 miles per day with a standard deviation of 23 miles per day. The mileage per day is distributed normally. Find the probability that a truck drives between 86 and 125 miles in a day. Round your answer to four decimal places.
Answer:
The probability that a truck drives between 86 and 125 miles in a day.
P(86≤ X≤125) = 0.5890 miles
Step-by-step explanation:
Step(i):-
Given mean of the Population = 100 miles per day
Given standard deviation of the Population = 23 miles per day
Let 'X' be the normal distribution
Let x₁ = 86
[tex]Z_{1} = \frac{x_{1} -mean}{S.D} = \frac{86-100}{23} =-0.61[/tex]
Let x₂= 86
[tex]Z_{2} = \frac{x_{2} -mean}{S.D} = \frac{125-100}{23} = 1.086[/tex]
Step(ii):-
The probability that a truck drives between 86 and 125 miles in a day.
P(86≤ X≤125) = P(-0.61 ≤ Z≤ 1.08)
= P(Z≤ 1.08) - P(Z≤ -0.61)
= 0.5 +A(1.08) - ( 0.5 - A(-0.61))
= A(1.08) + A(0.61) ( A(-Z)= A(Z)
= 0.3599 + 0.2291
= 0.5890
Conclusion:-
The probability that a truck drives between 86 and 125 miles in a day.
P(86≤ X≤125) = 0.5890 miles per day
Which steps would be used to solve the equation? Check all that apply. 2 and two-thirds + r = 8 Subtract 2 and two-thirds from both sides of the equation. Add 2 and two-thirds to both sides of the equation. 8 minus 2 and two-thirds = 5 and one-third 8 + 2 and two-thirds = 10 and two-thirds Substitute the value for r to check the solution.
Answer:
Subtract 2 and two-thirds from both sides of the equation
8 minus 2 and two-thirds = 5 and one-third
Substitute the value for r to check the solution.
Step-by-step explanation:
2 2/3 + r = 8
Subtract 2 2/3 from each side
2 2/3 + r - 2 2/3 = 8 - 2 2/3
r = 5 1/3
Check the solution
2 2/3 +5 1/3 =8
8 =8
Answer:
1, 3, 5
Step-by-step explanation:
edge
The tread life of a particular brand of tire is normally distributed with mean 60,000 miles and standard deviation 3800 miles. Suppose 35 tires are randomly selected for a quality assurance test. Find the probability that the mean tread life from this sample of 35 tires is greater than 59,000 miles. You may use your calculator, but show what you entered to find your answer. Round decimals to the nearest ten-thousandth (four decimal places).
Answer:
P [ x > 59000} = 0,6057
Step-by-step explanation:
We assume Normal Distribution
P [ x > 59000} = (x - μ₀ ) /σ/√n
P [ x > 59000} = (59000 - 60000)/ 3800
P [ x > 59000} = - 1000/3800/√35
P [ x > 59000} = - 1000*5,916 /3800
P [ x > 59000} = - 5916/3800
P [ x > 59000} = - 1,55
We look for p value for that z score n z-table and find
P [ x > 59000} = 0,6057
PLS HELP ASAP!!!!........
Answer:
aaaaha pues
Step-by-step explanation:
Answer:
what happened
Step-by-step explanation:
What number must you add to complete the square?
X^2 + 8x= 11
A. 12
B. 16
c. 8
D. 4
Answer:
16
Step-by-step explanation:
X^2 + 8x= 11
Take the coefficient of x
8
Divide by 2
8/2 =4
Square it
4^2 = 16
Add 16 to each side
F (X) = x² - 2x and 6(x) = 3x+1
A) Find F(g(-4))
B) Find F(g(x)) simply
C) find g^-1 (x)
Part A
g(x) = 3x+1
g(-4) = 3(-4)+1 ... every x replaced with -4
g(-4) = -12+1
g(-4) = -11
Plug this into the f(x) function
f(x) = x^2 - 2x
f( g(-4) ) = (g(-4))^2 - 2( g(-4) )
f( g(-4) ) = (-11)^2 - 2(-11)
f( g(-4) ) = 121 + 22
f( g(-4) ) = 143 is the answer====================================================
Part B
Plug the g(x) function into the f(x) function
f(x) = x^2 - 2x
f( g(x) ) = ( g(x) )^2 - 2( g(x) ) ... replace every x with g(x)
f( g(x) ) = (3x+1)^2 - 2(3x+1)
f( g(x) ) = (9x^2+6x+1) + (-6x-2)
f( g(x) ) = 9x^2+6x+1-6x-2
f( g(x) ) = 9x^2-1 is the answerNote that we can plug x = -4 into this result and we would get
f( g(x) ) = 9x^2-1
f( g(-4) ) = 9(-4)^2-1
f( g(-4) ) = 9(16)-1
f( g(-4) ) = 144-1
f( g(-4) ) = 143 which was the result from part A
====================================================
Part C
Replace g(x) with y. Then swap x and y. Afterward, solve for y to get the inverse.
[tex]g(x) = 3x+1\\\\y = 3x+1\\\\x = 3y+1\\\\3y+1 = x\\\\3y = x-1\\\\y = \frac{1}{3}(x-1)\\\\y = \frac{1}{3}x-\frac{1}{3}\\\\g^{-1}(x) = \frac{1}{3}x-\frac{1}{3}\\\\[/tex]
Researchers wanted to know whether it is better to give the diphtheria, tetanus and pertussis (DTaP) vaccine in the thigh or the arm. They collect data on severe reactions to this vaccine in children aged 3 to 6 years old. What would be the best statistical test for them to utilize?
A. One-sample chi-square
B. Linear regression
C. T-test
D. Two-sample chi-square
Answer:
D. Two-sample chi-square
Step-by-step explanation:
A chi-square test is a test used to compare the data that is observed, from the data that is expected.
In a two-sample chi-square test the observed data should be similar to the expected data if the two data samples are from the same distribution.
The hypotheses of the two-sample chi-square test is given as:
H0: The two samples come from a common distribution.
Ha: The two samples do not come from a common distribution
Therefore, in this case, the best statistical test to utilize is the two-sample chi-square test.
please i need this answer right now !!!! Dx
Answer: the answer is d sin30degrees equal 5/x because sin is opposite over hyponuese
Find the value of c such that the three points (5,5), (-3,1), and (6,c) lie on the same line. Note: Three points are on the same line if the slope of the line through any two points is always the same.
Answer:
c = 5.5
Step-by-step explanation:
We can find the slope of the line using the given points (5,5) and (-3,1) using rise over run:
-4/-8 = 1/2
Now, we can plug in the slope and a point into the equation y = mx + b to find b:
5 = 1/2(5) + b
5 = 2.5 + b
2.5 = b
Then, we can plug in 6 in the point (6,c) to find c:
y = (1/2)(6) + 2.5
y = 3 + 2.5
y = 5.5
c = 5.5
Answer:
c = 5.5
Step-by-step explanation:
Find the slope with two points
m = (y2-y1)/(x2-x1)
m = (1-5)/(-3-5)
= -4/-8
= 1/2
If all the points are on the same line, then they have the same slope
m = (y2-y1)/(x2-x1)
Using the first and third points
1/2 = (c-5)/(6-5)
1/2 = (c-5)/1
1/2 = c-5
Add 5 to each side
5+1/2 = c
5.5 =c
The following observations were made on fracture toughness of a base plate of 18% nickel maraging steel (in ksi √in, given in increasing order)].
68.6 71.9 72.6 73.1 73.3 73.5 75.5 75.7 75.8 76.1 76.2
76.2 77.0 77.9 78.1 79.6 79.8 79.9 80.1 82.2 83.7 93.4
Calculate a 90% CI for the standard deviation of the fracture toughness distribution. (Give answer accurate to 2 decimal places.)
Answer:
A 90% confidence interval for the standard deviation of the fracture toughness distribution is [4.06, 6.82].
Step-by-step explanation:
We are given the following observations that were made on fracture toughness of a base plate of 18% nickel maraging steel below;
68.6, 71.9, 72.6, 73.1, 73.3, 73.5, 75.5, 75.7, 75.8, 76.1, 76.2, 76.2, 77.0, 77.9, 78.1, 79.6, 79.8, 79.9, 80.1, 82.2, 83.7, 93.4.
Firstly, the pivotal quantity for finding the confidence interval for the standard deviation is given by;
P.Q. = [tex]\frac{(n-1) \times s^{2} }{\sigma^{2} }[/tex] ~ [tex]\chi^{2} __n_-_1[/tex]
where, s = sample standard deviation = [tex]\sqrt{\frac{\sum (X - \bar X^{2}) }{n-1} }[/tex] = 5.063
[tex]\sigma[/tex] = population standard deviation
n = sample of observations = 22
Here for constructing a 90% confidence interval we have used One-sample chi-square test statistics.
So, 90% confidence interval for the population standard deviation, [tex]\sigma[/tex] is ;
P(11.59 < [tex]\chi^{2}__2_1[/tex] < 32.67) = 0.90 {As the critical value of chi at 21 degrees
of freedom are 11.59 & 32.67}
P(11.59 < [tex]\frac{(n-1) \times s^{2} }{\sigma^{2} }[/tex] < 32.67) = 0.90
P( [tex]\frac{ 11.59}{(n-1) \times s^{2}}[/tex] < [tex]\frac{1}{\sigma^{2} }[/tex] < [tex]\frac{ 32.67}{(n-1) \times s^{2}}[/tex] ) = 0.90
P( [tex]\frac{(n-1) \times s^{2} }{32.67 }[/tex] < [tex]\sigma^{2}[/tex] < [tex]\frac{(n-1) \times s^{2} }{11.59 }[/tex] ) = 0.90
90% confidence interval for [tex]\sigma^{2}[/tex] = [ [tex]\frac{(n-1) \times s^{2} }{32.67 }[/tex] , [tex]\frac{(n-1) \times s^{2} }{11.59 }[/tex] ]
= [ [tex]\frac{21 \times 5.063^{2} }{32.67 }[/tex] , [tex]\frac{21 \times 5.063^{2} }{11.59 }[/tex] ]
= [16.48 , 46.45]
90% confidence interval for [tex]\sigma[/tex] = [[tex]\sqrt{16.48}[/tex] , [tex]\sqrt{46.45}[/tex] ]
= [4.06 , 6.82]
Therefore, a 90% confidence interval for the standard deviation of the fracture toughness distribution is [4.06, 6.82].
a silver coin is dropped from the top of a building that is 64 feet tall. the position function of the coin at time t seconds is represented by
Question:
A silver coin is dropped from the top of a building that is 64 feet tall. the position function of the coin at time t seconds is represented by
s(t) = -16t² + v₀t + s₀
Determine the position and velocity functions for the coin.
Answer:
position function: s(t) = (-16t² + 64) ft
velocity function: v(t) = (-32t) ft/s
Step-by-step explanation:
Given position equation;
s(t) = -16t² + v₀t + s₀ ---------(i)
v₀ and s₀ are the initial values of the velocity and position of the coin respectively.
(a) Since the coin is dropped, the initial velocity, v₀, of the coin is 0 at t = 0. i.e
v₀ = 0.
Also since the drop is from the top of a building that is 64 feet tall, this implies that the initial position, s₀, of the coin is 64 ft at t=0. i.e
s₀ = 64ft
Substitute the values of v₀ = 0 and s₀ = 64 into equation (i) as follows;
s(t) = -16t² + (0)t + 64
s(t) = -16t² + 64
Therefore, the position function of the coin is;
s(t) = (-16t² + 64) ft
(b) To get the velocity function, v(t), the position function, s(t), calculated above is differentiated with respect to t as follows;
v(t) = [tex]\frac{ds(t)}{dt}[/tex]
v(t) = [tex]\frac{d(-16t^2 + 64)}{dt}[/tex]
v(t) = -32t + 0
v(t) = -32t
Therefore, the velocity function of the coin is;
v(t) = (-32t) ft/s
solve and find the value of (1.7)^2
Answer:
2.89
Step-by-step explanation:
just do 1.7×1.7=2.89
1. Lisa is a regional manager for a restaurant chain that has locations in the towns of Berwick, Milton, and Leesburg. She would like to investigate if a difference exists in the proportion of customers who rate their experience as satisfactory or better between the three locations. The following data represent the number of customers who indicated they were satisfied from random samples taken at each location.
Berwick Milton Leesburg
Number Satisfied 80 85 60
Sample Size 100 120 80
The expected frequency of satisfied customers from the Berwick sample is________.
a. 60
b. 75
c. 80
d. 90
2. Lisa is a regional manager for a restaurant chain that has locations in the towns of Berwick, Milton, and Leesburg. She would like to investigate if a difference exists in the proportion of customers who rate their experience as satisfactory or better between the three locations. The following data represent the number of customers who indicated they were satisfied from random samples taken at each location.
Berwick Milton Leesburg
Number Satisfied 80 85 60
Sample Size 100 120 80
The expected frequency of satisfied customers from the Milton sample is________.
a. 60
b. 75
c. 80
d. 90
3. Lisa is a regional manager for a restaurant chain that has locations in the towns of Berwick, Milton, and Leesburg. She would like to investigate if a difference exists in the proportion of customers who rate their experience as satisfactory or better between the three locations. The following data represent the number of customers who indicated they were satisfied from random samples taken at each location.
Berwick Milton Leesburg
Number Satisfied 80 85 60
Sample Size 100 120 80
The expected frequency of satisfied customers from the Leesburg sample is________.
a. 60
b. 75
c. 80
d. 90
4. Lisa is a regional manager for a restaurant chain that has locations in the towns of Berwick, Milton, and Leesburg. She would like to investigate if a difference exists in the proportion of customers who rate their experience as satisfactory or better between the three locations. The following data represent the number of customers who indicated they were satisfied from random samples taken at each location.
Berwick Milton Leesburg
Number Satisfied 80 85 60
Sample Size 100 120 80
The chi-square test statistic for these samples is_______.
a. 1.49
b. 2.44
c. 4.15
d. 5.33
5. Lisa is a regional manager for a restaurant chain that has locations in the towns of Berwick, Milton, and Leesburg. She would like to investigate if a difference exists in the proportion of customers who rate their experience as satisfactory or better between the three locations. The following data represent the number of customers who indicated they were satisfied from random samples taken at each location.
Berwick Milton Leesburg
Number Satisfied 80 85 60
Sample Size 100 120 80
The degrees of freedom for the chi-square critical value is_______.
a. 1
b. 2
c. 3
d. 4
6. Lisa is a regional manager for a restaurant chain that has locations in the towns of Berwick, Milton, and Leesburg. She would like to investigate if a difference exists in the proportion of customers who rate their experience as satisfactory or better between the three locations. The following data represent the number of customers who indicated they were satisfied from random samples taken at each location.
Berwick Milton Leesburg
Number Satisfied 80 85 60
Sample Size 100 120 80
The chi-square critical value using alpha = 0.05 is_______.
a. 2.706
b. 3.841
c. 5.991
d. 7.815
7. Lisa is a regional manager for a restaurant chain that has locations in the towns of Berwick, Milton, and Leesburg. She would like to investigate if a difference exists in the proportion of customers who rate their experience as satisfactory or better between the three locations. The following data represent the number of customers who indicated they were satisfied from random samples taken at each location.
Berwick Milton Leesburg
Number Satisfied 80 85 60
Sample Size 100 120 80
Using alpha = 0.05, the conclusion for this chi-square test would be that because the test statistic is
A. More than the critical value, we can reject the null hypothesis and conclude that there is a difference in proportion of satisfied customers between these three locations.
B. Less than the critical value, we can reject the null hypothesis and conclude that there is a difference in proportion of satisfied customers between these three locations.
C. More than the critical value, we fail to reject the null hypothesis and conclude that there is no difference in proportion of satisfied customers between these three locations.
D. Less than the critical value, we fail to reject the null hypothesis and conclude that there is no difference in proportion of satisfied customers between these three locations.
Answer:
1) Option B is correct.
Expected frequency of satisfied customers from the Berwick sample = 75
2) Option D is correct.
Expected frequency of satisfied customers from the Milton sample = 90
3) Option A is correct.
Expected frequency of satisfied customers from the Leesburg sample = 60
4) Option B is correct.
The chi-square test statistic for these samples = 2.44
5) Option B is correct.
The degrees of freedom for the chi-square critical value = 2
6) Option C is correct.
The chi-square critical value using alpha = 0.05 is 5.991
7) Option D is correct.
The conclusion for this chi-square test would be that because the test statistic is less than the critical value, we fail to reject the null hypothesis and conclude that there is no difference in proportion of satisfied customers between these three locations.
Step-by-step explanation:
Berwick Milton Leesburg
Number Satisfied 80 85 60
Unsatisfied 20 35 20
Sample Size 100 120 80
Since this is a chi test that aims to check if there are differences in the proportion of expected number of customers for each location, we state the null and alternative hypothesis first.
The null hypothesis usually counters the claim we hope to test and would be that there is no difference between the proportion of expected frequency of satisfied customers at the three locations.
The alternative hypothesis confirms the claim we want to test and is that there is a significant difference between the proportion of expected frequency of satisfied customers at the three locations.
So, the total proportion of satisfied customers is used to calculate the expected number of satisfied customers for each of the three locations.
80+85+60= 225
Total number of customers = 100 + 120 + 80 = 300
Proportion of satisfied customers = (225/300) = 0.75
1) Expected frequency of satisfied customers from the Berwick sample = np = 100 × 0.75 = 75
2) Expected frequency of satisfied customers from the Milton sample = np = 120 × 0.75 = 90
3) Expected frequency of satisfied customers from the Leesburg sample = np = 80 × 0.75 = 60
4) Berwick Milton Leesburg
Number Satisfied 80 85 60
Unsatisfied 20 35 20
Sample Size 100 120 80
Proportion for unsatisfied ccustomers = 0.25
So, expected number of unsatisfied customers for the three branches are 25, 30 and 20 respectively.
Chi square test statistic is a sum of the square of deviations from the each expected value divided by the expected value.
χ² = [(X₁ - ε₁)²/ε₁] + [(X₂ - ε₂)²/ε₂] + [(X₃ - ε₃)²/ε₃] + [(X₄ - ε₄)²/ε₄] + [(X₅ - ε₅)²/ε₅] + [(X₆ - ε₆)²/ε₆]
X₁ = 80, ε₁ = 75
X₂ = 85, ε₂ = 90
X₃ = 60, ε₃ = 60
X₄ = 20, ε₄ = 25
X₅ = 35, ε₅ = 30
X₆ = 20, ε₆ = 20
χ² = [(80 - 75)²/75] + [(85 - 90)²/90] + [(60 - 60)²/60] + [(20 - 25)²/25] + [(35 - 30)²/30] + [(20 - 20)²/20]
χ² = 0.3333 + 0.2778 + 0 + 1 + 0.8333 + 0 = 2.4444 = 2.44
5) The degree of freedom for a chi-square test is
(number of rows - 1) × (number of columns - 1)
= (2 - 1) × (3 - 1) = 1 × 2 = 2
6) Using the Chi-square critical value calculator for a degree of freedom of 2 and a significance level of 0.05, the chi-square critical value is 5.991.
7) Interpretation of results.
If the Chi-square test statistic is less than the critical value, we fail to reject the null hypothesis.
If the Chi-square test statistic is unusually large and is greater than the critical value, we reject the null hypothesis.
For this question,
Chi-square test statistic = 2.44
Critical value = 5.991
2.44 < 5.991
test statistic < critical value
The test statistic is Less than the critical value, we fail to reject the null hypothesis and conclude that there is no difference in proportion of satisfied customers between these three locations.
Hope this Helps!!!
forex is the name of the U.S. stock exchange.
-true
-false
Answer:
false
Step-by-step explanation:
hello
this is false
FOREX means Foreign Exchange
it refers to the foreign exchange market
hope this helps
Answer:
true, forex trading is a profitable than staking cryptocurrency. forex trading is the best thing I will refer someone I love because learning never stops and no on is above blowing accounts when beginning Forex