Answer:
6 ties
Step-by-step explanation:
At least 37 but fewer than 63 makes it 37-62
So,
37-62 => 6 ties
A homogeneous second-order linear differential equation, two functions y 1y1 and y 2y2, and a pair of initial conditions are given. First verify that y 1y1 and y 2y2 are solutions of the differential equation. Then find a particular solution of the form y = c1y1 + c2y2 that satisfies the given initial conditions. Primes denote derivatives with respect to x.
y'' + 49y = 0; y1 = cos(7x) y2 = sin(7x); y(0) = 10 y(0)=-4
1.Why is the function y, = e * a solution to the differential equation?
A. The function y1 =e 4X is a solution because when the function and its indefinite integral, , are substituted into the equation, the result is a true statement.
B. The function y1 = e 4X is a solution because when the function and its second derivative, y1" = 16 e 4x, are substituted into the equation, the result is a true statement.
2. Why is the function y2 solution the differential equation?
A. The function y2 = e 4x is a solution because when the function and its indefinite integral, are substituted into the equation, the result a true statement. The function y2 = e 4X is a solution because when the function and its second derivative, y2" = 16 e -4x are substituted into the equation, the result is a true statement. The particular solution of the form y = c, y, +c,y2 that satisfies the initial conditions y(0) 2 and y'(0) = 9 is y =.
Answer:
[tex]y = 10cos (7x) - \frac{4}{7}sin ( 7x )[/tex]
B.
B.
[tex]y = \frac{17}{8}e^4^x - \frac{1}{8}e^-^4^x[/tex]
Step-by-step explanation:
Question 1:
- We are given a homogeneous second order linear ODE as follows:
[tex]y'' + 49y = 0[/tex]
- A pair of independent functions are given as ( y1 ) and ( y2 ):
[tex]y_1 = cos ( 7x )\\\\y_2 = sin ( 7x )[/tex]
- The given ODE is subjected to following initial conditions as follows:
[tex]y ( 0 ) = 10\\\\y ' ( 0 ) = -4[/tex]
- We are to verify that the given independent functions ( y1 ) and ( y2 ) are indeed the solution to the given ODE. If the functions are solutions then find the complete solution of the homogeneous ODE of the form:
[tex]y = c_1y_1 + c_2y_2[/tex]
Solution:-
- To verify the functions are indeed the solution to the given ODE. We will plug the respective derivatives of each function [ y1 and y2 ] into the ODE and prove whether the equality holds true or not.
- Formulate the second derivatives of both functions y1 and y2 as follows:
[tex]y'_1 = -7sin(7x) , y''_1 = -49cos(7x)\\\\y'_2 = -7cos(7x) , y''_2 = -49sin(7x)\[/tex]
- Now plug the second derivatives of each function and the functions itself into the given ODE and verify whether the equality holds true or not.
[tex]y''_1 + 49y_1 = 0\\\\-49cos(7x) + 49cos ( 7x ) = 0\\0 = 0\\\\y''_2 + 49y_2 = 0\\\\-49sin(7x) + 49sin ( 7x ) = 0\\0 = 0\\\\[/tex]
- We see that both functions [ y1 and y2 ] holds true as the solution to the given homogeneous second order linear ODE. Hence, are the solution to given ODE.
- The complete solution to a homogeneous ODE is given in the form as follows:
[tex]y = c_1y_1 + c_2y_2\\\\y = c_1*cos(7x) + c_2*sin(7x)\\[/tex]
- To complete the above solution we need to determine the constants [ c1 and c2 ] using the initial conditions given. Therefore,
[tex]y (0) = c_1cos ( 0 ) + c_2sin ( 0 ) = 10\\\\y'(0) = -7c_1*sin(0) + 7c_2*cos(0) = -4\\\\c_1 ( 1 ) + c_2 ( 0 ) = 10, c_1 = 10\\\\-7c_1(0) + 7c_2( 1 ) = -4 , c_2 = -\frac{4}{7}[/tex]
- Now we can write the complete solution to the given homogeneous second order linear ODE as follows:
[tex]y = 10cos (7x) - \frac{4}{7}sin ( 7x )[/tex] .... Answer
Question 2
- We are given a homogeneous second order linear ODE as follows:
[tex]y'' -16y =0[/tex]
- A pair of independent functions are given as ( y1 ) and ( y2 ):
[tex]y_1 = e^4^x\\\\y_2 = e^-^4^x[/tex]
- The given ODE is subjected to following initial conditions as follows:
[tex]y( 0 ) = 2\\y'( 0 ) = 9[/tex]
- We are to verify that the given independent functions ( y1 ) and ( y2 ) are indeed the solution to the given ODE. If the functions are solutions then find the complete solution of the homogeneous ODE of the form:
[tex]y = c_1y_1 + c_2y_2[/tex]
Solution:-
- To verify the functions are indeed the solution to the given ODE. We will plug the respective derivatives of each function [ y1 and y2 ] into the ODE and prove whether the equality holds true or not.
- Formulate the second derivatives of both functions y1 and y2 as follows:
[tex]y'_1 = 4e^4^x , y''_1 = 16e^4^x\\\\y'_2 = -4e^-^4^x , y''_2 = 16e^-^4^x[/tex]
- Now substitute the second derivatives of each function and the functions itself into the given ODE and verify whether the equality holds true or not.
[tex]y''_1 - 16y_1 = 0\\\\16e^4^x - 16e^4^x = 0\\\\0 = 0\\\\y''_2 - 16y_2 = 0\\\\16e^-^4^x - 16e^-^4^x = 0\\\\0 = 0[/tex]
- We see that both functions [ y1 and y2 ] holds true as the solution to the given homogeneous second order linear ODE. Hence, are the solution to given ODE.
- The complete solution to a homogeneous ODE is given in the form as follows:
[tex]y = c_1y_1 + c_2y_2\\\\y = c_1*e^4^x + c_2*e^-^4^x[/tex]
- To complete the above solution we need to determine the constants [ c1 and c2 ] using the initial conditions given. Therefore,
[tex]y ( 0 ) = c_1 * e^0 + c_2 * e^0 = 2\\\\y' ( 0 ) = 4 c_1 * e^0 - 4c_2 * e^0 = 9\\\\c_1 + c_2 = 2 , 4c_1 - 4c_2 = 9\\\\c_1 = \frac{17}{8} , c_2 = -\frac{1}{8}[/tex]
- Now we can write the complete solution to the given homogeneous second order linear ODE as follows:
[tex]y = \frac{17}{8} e^4^x - \frac{1}{8}e^-^4^x[/tex] .... Answer
g In a large midwestern university (the class of entering freshmen is 6000 or more students) respectively, who graduated in the bottom third of their high school class. A 99% confidence interval for p1 – p2 is:
Answer:
–0.029 to 0.229.
Step-by-step explanation:
So, we are given the following data or information or values/parameters which are going to help us in solving this particular equation:
=>" A class of entering freshmen = 6000 or more students) respectively"
=> "The class of entering freshmen graduated in the bottom third of their high school class."
=>" 99% confidence interval for p1 – p2"
Let p1 = k1 and p2 = k2
Here, we can deduce that p1 > p2; k1 > k2. Hence,
a = (1 - 0.99)/2 = 0.005.
b = 513 × 0.005 = 2.6.
c = standard deviation = ✓ [ k1 (1 - k1) / j1 + k2 (1 - k2) / j2] = 0.05.
99% confidence interval for p1 – p2 =
k1 - k2 - b × c = –0.029
Also, k1 - k2 + b × c = 0.029.
Which are the lower and upper boundaries respectively.
Find an explicit formula for the following sequence, an which starts with a1=−1. −1,1/2,−1/3,1/4,−1/5,…
Answer:
The sequence can be represented by the formula of its nth term:
[tex]a_n=\frac{(-1)^n}{n}[/tex]
Step-by-step explanation:
Notice that we are in the presence of an alternate sequence (the values alternate from negative to positive. Therefore we need to take into account that there should be a factor "-1" raised to the "n" value for the sequence. Also, given that the sequence looks in absolute value like the harmonic sequence, we conclude upon the following general form for the "nth" term of the sequence:
[tex]a_n=\frac{(-1)^n}{n}[/tex]
What is the slope of a line that is perpendicular to the line 2y – 3x = 8?
Answer:
[tex] = \frac{3}{2} [/tex]
Step-by-step explanation:
[tex]y = mx + c[/tex]
Here,
m => slopec => interceptIn this equation ,
[tex]2y - 3x = 8[/tex]
to find the value of m or the value of slope we have to solve for y
Let's solve,
[tex]2y - 3x = 8 \\ 2y = 8 + 3x \\ \frac{2y}{2} = \frac{8 + 3x}{2} \\ y = 4 + \frac{3x}{2} \\ y = \frac{3x}{2} + 4[/tex]
So, the slope is,
[tex] = \frac{3}{2}[/tex]
A college surveys 300 graduates and finds 98 graduated with honors and 207 had one or both parents graduate from college. Of the 98 students with honors, 79 had one or both parents graduate from college. Find the probability that a randomly chosen graduate from these 300 graduated with honors given that neither parent graduated from college.
Answer:
20.43% probability that a randomly chosen graduate from these 300 graduated with honors given that neither parent graduated from college.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
Graduated with honors:
98 students graduated with honors. Of those, 79 had at least one parent graduating from college. So 98 - 79 = 19 did not.
Of 300 students, 207 had one or both parents graduate from college. So 300 - 207 = 93 did not have at least one parent graduating.
Find the probability that a randomly chosen graduate from these 300 graduated with honors given that neither parent graduated from college.
Of the 93 with no graduated parent, 19 earned honors
19/93 = 0.2043
20.43% probability that a randomly chosen graduate from these 300 graduated with honors given that neither parent graduated from college.
The owner of a small machine shop has just lost one of his largest customers. The solution to his problem,he says, is to fire three machinists to balance his workforce with his current level of business. The owner says that it is a simple problem with a simple solution. The three machinists disagree. Why
Answer:
It may look simple to the owner because he is not the one losing a job. For the three machinists it represents a major event with major consequences
Which lists the ratios in order from least to greatest 2:9,1:6,7:36
Answer:
1:6, 7:36, 2:9
Step-by-step explanation:
2 : 9 → 8 : 36
1 : 6 → 6 : 36
7 : 36
Least → Greatest
1:6, 7:36, 2:9
Decide whether the method of undetermined coefficients together with superposition can be applied to find a particular solution of the given equation. Do not solve the equation Can the method of undetermined coefficients together with superposition be applied to find a particular solution of the given equation?
A. No, because the right side of the given equation is not the correct type of function
B, Yes °
C. No, because the differential equation is not linear.
D. No, because the differential equation does not have constant coefficients.
Answer:
D. No, because the differential equation does not have constant coefficients.
Step-by-step explanation:
The undetermined coefficient method cannot be applied to non homogeneous variables. The differential equation does not have constant variables therefore the method of undetermined superposition can not be applied. To complete a solution of non homogeneous equation the particular solution must be added to the homogeneous equation.
Consider the diagram and the proof below.
Given: In △ABC, AD ⊥ BC
Prove: StartFraction sine (uppercase B) Over b EndFraction = StartFraction sine (uppercase C) Over c EndFraction
Triangle A B C is shown. A perpendicular bisector is drawn from point A to point D on side C B forming a right angle. The length of A D is h, the length of C B is a, the length of C A is b, and the length of A B is c.
A 2-column table has 7 rows. The first column is labeled Statement with entries In triangle A B C line segment A D is perpendicular to line segment B C, In triangle A D B sine (uppercase B) = StartFraction h Over c EndFraction, c sine (uppercase B) = h, In triangle A C D, sine (uppercase C) = StartFraction h Over b EndFraction, b sine (uppercase C) = h, question mark, StartFraction sine (uppercase B) Over b EndFraction = StartFraction sine (uppercase C) Over c EndFraction. The second column is labeled Reason with entries given, definition of sine, multiplication property of equality, definition of sine, multiplication property of equality, substitution, and division property of equality.
What is the missing statement in Step 6?
b = c
StartFraction h Over b EndFraction = StartFraction h Over c EndFraction
csin(B) = bsin(C)
bsin(B) = csin(C)
Answer:
c- the right triangle altitude theorem
Step-by-step explanation:
i did it on edge! ; )
The missing statement in Step 6 is ,c- The right triangle altitude theorem.
We have given that,
In △ABC, AD ⊥ BC
Prove: StartFraction sine (uppercase B) Over b EndFraction = StartFraction sine (uppercase C) Over c EndFraction.
Triangle A B C is shown.
What is the right triangle altitude theorem?The right triangle altitude theorem or geometric mean theorem is a result in elementary geometry that describes a relation between the altitude on the hypotenuse in a right triangle and the two-line segments it creates on the hypotenuse
Therefore we have,
A perpendicular bisector is drawn from point A to point D on side C B forming a right angle.
The length of A D is h, the length of C B is a, the length of C A is b, and the length of A B is c.
So the missing statement in Step 6
b = c
c=The right triangle altitude theorem.
To learn more about the right triangle altitude theorem visit:
https://brainly.com/question/723406
Points a, b, and c are midpoints of the sides of right triangle def. Which statements are true select three options. A B C D E
Answer : The correct statements are,
AC = 5 cm
BA = 4 cm
The perimeter of triangle ABC is 12 cm.
Step-by-step explanation :
As we know that a, b, and c are midpoints of the sides of right triangle that means midpoint divide the side in equal parts.
Now we have to calculate the sides of triangle ABC by using Pythagoras theorem.
Using Pythagoras theorem in ΔACF :
[tex](AC)^2=(FA)^2+(CF)^2[/tex]
Now put all the values in the above expression, we get the value of side AC.
[tex](AC)^2=(3)^2+(4)^2[/tex]
[tex]AC=\sqrt{(9)^2+(16)^2}[/tex]
[tex]AC=5cm[/tex]
Using Pythagoras theorem in ΔDAB :
[tex](Hypotenuse)^2=(Perpendicular)^2+(Base)^2[/tex]
[tex](BD)^2=(AD)^2+(BA)^2[/tex]
Now put all the values in the above expression, we get the value of side BA.
[tex](5)^2=(3)^2+(BA)^2[/tex]
[tex]BA=\sqrt{(5)^2-(3)^2}[/tex]
[tex]BA=4cm[/tex]
Using Pythagoras theorem in ΔBEC :
[tex](Hypotenuse)^2=(Perpendicular)^2+(Base)^2[/tex]
[tex](BE)^2=(CE)^2+(CB)^2[/tex]
Now put all the values in the above expression, we get the value of side CB.
[tex](5)^2=(4)^2+(CB)^2[/tex]
[tex]CB=\sqrt{(5)^2-(4)^2}[/tex]
[tex]CB=3cm[/tex]
Now we have to calculate the perimeter of ΔABC.
Perimeter of ΔABC = Side AB + Side CB+ Side AC
Perimeter of ΔABC = 4 + 3 + 5
Perimeter of ΔABC = 12 cm
Now we have to calculate the area of ΔABC.
Area of ΔABC = [tex]\frac{1}{2}\times 4\times 3=6cm^2[/tex]
Now we have to calculate the area of ΔDEF.
Area of ΔDEF = [tex]\frac{1}{2}\times 8\times 6=24cm^2[/tex]
Area of ΔABC = [tex]\frac{6}{24}\times[/tex] Area of ΔDEF
Area of ΔABC = [tex]\frac{1}{4}[/tex] Area of ΔDEF
A real estate agent has 1313 properties that she shows. She feels that there is a 40%40% chance of selling any one property during a week. The chance of selling any one property is independent of selling another property. Compute the probability of selling at least 11 property in one week. Round your answer to four decimal places.
Answer:
0.0013
Step-by-step explanation:
The probability of selling a property is 40%, so the probability of not selling it is 60%.
To find the probability of selling at least 11 properties, we can calculate the following cases:
Selling 11:
P(11) = C(13,11) * P(sell)^11 * P(not sell)^2
P(11) = (13! / (11! * 2!)) * 0.4^11 * 0.6^2
P(11) = 13*12/2 * 0.4^11 * 0.6^2 = 0.001178
Selling 12:
P(12) = C(13,12) * P(sell)^12 * P(not sell)^1
P(11) = (13! / (12! * 1!)) * 0.4^12 * 0.6^1
P(11) = 13 * 0.4^12 * 0.6 = 0.000131
Selling 13:
P(13) = C(13,13) * P(sell)^13 * P(not sell)^0
P(11) = 1 * 0.4^13 * 0.6^0
P(11) = 1 * 0.4^13 * 1 = 0.000007
Final probability:
P(at least 11) = P(11) + P(12) + P(13)
P(at least 11) = 0.001178 + 0.000131 + 0.000007 = 0.001316
P(at least 11) = 0.0013
Find the volume of the cone.
Diameter: 20 m, Slant Height: 26 m
Round to the nearest whole number.
Volume
=
[?] m3
Answer:
2513the step-by-step explanation for height first :
[tex]h=\sqrt{h^{2} } +r^{2} =26[/tex]
[tex]h=\sqrt{h^{2} } +10^{2} =676[/tex]
[tex]h=\sqrt{h^{2} } + 100 = 676[/tex]
[tex]100-100 = 0[/tex]
[tex]676-100=576[/tex]
[tex]\sqrt{576}[/tex]
[tex]height =[/tex] 24 m
________________
step-by-step explanation for the problem :
FORMULA : [tex]v = \frac{1}{3}[/tex] · [tex]\pi[/tex] · [tex]r^{2}[/tex] · [tex]h[/tex]
v = [tex]\frac{1}{3}[/tex] · [tex]\pi[/tex] · [tex]10^{2}[/tex] · [tex]24[/tex] = [tex]800\pi[/tex] = [tex]2513.27412[/tex] = 2513
Consider the set of sequences of seven letters chosen from W and L. We may think of these sequences as representing the outcomes of a match of seven games, where W means the first team wins the game and L means the second team wins the game. The match is won by the first team to win four games (thus, some games may never get played, but we need to include their hypothetical outcomes in the points in order that we have a probability space of equally likely points).A. What is the probability that a team will win the match, given that it has won the first game?B. What is the probability that a team will win the match, given that it has won the first two games? C. What is the probability that a team will win the match, given that it has won two out of the first three games?
Answer:
a) Probability that a team will win the match given that it has won the first game = 0.66
b) Probability that a team will win the match given that it has won the first two games= 0.81
c) Probability that a team will win the match, given that it has won two out of the first three games = 0.69
Step-by-step explanation:
There are a total of seven games to be played. Therefore, W and L consists of 2⁷ equi-probable sample points
a) Since one game has already been won by the team, there are 2⁶ = 64 sample points left. If the team wins three or more matches, it has won.
Number of ways of winning the three or more matches left = [tex]6C3 + 6C4 + 6C5 + 6C6[/tex]
= 20 + 15 + 6 + 1 = 42
P( a team will win the match given that it has won the first game) = 42/64 = 0.66
b) Since two games have already been won by the team, there are 2⁵ = 32 sample points left. If the team wins two or more matches, it has won.
Number of ways of winning the three or more matches left = [tex]5C2 + 5C3 + 5C4 + 5C5[/tex] = 10 + 10 + 5 +1 = 26
P( a team will win the match given that it has won the first two games) = 26/32 = 0.81
c) Probability that a team will win the match, given that it has won two out of the first three games
They have played 3 games out of 7, this means that there are 4 more games to play. The sample points remain 2⁴ = 16
They have won 2 games already, it means they have two or more games to win.
Number of ways of winning the three or more matches left = [tex]4C2 + 4C3 + 4C4[/tex] = 6 + 4 + 1 = 11
Probability that a team will win the match, given that it has won two out of the first three games = 11/16
Probability that a team will win the match, given that it has won two out of the first three games = 0.69
Insurance companies track life expectancy information to assist in determining the cost of life insurance policies. AIB Insurance randomly sampled 100 recently paid policies and determined the average age of clients in this sample to be 77.7 years with a standard deviation of 3.6. The 90% confidence interval for the true mean age of its life insurance policy holders is
A. (76.87, 80.33)
B. (72.5, 82.9)
C. (77.1, 78.3)
D. (74.1, 81.3)
E. (74.5, 80)
Answer:
[tex]77.7-1.66\frac{3.6}{\sqrt{100}} =77.102[/tex]
[tex]77.7+1.66\frac{3.6}{\sqrt{100}} =78.30[/tex]
And the best option would be:
C. (77.1, 78.3)
Step-by-step explanation:
Information given
[tex]\bar X=77.7[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=3.6 represent the sample standard deviation
n=100 represent the sample size
Confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The degrees of freedom are given by:
[tex]df=n-1=100-1=99[/tex]
Since the Confidence is 0.90 or 90%, the significance would be [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and the critical value for this case would be [tex]t_{\alpha/2}=1.66[/tex]
And replacing we got:
[tex]77.7-1.66\frac{3.6}{\sqrt{100}} =77.10[/tex]
[tex]77.7+1.66\frac{3.6}{\sqrt{100}} =78.30[/tex]
And the best option would be:
C. (77.1, 78.3)
The average lifetime of a set of tires is 3.4 years. The manufacturer will replace any set of tires failing within three years of the date of purchase. The lifetime of these tires is known to follow an exponential distribution. What is the probability that the tires will fail within three years of the date of purchase?
Answer: the probability that the tires will fail within three years of the date of purchase is 0.12
Step-by-step explanation:
The average lifetime of a set of tires is 3.4 years. It means that μ = 3.4
Decay parameter, m = 1/3.4 = 0.294
The probability density function is
f(x) = me^-mx
Where x is a continuous random variable representing the time interval of interest(the reliability period that we are testing)
Since x = 3 years,
Therefore, the probability that the tires will fail within three years of the date of purchase is
f(3) = 0.294e^-(0.294 × 3)
f(3) = 0.294e^- 0.882
f(3) = 0.12
What is the product? (3x-b)(2x^2-7x+1) A. -12x^2+42x-6 B. -12x^2+21x+6 C. 6x^3-33x^2+45x-6 D. 6x^3-27x^2-39x+6
Answer:
C.6x³-33x² + 45x-6
Step-by-step explanation:
(3x-6)(2x^2-7x+1)
= 3x(2x² - 21x +1) -6(2x² - 7x+1)
= (6x³ - 21x² + 3x) - (12x² - 42x+6)
= 6x³ - 21x² + 3x -12x² + 42x -6
= 6x³-33x² + 45x-6
Kaya figured out that she will need $47,592 to attend college. What is the amount rounded to the nearest ten thousand? Help meeee
Answer:
50,000
Step-by-step explanation:
ten thousand thousand hundreds tens ones
4 7 5 9 2
When rounding to the ten thousands, we look at the thousands place
If it is 5 or higher we round the ten thousands place up
7 is five or higher so we round the 4 up one place 4 becomes 5 and the rest becomes 0
5 0 0 0 0
Answer:
$50,000
Step-by-step explanation:
=> $47,592
While rounding off to the nearest thousand, we check the thousands place. If the digit in the thousands place is greater than 5, 1 will be added to the T. Th. place while if its less than 5, there will be no change and The digits except the ten thousands place will all become zero.
So,
=> $50,000
Marie plants 12 packages of vegetable seeds in a community garden. Each package costs $1.97. What is the total cost of the seeds?
Answer:
$23.64
Step-by-step explanation:
12 * $1.97 = $23.64
Click an item in the list or group of pictures at the bottom of the problem and, holding the button down,drag into the correct position in the answer box.Release your mouse button when the item is place. If you change your mind,drag the item to the trash can,click the trash and to clear all your answers. Divide the following polynomials,then place the answer in the proper location on the grid. Write the answer in descending powers of x. 9x^2-18x-7 divided by (3x+1)
Answer:
[tex]\boxed{\sf \ \ 9x^2-18x-7 \ \ divided \ by \ (3x+1) \ is \ (3x-7) \ }[/tex]
Step-by-step explanation:
Hello,
let's find a and b reals so that
[tex]9x^2-18x-7=(3x+1)(ax+b)[/tex]
[tex](3x+1)(ax+b)=3ax^2+(3b+a)x+b[/tex]
we identify the terms in [tex]x^2[/tex]
9 = 3a
we identify the terms in x
-18 = 3b + a
we identify the constant terms
-7 = b
so ti goes with a = 9/3 = 3, b = -7
so we can write
[tex]9x^2-18x-7=(3x+1)(3x-7)[/tex]
so [tex]9x^2-18x-7 \ divided \ by \ (3x+1) \ is \ (3x-7)[/tex]
hope this helps
The weight of high school football players is normally distributed with a mean of 195 pounds and a standard deviation of 20 pounds.The probability of a player weighing more than 238 pounds is a.0.0334 b.0.0486 c.0.0158 d.0.9842
Answer:
c)
The probability of a player weighing more than 238
P( X > 238) = 0.0174
Step-by-step explanation:
Step(i):-
Given mean of the normally distribution = 195 pounds
Given standard deviation of the normally distribution
= 20 pounds.
Let 'x' be the random variable of the normally distribution
Let X = 238
[tex]Z = \frac{x-mean}{S.D} = \frac{238-195}{20} = 2.15[/tex]
Step(ii):-
The probability of a player weighing more than 238
P( X > 238) = P( Z> 2.15)
= 1 - P( Z < 2.15)
= 1 - ( 0.5 + A(2.15)
= 1 - 0.5 - A(2.15)
= 0.5 - 0.4821 ( from normal table)
= 0.0174
The probability of a player weighing more than 238
P( X > 238) = 0.0174
State whether the data described below are discrete or continuous, and explain why.
The exact lengths (in kilometers) of the ocean coastlines of different countries.
a. The data are continuous because the data can only take on specific values.
b. The data are discrete because the data can only take on specific values.
c. The data are continuous because the data can take on any value in an interval.
d. The data are discrete because the data can take on any value in an interval.
Answer:
c. The data are continuous because the data can take on any value in an interval.
Step-by-step explanation:
A variable is said to be continuous if it can take on any value in an interval. Examples are lengths, temperature, etc
A discrete variable, on the other hand, can only take on specific values. Examples of discrete variables are the number of students and age.
The exact lengths (in kilometers) of the ocean coastlines of different countries is a continuous variable because it can take on any value in an interval.
A stated earlier, Lengths are in general, continuous variables.
1/5divided by (-5/7)
Answer:
-0.28
Step-by-step explanation:
(1/5) : (-5/7)=(1*5)/(5*(-5))=-(7/25)=-0.28
Answer:
[tex]-7/25[/tex]
Step-by-step explanation:
[tex]1/5 \div -5/7[/tex]
Do the reciprocal of the second fraction.
[tex]1/5 \times 7/-5[/tex]
Multiply the first fraction by the reciprocal of the second fraction.
[tex]7/-25=-0.28[/tex]
The answer in decimal form is -0.28.
Identify the Type II error if the null hypothesis, H0, is: The capacity of Anna's car gas tank is 10 gallons. And, the alternative hypothesis, Ha, is: Anna believes the capacity of her car's gas tank is not 10 gallons.
Answer:
20gallons
Step-by-step explanation:
Need help ASAP!! thank you sorry if u can’t see it good :(
Answer/Step-by-step explanation:
==>Given:
=>Rectangular Pyramid:
L = 5mm
W = 3mm
H = 4mm
=>Rectangular prism:
L = 5mm
W = 3mm
H = 4mm
==>Required:
a. Volume of pyramid:
Formula for calculating volume of a rectangular pyramid us given as L*W*H
V = 5*3*4
V = 60 mm³
b. Volume of prism = ⅓*L*W*H
thus,
Volume of rectangular prism given = ⅓*5*3*4
= ⅓*60
= 20mm³
c. Volume of the prism = ⅓ x volume of the pyramid
Thus, 20 = ⅓ × 60
As we can observe from our calculation of the solid shapes given, the equation written above is true for all rectangular prism and rectangular pyramid of the same length, width and height.
Which of the following is the equation of the function below?
Answer:
Step-by-step explanation:
its B
Answer:
the answer is B
Step-by-step explanation:
Find the general solution to 3y′′+12y=0. Give your answer as y=... . In your answer, use c1 and c2 to denote arbitrary constants and x the independent variable. Enter c1 as c1 and c2 as c2.
Answer:
[tex]y(x)=c_1e^{2ix}+c_2e^{-2ix}[/tex]
Step-by-step explanation:
You have the following differential equation:
[tex]3y''+12y=0[/tex] (1)
In order to find the solution to the equation, you can use the method of the characteristic polynomial.
The characteristic polynomial of the given differential equation is:
[tex]3m^2+12=0\\\\m^2=-\frac{12}{3}=-4\\\\m_{1,2}=\pm2\sqrt{-1}=\pm2i[/tex]
The solution of the differential equation is:
[tex]y(x)=c_1e^{m_1x}+c_2e^{m_2x}[/tex] (2)
where m1 and m2 are the roots of the characteristic polynomial.
You replace the values obtained for m1 and m2 in the equation (2). Then, the solution to the differential equation is:
[tex]y(x)=c_1e^{2ix}+c_2e^{-2ix}[/tex]
What is the simplified form of square root of 10,000x64 ?
Answer:
800
Step-by-step explanation:
10,000 x 64 = 640,000
Square Root It Makes It
800
Answer:
6,400
Step-by-step explanation:
The square root of 10,000 times 64 is simplified to 6,400
How can you use an equilateral triangle to find the lengths of the sides in a 30-60-90 triangle?
Answer:
Step-by-step explanation:
1) divide equilateral tri from the middle you will get two 30-60-90 triangles
2) by using pythagorean law & trigimintory, you will get two unknowns (height and side length) and two functions
A study of women’s weights found that a randomly selected sample of 234 women had a mean weight of 157.3 lb. Assuming that the population standard deviation is 15.6 lb., construct a 95% confidence interval estimate of the mean weight of all women.
A. (145.3, 160.5)
B. (155.3, 159,3)
C. (165.5, 173.5)
D. (185.7, 199.3)
Answer:
[tex]157.3-1.96\frac{15.6}{\sqrt{234}}=155.301[/tex]
[tex]157.3+1.96\frac{15.6}{\sqrt{234}}=159.299[/tex]
So on this case the 95% confidence interval would be given by (155.301;159.299)
And the best option would be:
B. (155.3, 159,3)
Step-by-step explanation:
Information given
[tex]\bar X=157.3[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma =15.6[/tex] represent the population standard deviation
n=234 represent the sample size
Confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
The Confidence level is is 0.95 or 95%, the significance is [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], the critical value for this case would be [tex]z_{\alpha/2}=1.96[/tex]
And replacing we got:
[tex]157.3-1.96\frac{15.6}{\sqrt{234}}=155.301[/tex]
[tex]157.3+1.96\frac{15.6}{\sqrt{234}}=159.299[/tex]
So on this case the 95% confidence interval would be given by (155.301;159.299)
And the best option would be:
B. (155.3, 159,3)
A robotic machine fills containers with an acid solution at the rate of 50 + 5 t milliliters (mL) per second, where t is in seconds and 0 ≤ t ≤ 60 . How many mL are put into a container in 60 seconds? Evaluate your answer to a whole number.
Answer:
12,000
Step-by-step explanation:
The machine fills the containers at a rate of 50+5t milliliters (mL) per second.
Therefore, the rate of change of the number of containers, N is:
[tex]\dfrac{dN}{dt}=50+5t, 0\leq t\leq 60[/tex]
[tex]dN=(50+5t)dt\\$Taking integrals of both sides\\\int dN=\int (50+5t)dt\\N(t)=50t+\frac{5t^2}{2}+C $(C a constant of integration)\\\\When t=0, , No containers are filled, therefore:$ N(t)=0\\0=50(0)+\frac{5(0)^2}{2}+C\\C=0\\$Therefore, N(t)=50t+2.5t^2[/tex]
When t=60 seconds
[tex]N(60)=50(60)+2.5(60)^2\\N(60)=12000$ mL[/tex]
Therefore, 12,000 milliliters of acid solution are put into a container in 60 seconds.