Divide up the number 480 in a ratio of 3:5.

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

Answer:

50

Step-by-step explanation:

The probability of it landing on yellow or blue is 5 out of 7 total possibilities.

5/7

Multiply by 70.

5/7 × 70

350/7

= 50

Answer:

50 times

Step-by-step explanation:

Yellow or blue is 5/7 of the spinner

Multiply 70 by 5/7 to find the prediction of the number of times it will land there

70(5/7) = 50

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)

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]

Answer:

20gallons

Step-by-step explanation:

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.

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.

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.

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.

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.

Answer:

$23.64

Step-by-step explanation:

12 * $1.97 = $23.64

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

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.

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

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

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

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

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.

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

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

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

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

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

Answer:

Step-by-step explanation:

its B

Answer:

the answer is B

Step-by-step explanation:

Answer:

[tex] = \frac{3}{2} [/tex]

Step-by-step explanation:

[tex]y = mx + c[/tex]

Here,

In 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]

Answer:

the 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

Answer:

that means each side equals 8

Step-by-step explanation:

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)

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

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