The score on a trivia game is obtained by subtracting number of incorrect answers from twice the number of correct answers. If a player answered 40 questions and obtained a score of 50, how mary questions did the player answer correctly?

Answer:

a) There are 11 balls left in the bag.

b) 4 of the balls in the bag are red.

c) 3 brown, 2 green, 2 white.

d) [tex]P = \frac{2}{11}[/tex]

e) The probability that the first ball is red and the second is green is:

[tex]\frac{5}{66}[/tex]

The probability that both balls are red is:

[tex]\frac{5}{33}[/tex]

f) The probability that both balls are green is:

[tex]\frac{1}{66}[/tex]

g) The probability that the first ball is brown and the second is white is:

[tex]\frac{1}{22}[/tex]

The probability that the first ball is brown and the second is white is:

[tex]\frac{1}{22}[/tex]

They are the same probabilities.

Step-by-step explanation:

There are 12 balls in the bag

5 red

3 brown

2 green

2 white

If you draw a red ball and put it aside:

a) How many balls are left in the bag?

There were 12 balls in the bag, and one was put aside.

So, there are 11 balls left in the bag.

b) How many of the balls in the bag are red?

There were 5 red balls in the bag, and one was put aside.

So, there are 4 red balls in the bag.

c) How many are brown? Green? White?

The ball put aside was red, so we still have the same number of the balls of the other colors.

3 brown, 2 green, 2 white.

d) Draw a second ball from the bag. What is the probability that it is green?

There are 11 balls in the bag, 2 of which are green. So the probability that the second ball is green is:

[tex]P = \frac{2}{11}[/tex]

Now put both balls back in the bag and draw two balls without replacing them.

e) What is the probability that the first ball is red and the second ball is green (without replacement)?

[tex]P = P_{1}*P_{2}[/tex]

[tex]P_{1}[/tex] is the probability that the first ball is red. There are 12 balls, 5 of which are red. So:

[tex]P_{1} = \frac{5}{12}[/tex]

Since there are no replacements, now there are 11 balls in the bag, 2 of which are green. So:

[tex]P_{2} = \frac{2}{11}[/tex]

The probability that the first ball is red and the second is green is:

[tex]P = P_{1}*P_{2} = \frac{5}{12}*\frac{2}{11} = \frac{5}{66}[/tex]

What is the probability that the first ball is red and the second ball is also red (without replacement)?

[tex]P = P_{1}*P_{2}[/tex]

[tex]P_{1}[/tex] is the probability that the first ball is red. There are 12 balls, 5 of which are red. So:

[tex]P_{1} = \frac{5}{12}[/tex]

Since there are no replacements, now there are 11 balls in the bag, 4 of which are red. So:

[tex]P_{2} = \frac{4}{11}[/tex]

The probability that both balls are red is:

[tex]P = P_{1}*P_{2} = \frac{5}{12}*\frac{4}{11} = \frac{5}{33}[/tex]

f) What is the probability that the first ball is green and the second ball is also green (without replacement)?

[tex]P = P_{1}*P_{2}[/tex]

[tex]P_{1}[/tex] is the probability that the first ball is green. There are 12 balls, 2 of which are green. So:

[tex]P_{1} = \frac{2}{12}[/tex]

Since there are no replacements, now there are 11 balls in the bag, 1 of which is green. So:

[tex]P_{2} = \frac{1}{11}[/tex]

The probability that both balls are green is:

[tex]P = P_{1}*P_{2} = \frac{2}{12}*\frac{1}{11} = \frac{1}{66}[/tex]

g) What is the probability that the first ball is brown and the second ball is white (without replacement)?

[tex]P = P_{1}*P_{2}[/tex]

[tex]P_{1}[/tex] is the probability that the first ball is brown. There are 12 balls, 3 of which are brown. So:

[tex]P_{1} = \frac{3}{12}[/tex]

Since there are no replacements, now there are 11 balls in the bag, 2 of which are white. So

[tex]P_{2} = \frac{2}{11}[/tex]

The probability that the first ball is brown and the second is white is:

[tex]P = P_{1}*P_{2} = \frac{3}{12}*\frac{2}{11} = \frac{1}{22}[/tex]

Would this be different than the probability that the first ball is white and the second ball is brown?

[tex]P = P_{1}*P_{2}[/tex]

[tex]P_{1}[/tex] is the probability that the first ball is white. There are 12 balls, 2 of which are brown. So:

[tex]P_{1} = \frac{2}{12}[/tex]

Since there are no replacements, now there are 11 balls in the bag, 3 of which are brown. So

[tex]P_{2} = \frac{3}{11}[/tex]

The probability that the first ball is brown and the second is white is:

[tex]P = P_{1}*P_{2} = \frac{2}{12}*\frac{3}{11} = \frac{1}{22}[/tex]

They are the same probabilities.

Answer:

Percentage of total net sales = 91.17%

The percentage of cost of merchandise sold = 44.11 %

Percentage of First year expenses = 41.17 %

Percentage of Overall profit = 5.88 %

Step-by-step explanation:

Given:

Gross sales= $204,000

Customer returns and allowances = $18,000

Cost of the merchandise they sold = $90,000

overall profit before taxes = $12,000

Now,

The Net sales = Gross sales - sales returns  

or

The net sales = $204,000 - $18,000 = $186,000

Thus,

Percentage of total net sales = [tex]\frac{\textup{Net sales}}{\textup{Gross sales}}\times100[/tex]

or

Percentage of total net sales = [tex]\frac{186,000}{204000}\times100[/tex]

or

Percentage of total net sales = 91.17%

Now,

The percentage of cost of merchandise sold = [tex]\frac{\textup{cost of the merchandise sold }}{\textup{Gross sales}}\times100[/tex]

or

The percentage of cost of merchandise sold = [tex]\frac{\textup{90,000}}{\textup{204,000}}\times100[/tex]

or

The percentage of cost of merchandise sold = 44.11 %

And,

Percentage of First year expenses = [tex]\frac{\textup{Expenses}}{\textup{Gross sales}}\times100[/tex]

or

Percentage of First year expenses = [tex]\frac{\textup{84000}}{\textup{204000}}\times100[/tex]

or

Percentage of First year expenses = 41.17 %

also,

Percentage of Overall profit = [tex]\frac{\textup{Overall profit}}{\textup{Gross sales}}\times100[/tex]

or

Percentage of Overall profit = [tex]\frac{\textup{12,000}}{\textup{204,000}}\times100[/tex]

or

Percentage of Overall profit = 5.88 %

Answer:  p = -0.004n + 8

Step-by-step explanation:

Consider the number of shirts & the price as an x,y - coordinate

4000 shirts at $32  -->  (4000, 32)

5000 shirts at $28  --> (5000, 28)

Use the slope formula: [tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

[tex]m=\dfrac{28-32}{5000-4000}\quad =\dfrac{-4}{1000}\quad =-0.004[/tex]

Next, use the Point-Slope formula (replace x with n & replace y with p):

[tex]y-y_1=m(x-x_1)\implies p-p_1=m(n-n_1)\\\\\\p-28=-0.004(n-5000)\\\\p-28=-.004n-20\\\\\large\boxed{p=-0.004n+8}[/tex]

Answer:

3500

Step-by-step explanation:

Number of germs that a scientist can see under a microscope = 1000 germs

We need to find how many germs will there be in 7 hours if the germs double in number every 4 hours .

It's given that the germs double in number every 4 hours .

So, increase in number of germs in one hour = [tex]\frac{2}{4}=\frac{1}{2}[/tex]

Increase in number of germs in seven hours = [tex]\frac{7}{2}[/tex]

Therefore , number of germs in 7 hours = Increase in number of germs in seven hours × Number of germs initially

= [tex]\frac{7}{2}\times 1000=7\times 500=3500[/tex]

So, number of germs in 7 hours if the germs double in number every 4 hours = 3500

After 7 hours, there will be 2000 germs.

To calculate how many germs there will be in 7 hours, we need to understand the concept of exponential growth. In this scenario, the germs double every 4 hours.

Initial Number of Germs: You start with 1000 germs.

Doubling Time: The germs double every 4 hours. This means that after each 4-hour period, the population multiplies by 2.

Calculating How Many Doubling Periods in 7 Hours:

Final Count:

Answer:

12,345 tablets may be prepared from 1 kg of aspirin.

Step-by-step explanation:

The problem states that low-strength children’s/adult chewable aspirin tablets contains 81 mg of aspirin per tablet. And asks how many tablets may be prepared from 1 kg of aspirin.

Since the problem measures the weight of a tablet in kg, the first step is the conversion of 81mg to kg.

Each kg has 1,000,000mg. So

1kg - 1,000,000mg

xkg - 81mg.

1,000,000x = 81

[tex]x = \frac{81}{1,000,000}[/tex]

x = 0.000081kg

Each tablet generally contains 0.000081kg of aspirin. How many such tablets may be prepared from 1 kg of aspirin?

1 tablet - 0.000081kg

x tablets - 1kg

0.000081x = 1

[tex]x = \frac{1}{0.000081}[/tex]

x = 12,345 tablets

12,345 tablets may be prepared from 1 kg of aspirin.

Answer:

The solution to the system is [tex]x=1[/tex],[tex]y=-2[/tex] and [tex]z=-5[/tex]

Step-by-step explanation:

Cramer's rule defines the solution of a system of equations in the following way:

[tex]x= \frac{D_x}{D}[/tex], [tex]y= \frac{D_y}{D}[/tex] and [tex]z= \frac{D_z}{D}[/tex] where [tex]D_x[/tex], [tex]D_y[/tex] and [tex]D_z[/tex] are the determinants formed by replacing the x,y and z-column values with the answer-column values respectively. [tex]D[/tex] is the determinant of the system. Let's see how this rule applies to this system.

The system can be written in matrix form like:

[tex]\left[\begin{array}{ccc}5&-3&1\\0&2&-3\\7&10&0\end{array}\right]\times \left[\begin{array}{c}x&y&z\end{array}\right] = \left[\begin{array}{c}6&11&-13\end{array}\right][/tex]

Then each of the previous determinants are given by:

[tex]D_x = \left|\begin{array}{ccc}6&-3&1\\11&2&-3\\-13&10&0\end{array}\right|=199[/tex] Notice how the x-column has been substituted with the answer-column one.

[tex]D_y = \left|\begin{array}{ccc}5&6&1\\0&11&-3\\7&-13&0\end{array}\right|=-398[/tex] Notice how the y-column has been substituted with the answer-column one.

[tex]D_z = \left|\begin{array}{ccc}5&-3&6\\0&2&11\\7&10&-13\end{array}\right|=-995[/tex]

Then, substituting the values:

[tex]x= \frac{D_x}{D}=\frac{199}{199}\\ x=1[/tex]

[tex]x= \frac{D_y}{D}=\frac{-398}{199}\\ y=-2[/tex]

[tex]x= \frac{D_z}{D}=\frac{-995}{199}\\ x=-5[/tex]

Answer:

See explanation

Step-by-step explanation:

During daytime in New York City, UberRUSH has a base fare of $3.00 plus $2.50 per mile.

A. Let x be the number of miles Sergio uses the taxi, then he will pay $2.50x for x miles plus a base fare of $3.00. In total, the cost of using UberRUSH services during the day in New York City is

[tex]C=3.00+2.50x[/tex]

B. The rate of change for the cost of riding UberRUSH during the day in New York City is the slope of the line represented by linear function in part A. So, the rate of change is 2.5 and it shows the change in price per 1 mile driven.

C. The y-intercept is 3.00 (or simply 3) and it represents the initial cost (when 0 miles are driven)

D. The graph of the function is shown in attached diagram. The diagram shows the part of the line starting from point (0,3). This is because the number of miles driven cannot be negative.

The correct representation are as follows:

Part (a): [tex]\boxed{m\cdot a=}[/tex]

Part (b): [tex]\boxed{2*\left(\text{cos}(2\cdot \theta)\right)-1=}[/tex]

Part (c): [tex]\boxed{a*\left(\text{sin}(x)\right)+15=}[/tex]

Part (d): [tex]\boxed{\sqrt{\dfrac{2\cdot g\cdot \triangle y}{m}}=}[/tex]

Part (e): [tex]\boxed{N_{0}\cdot e^{-(\lambda\cdot t)}}[/tex]

Further explanation:

in the question it is given that while writing any mathematical expression the symbol to be used for multiplication is [tex]\cdot\text{ dot}[/tex].

For example: [tex](2\times 3)[/tex] is incorrect and [tex](2\cdot 3)[/tex] is correct.

If a mathematical expression is to be written in its explicit form then the symbol [tex]*\text{ aestrick}[/tex] is used.

Part (a):

The expression given in part (a) is as follows:

[tex]ma=[/tex]

The correct representation of the above expression is as follows:

[tex]\boxed{m\cdot a=}[/tex]

For multiplication [tex]\cdot\text{ dot}[/tex] symbol is used.

Part (b):

The expression given in part (b) is as follows:

[tex]2\text{cos}(2\theta)-1=[/tex]

The correct representation of the above expression is as follows:

[tex]\boxed{2*\left(\text{cos}(2\cdot \theta)\right)-1=}[/tex]

For multiplication [tex]\cdot\text{ dot}[/tex] symbol is used.

Part (c):

The expression given in part (c) is as follows:

[tex]a\text{sin}(x)+15=[/tex]

The correct representation of the above expression is as follows:

[tex]\boxed{a*\left(\text{sin}(x)\right)+15=}[/tex]

Part (d):

The expression given in part (d) is as follows:

[tex]\sqrt{{2g\triangle y/m}=[/tex]

The correct representation of the above expression is as follows:

[tex]\boxed{\sqrt{\dfrac{2\cdot g\cdot \triangle y}{m}}=}[/tex]

For multiplication [tex]\cdot\text{ dot}[/tex] symbol is used.

Part (e):

The expression given in part (e) is as follows:

[tex]N_{0}e^{-\lambda t}[/tex]

The correct representation of the above expression is as follows:

[tex]\boxed{N_{0}\cdot e^{-(\lambda\cdot t)}}[/tex]

For multiplication [tex]\cdot\text{ dot}[/tex] symbol is used.

Learn more:

1. A problem on greatest integer function https://brainly.com/question/8243712

2. A problem to find radius and center of circle https://brainly.com/question/9510228  

3. A problem to determine intercepts of a line https://brainly.com/question/1332667  

Answer details:

Grade: College

Subject: Mathematics

Chapter: NA

Keywords: Expression, representation, ma, cdot, aestrick, wcos(2theta)-1, asin(x)+15, sqraureroot(2gdeltay/m), N0e-lamda t, lowercase greek, multiplication dot.

Basically, here we want to just rewrite all the multiplications using the * symbol, which represents multiplication in most common programmation languages and math calculators.

We usually do not use it when a scalar multiplies a variable, like in:

2x

but just to be complete, let's also use it as 2*x.

for example, the first expression we get is:

ma = m*a

b) Completing the others is trivial, we just need to identify where we have multiplications and add the correspondent symbol there.

[tex]2cos^2(\theta) - 1 = 2*cos^2(\theta) - 1[/tex]

c) asin(x)+15 =

Note that while "Asin(x)" is a function, I assume that here we have "a times sin(x)", so we will write:

[tex]asin(x) + 15 = a*sin(x) + 15[/tex]

D)  (√(2gΔy)/m)

Here we have the term "Δy", this is not a multiplication, this means a "difference in the value of y", so we will leave it as it is.

[tex]\frac{\sqrt{2g\Delta y} }{m} = \frac{\sqrt{2*g*\Delta y} }{m}[/tex]

E)  N0e−λt,

Here N0 is a term in itself, and actually should be written as N₀, so this is not N times zero.

So we will get:

[tex]N_0e -\lambda t = N_0*e - \lambda*t[/tex]

If instead of that, the right part was an exponent (I can't tell because of how you wrote it) we would get:

[tex]N_0e^{ -\lambda t} = N_0*e^{ - \lambda*t}[/tex]

If you want to learn more, you can read:

https://brainly.com/question/21619770

Answer:  30

Step-by-step explanation:

The combination of n things taking r at a time is given by :-

[tex]^nC_r=\dfrac{n!}{r!(n-r)!}[/tex]

Given : At a restaurant a menu has 5 salads and 6 entrees.

Then, the number of ways you can order a dinner that contains 1 salad and 1 entree will be :_

[tex]^5C_1\times ^6C_1\\\\=5\times6=30[/tex]

Hence, the number of ways you can order a dinner that contains 1 salad and 1 entree = 30

Answer:

58 seconds

Step-by-step explanation:

Given:

Initial mass of water = 1 kg

Specific energy = 2297 kJ/kg

Heat supplied by the stove = 20 kJ/s

Now,

Half water is to be vaporized i.e 0.5 kg

Thus, heat required for vaporizing 0.5 kg water = mass × specific heat

or

heat required for vaporizing 0.5 kg water = 0.5 × 2297 = 1148.5 kJ

Therefore,

time taken to provide the required heat = [tex]\frac{\textup{Heat required}}{\textup{Heat supplied per second}}[/tex]

or

time taken to provide the required heat = [tex]\frac{\textup{1148.5 kJ}}{\textup{20 kJ/s}}[/tex]

or

time taken to provide the required heat = 57.425 ≈ 58 seconds

It will take approximately 58 seconds to vaporize half a kilogram of water with a heat supply of 20 kJ/s.

The question is asking how long it will take to vaporize half a kilogram of water with a heat supply of 20 kJ/s, assuming the water is at its boiling point and the specific energy required for the phase change is 2297 kJ/kg. To calculate the time required, we can use the formula:

Time (s) = Amount of energy required (kJ) / Energy supply rate (kJ/s).

Since it takes 2297 kJ to vaporize 1 kg, half of this amount is required to vaporize 0.5 kg, which is 1148.5 kJ. Hence, the time taken can be calculated as follows:

Time (s) = 1148.5 kJ / 20 kJ/s = 57.425 s.

So, it would take approximately 58 seconds to vaporize half of the water.

Step-by-step explanation:

Given that,

Radius of circle, r = 38 cm = 0.38 m

It rotates form 10 degrees to 100 degrees in 11 seconds i.e.

[tex]\theta_i=10^{\circ}=0.174\ rad[/tex]

[tex]\theta_f=100^{\circ}=1.74\ rad[/tex]

Let [tex]\omega[/tex] is the angular velocity of the particle such that, [tex]\omega=\dfrac{\omega_f-\omega_i}{t}[/tex]

[tex]\omega=\dfrac{1.74-0.174}{11}[/tex]

[tex]\omega=0.142\ rad/s[/tex]

We need to find the instantaneous velocity of the particle. The relation between the angular velocity and the linear velocity is given by :

[tex]v=r\times \omega[/tex]

[tex]v=0.38\times 0.142[/tex]

v = 0.053 m/s

So, the instantaneous velocity of the particle is 0.053 m/s. Hence, this is the required solution.  

Answer:

[tex]a.\hspace{3} \frac{6}{38} = 0.15789474\\\\b.\hspace{3} \frac{45}{55} = 0.81818181\\\\c.\hspace{3} \frac{4}{28} = 0.14285714\\\\d.\hspace{3} \frac{35}{28} = 1.25\\\\e.\hspace{3} \frac{1030}{2030} = 0.50738916[/tex]

Step-by-step explanation:

Rational numbers, expressed as decimal numbers, are obtained from the operation of division between the integer of the numerator and the integer of the denominator. Then:

[tex]a.\hspace{3} \frac{6}{38} = 6\div38 = 0.15789474\\\\ b.\hspace{3} \frac{45}{55} = 45\div55 = 0.81818181\\\\c.\hspace{3} \frac{4}{28} = 4\div28 = 0.14285714\\\\ d.\hspace{3} \frac{35}{28} = 35\div28 = 1.25\\\\ e.\hspace{3} \frac{1030}{2030} = 1030\div2030 = 0.50738916[/tex]

Answer:

1. 9 + 3 - ( 2 + 4) = 6

2. 4^2 - (5 x (2 + 1)) = 1

Step-by-step explanation:

Here we must follow order of operations - that is commonly expressed as PEDMAS - First do parenthesis, then exponents, then divisions and multiplications from left to right and finally addition and subtraction from left to right.

If we follow this rule on 1)

9+3-2+4= 12-2+4= 10+4 = 14

Sow lets do it by parts

9+3-2+4= 12-2+4

if we can subtract 6 from 12 we would arrive to 6. This can be done id 2 and 4 are added first by  12-(2+4). So the result would be at:

9 + 3 - ( 2 + 4) = 6

In 2)

4^2 - 5 x 2 + 1 = 16-5x2+1 = 16-10 + 1 = 6+1 = 7

4^2 is always the first operation

16-5x2+1

Now if from 16 we subtract 15 we would obtain 1 so  5x2+1 must be equal 15 that can be done if we express it as:

16- 5x2+1    

16- (5*(2+1)) = 5x3 = 15

So we have at the end:  

4^2 - (5 x (2 + 1)) = 16 - 15 = 1

Answer:

The probabilities are [tex]\frac{1}{10}[/tex] and [tex]\frac{9}{80}[/tex]

Step-by-step explanation:

There are 10 doors. 9 of wich have no prizes and 1 with the prize. So the probability to choose the winner one is 1 out of 10. So:

The probability of finding the prize behind your chosen door before the game show host reveals that one door is empty is [tex]\frac{1}{10}[/tex].

Now. If the game show host opens one of the other 9 doors to reveal that it is an empty door, there are 2 posibilities:

1) Do not change your chosen door: In this case the probability reamins the same, [tex]\frac{1}{10}[/tex].

2) Change your chosen door. Lets compute the probability to loose: There are two posibilities.

  2a) If your initial door is the one with the prize. In this case you are going to loose (because you will change your door). The probability for this to happen is [tex]\frac{1}{10}[/tex].

 2b) If your initial door is not the one with the prize (the probability of this is  [tex]\frac{9}{10}[/tex]). In this case we will loose if, after the game show host opens an empty door, we choose an empty door. The probability of choosing an empty door in this case is [tex]\frac{7}{8}[/tex].

So the probability to loose is:

[tex]\frac{1}{10}+\frac{7}{8}\frac{9}{10}=\frac{1}{10}+\frac{63}{80}=\frac{71}{80}[/tex]

Then, the probability to win is [tex]1-\frac{71}{80}=\frac{9}{80}>\frac{1}{10}[/tex]

In conclusion: Changing the door improves the probability to win.

Answer:

[tex](a)\ y(t) =\ 4.e^{2(1-t)}\ +\ \dfrac{t^2e^{-2t}}{4}[/tex]

[tex](b)\ y(t)=\ (1-t)e^{-t}\ -\ 2e[/tex]

Step-by-step explanation:

(a) [tex]y'\ +\ 2y\ =\ te^{-2t},\ y(1)\ =\ 0[/tex]

 [tex]=>\ (D+2)y\ =\ te^{-2t}[/tex]

To find the complementary function

   D+2 = 0

=> D = -2

So, the complementary function can by given by

[tex]y_c(t)\ =\ C.e^{-2t}[/tex]

Now, to find particular integral

  [tex](D+2)y_p(t)\ =\ te^{-2t}[/tex]

[tex]=>y_p(t)\ =\ \dfrac{ te^{-2t}}{D+2}[/tex]

              [tex]=\ \dfrac{ te^{-2t}}{-2+2}[/tex]

               = not defined

So,

[tex]y_p(t)\ =\ \dfrac{ t^2e^{-2t}}{D^2}[/tex]

           [tex]=\ \dfrac{t^2e^{-2t}}{(-2)^2}[/tex]

           [tex]=\ \dfrac{t^2e^{-2t}}{4}[/tex]

So, complete solution can be given by

    [tex]y(t)\ =\ y_c(t)\ +\ y_p(t)[/tex]

[tex]=> y(t) =\ C.e^{-2t}\ +\ \dfrac{t^2e^{-2t}}{4}[/tex]

As given in question

[tex]=>\ y(1)\ =\ C.e^{-2}\ +\ \dfrac{1^2e^{-2}}{4}[/tex]

[tex]=>\ 0\ =\ C.e^{-2}\ +\ \dfrac{1^2e^{-2}}{4}[/tex]

[tex]=>\ C\ =\ 4e^2[/tex]

Hence, the complete solution can be give by

[tex]=>\ y(t) =\ 4e^2.e^{-2t}\ +\ \dfrac{t^2e^{-2t}}{4}[/tex]

[tex]=>\ y(t) =\ 4.e^{2(1-t)}\ +\ \dfrac{t^2e^{-2t}}{4}[/tex]

(b) [tex]t^{3}y'\ +\ 4t^{2}y\ =\ e^{-t},\ y(-1)\ =\ 0[/tex]

[tex]=>\ y'\ +\ 4t^{-1}y\ =\ t^{-3}e^{-t}[/tex]

Integrating factor can be given by

[tex]I.F\ =\ e^{\int (4t^{-1})dt}[/tex]

     [tex]=\ e^{log\ t^4}[/tex]

     [tex]=\ t^4[/tex]

Now , the solution of the given differential equation can be given by

[tex]y(t)\times t^4\ =\ \int t^{-3}e^{-t}t^4dt\ +\ C[/tex]

[tex]=>\ y(t)\ =\ \int t.e^{-t}dt\ +\ C[/tex]

         [tex]=\ (1-t)e^{-t}\ +\ C[/tex]

According to question

[tex]y(-1)\ =\ (1-(-1))e^1\ +\ C[/tex]

[tex]=>\ 0\ =\ 2e\ +\ C[/tex]

[tex]=>\ C\ =\ -2e[/tex]

Now, the complete solution of the given differential equation cab be given by

[tex]y(t)\ =\ (1-t)e^{-t}\ -\ 2e[/tex]

Answer:

a. [tex]y(t)=\frac{t^2e^{-2t}}{2}-\frac{1}{2}e^{-2t}[/tex]

b.[tex]y=-t^{-3}e^{-t}-t^{-4}e^{-t}[/tex]

Step-by-step explanation:

We are given that

a.[tex]y'+2y=te^{-2t},y(1)=0[/tex]

Compare with [tex]y'+P(t)y=Q(t)[/tex]

We have P(t)=2,Q(t)=[tex]te^{-2t}[/tex]

Integration factor=[tex]\int e^{2dt}=e^{2t}[/tex]

[tex]y\cdot I.F=\int Q(t)\cdot I.F dt+C[/tex]

Substitute the values then, we get

[tex]y\cdot e^{2t}=\int te^{-2t}\cdot e^{2t} dt+C[/tex]

[tex]y\cdot e^{2t}=\int tdt+C[/tex]

[tex]ye^{2t}=\frac{t^2}{2}+C[/tex]

Substitute the values x=1 and y=0

Then, we get [tex]0\cdot e^2=\frac{1}{2}+C[/tex]

[tex]C=-\frac{1}{2}[/tex]

Substitute the value in the given function

[tex]ye^{2t}=\frac{t^2}{2}-\frac{1}{2}[/tex]

[tex]y=\frac{t^2}{2}e^{-2t}-\frac{1}{2}e^{-2t}[/tex]

Hence, [tex]y(t)=\frac{t^2e^{-2t}}{2}-\frac{1}{2}e^{-2t}[/tex]

b.[tex]t^3y'+4t^2y=e^{-t},y(-1)=0[/tex]

[tex]y'+\frac{4}{t}y=\frac{e^{-t}}{t^3}[/tex]

[tex]P(t)=\frac{4}{t},Q(t)=\frac{e^{-t}}{t^3}[/tex]

I.F=[tex]\int e^{\frac{4}{t}dt}=e^{4lnt}=e^{lnt^4}=t^4[/tex]

[tex]y\cdot \frac{t^4}=\int e^{-t}\frac{t^4}{t^3} dt+C[/tex]

[tex]y\cdot t^4=\int te^{-t}dt+C[/tex]

[tex]yt^4=-te^{-t}+\int e^{-t} dt+C[/tex]

[tex]u\cdot v dt=u\int vdt-\int (\frac{du}{dt}\cdot \int vdt)dt[/tex]

[tex]yt^4=-te^{-t}-e^{-t}+C[/tex]

Substitute the values x=-1,y=0 then, we get

[tex]0=-(-1)e-e+C[/tex]

[tex]C+e-e=0[/tex]

C=0

Substitute the value of C then we get

[tex]yt^4=-te^{-t}-e^{-t}[/tex]

[tex]y=-t^{-3}e^{-t}-t^{-4}e^{-t}[/tex]

Step-by-step explanation:

We are picking 6 numbers from the numbers 1,2,3,4,5,6,7,8,9,10. Since we care about numbers being next to each other, we might think of the 10 numbers as being distributed in 5 boxes (which you can think of as the holes):

|   1 2   |   3 4   |   5 6   |   7 8   |   9 10  |

So on the first box we have the numbers 1 and 2, on the second box we have the numbers 3 and 4, and so on. Since we are picking 6 numbers from those 10 numbers, that means we'll have to pick 6 boxes (and inside each box we pick a number), but we only have 5 available boxes, so by the pigeonhole principle, we'll have to pick 1 same box at least two times. Since on each picked box we'll need to pick a number, on this box which was picked two times, we will have to pick both of its numbers. And so those 2 numbers inside that box will be next to each other (meaning they're consecutive numbers).

Answer:

The standard deviation of x is 2.8867

Step-by-step explanation:

The standard deviation of variable x that follows a uniform distribution is calculated as:

[tex]s = \sqrt{\frac{(b-a)^{2} }{12} }[/tex]

Where (a,b) is the interval where x is defined.

So, replacing a by -4 and b by 6, the standard deviation is:

[tex]s = \sqrt{\frac{(6-(-4))^{2} }{12} }[/tex]

[tex]s = \sqrt{\frac{(10)^{2} }{12} }[/tex]

[tex]s=\sqrt{\frac{100}{12} }[/tex]

[tex]s=\sqrt{8.3333}[/tex]

[tex]s=2.8867[/tex]

To find when Ima Neworker's rate will exceed 12 units per hour, given a learning curve rate of 65%, we analyze the improvement in production rate from the initial 2 units per hour up to the target, using the learning curve concept.

The question relates to the concept of a learning curve, which represents how new workers or processes improve in efficiency as experience is gained. Ima Neworker can produce her first unit in 30 minutes (which is half an hour), so when she starts, her production rate is 2 units per hour. The question asks how many units will be produced before her production rate exceeds 12 units per hour, given a learning curve rate of 65%. This means that each time the cumulative production doubles, the time taken to produce each unit falls to 65% of the previous time.

Since the initial production rate is 2 units per hour, we want to know how many units she has to produce before her production rate exceeds 12 units per hour. 12 units per hour is 6 times faster than her initial rate, and we can reference a learning curve table or use the formula to calculate the necessary doubling periods required to achieve this.

To determine when Ima Neworker's production rate exceeds 12 units per hour, we use a 65% learning curve. By calculations, production time per unit drops below 5 minutes per unit between producing 8 and 16 units, indicating she exceeds the rate at around 12 units. Thus, she will need to produce approximately 12 units before reaching this threshold.

Calculating Production Using a Learning Curve

Ima Neworker requires 30 minutes to produce her first unit, which translates to 2 units per hour initially. The learning curve rate of 65% indicates that with each doubling of previously produced units, the time required to produce another unit will be 65% of the time it took for the previous set.

Step-by-Step Calculation

We need to produce units such that Ima's production rate exceeds 12 units per hour, meaning she should produce a unit in less than 1/12 hours (5 minutes).

Thus, Ima will need to produce more than 8 but fewer than 16 units. By interpolation, it will be close to 12 units when her rate exceeds 12 units per hour.

Answer:

36.3255814 or 36 (when rounded)

Step-by-step explanation:

Calculator

Answer:

In the step-by-step explanation, the verifications are made.

Step-by-step explanation:

a) [tex]y' = -5y[/tex]

This one can be solved by the variable separation method

[tex]y' = -5y[/tex]

[tex]\frac{dy}{dx} = -5y[/tex]

[tex]\frac{dy}{y} = -5dx[/tex]

[tex]\int \frac{dy}{y}  = \int {-5} \, dx[/tex]

[tex]ln y = -5x + C[/tex]

[tex]e^{ln y} = e^{-5x + C}[/tex]

[tex]y = Ce^{-5x}[/tex]

The value of C is the value of y when x = 0. If [tex]y(0) = 3[/tex], then we have the following solution:

[tex]y = 3e^{-5x}[/tex]

b) [tex]y' = cos(3x)[/tex]

This one can also be solved by the variable separation method

[tex]y' = cos(3x)[/tex]

[tex]\int y' \,dy  = \int {cos(3x)} \, dx[/tex]

[tex]y = \frac{sin(3x)}{3} + K[/tex]

K is also the value of y, when x = 0. So, if [tex]y(0) = 7[/tex], we have the following solution.

[tex]y = \frac{sin(3x)}{3} + 7[/tex]

c) [tex]y' = 2y[/tex]

Another one that can be solved by the variable separation method

[tex]y' = 2y[/tex]

[tex]\frac{dy}{dx} = 2y[/tex]

[tex]\frac{dy}{y} = 2dx[/tex]

[tex]\int \frac{dy}{y}  = \int {2} \, dx[/tex]

[tex]ln y = 2x + C[/tex]

[tex]e^{ln y} = e^{2x + C}[/tex]

[tex]y = Ce^{2x}[/tex]

C is any real number depending on the initial conditions.

d) [tex]y'' + y' - 6y = 0[/tex]

Here, the solution depends on the roots of the following equation:

[tex]r^{2} + r - 6 = 0[/tex]

[tex]r = \frac{-1 \pm 5}{2}[/tex]

[tex]r = -3[/tex] or [tex]r = 2[/tex].

So the solution is

[tex]y(t) = c_{1}e^{-3t} + c2e^{2t}[/tex]

The values of [tex]c_{1}, c_{2}[/tex] depends on the initial conditions.

e) [tex]y'' + 16y = 0[/tex]

Again, we find the roots of the following equation:

[tex]r^{2} + 16 = 0[/tex]

[tex]r^{2} = -16[/tex]

[tex]r = \pm 4i[/tex]

So we have the following solution

[tex]y(t) = c_{1}cos(4t) + c_{2}sin(4t)[/tex]

The values of [tex]c_{1}, c_{2}[/tex] depends on the initial conditions.

[tex]f'(x)=\dfrac{4x}{1+7x^2}[/tex]

Integrating gives

[tex]f(x)=\displaystyle\int\frac{4x}{1+7x^2}\,\mathrm dx[/tex]

To compute the integral, substitute [tex]u=1+7x^2[/tex], so that [tex]\frac27\,\mathrm du=4x\,\mathrm dx[/tex]. Then

[tex]f(x)=\displaystyle\frac27\int\frac{\mathrm du}u=\frac27\ln|u|+C[/tex]

Since [tex]u=1+7x^2>0[/tex] for all [tex]x[/tex], we can drop the absolute value, so we end up with

[tex]f(x)=\dfrac27\ln(1+7x^2)+C[/tex]

Given that [tex]f(0)=10[/tex], we have

[tex]10=\dfrac27\ln1+C\implies C=10[/tex]

so that

[tex]\boxed{f(x)=\dfrac27\ln(1+7x^2)+10}[/tex]

Answer:

The nurse infuse [tex]26ml/hr[/tex] to provide the dose.

Step-by-step explanation:

Argatroban was ordered at a dose of 2 mcg/kg/min.

JY weighs 85 kg.

So, Argatroban was ordered= [tex]2 \times 85[/tex]

                                              = [tex]170mcg/min.[/tex]

Convert the dose in mg/hr

1 hr = 60 minutes and 1 mg = 1000 mcg

So, Dose in ml/hr = [tex]170 \times \frac{60}{1000}[/tex]

                             = [tex]10.2 mg/hr[/tex]

Now to find in 250 mL of DSW. JY weighs 85 kg. How many mL/hour should the nurse infuse to provide the dose?

The nurse infuse to provide the dose = [tex]\text{Dose ordered} \times \frac{\text{volume available}}{\text{Dose available}}[/tex]

The nurse infuse to provide the dose = [tex]10.2 mg/hr \times \frac{250 ml}{100 mg}[/tex]

The nurse infuse to provide the dose = [tex]26ml/hr[/tex]

Hence The nurse infuse [tex]26ml/hr[/tex] to provide the dose.

To find the surface area of the resulting surface when rotating the arc of a parabola about the y-axis, you can integrate using either x or y as the variable. Both methods yield the same result.

The question asks for the area of the surface created by rotating the arc of the parabola y = 3x^2 from (5, 75) to (10, 300) about the y-axis.

There are two different solutions provided, both utilizing different methods of integration.

The first solution uses the given equation y = 3x^2 and integrates with respect to x, while the second solution uses the equation x = y^(1/3) and integrates with respect to y.

Both solutions arrive at the same answer to find the surface area.

Answer:

(a) 0.704

(b) 0.8475

Step-by-step explanation:

(a) Let 'A' be the event that you travel by car and late

Let 'B' be the event that you travel by bus and late

Let 'C' be the event that you travel by Bicycle and late

Then, P (A)  = 50% = [tex]\frac{50}{100}[/tex] = [tex]\frac{1}{2}[/tex]

         P (B)   = 20% = [tex]\frac{20}{100}[/tex] = [tex]\frac{1}{5}[/tex]

         P (C)   = 1%  = [tex]\frac{1}{100}[/tex]  = [tex]\frac{1}{100}[/tex]

A₁ = Student travels by car

B₁ = Student travels by bus

C₁ = Student travels by bicycle

Then according to teacher P(A₁) = [tex]\frac{1}{3}[/tex], P(B₁) = [tex]\frac{1}{3}[/tex], P(C₁) = [tex]\frac{1}{3}[/tex]

Now we have to find "Student is already late and traveled to school that day by car." which will be given as [tex]P(\frac{A}{L})[/tex]

where L : student is late

By using Bay's Theorem :

[tex]P(\frac{A}{L})[/tex]  = [tex]\frac{P(A)\times P(A_1)}{P(A)\times P(A_1)+P(B)\times P(B_1)+P(C)\times P(C_1)}[/tex]

= [tex]\frac{\frac{1}{2}\times \frac{1}{3}}{\frac{1}{2}\times \frac{1}{3}+\frac{1}{5}\times \frac{1}{3}+\frac{1}{100}\times \frac{1}{3}}[/tex]

= [tex]\frac{\frac{1}{6}}{\frac{1}{6}+\frac{1}{15}+\frac{1}{300}}[/tex]

= [tex]\frac{\frac{1}{6}}{\frac{50+20+1}{300}}[/tex]

= [tex]\frac{1}{6}\times \frac{300}{71}[/tex]

= [tex](\frac{50}{71})[/tex]

= 0.704

(b) Here P(A₁) = [tex](\frac{10}{100})[/tex]

              P(C₁) = [tex](\frac{90}{100})[/tex]

            [tex]P(\frac{A}{L})[/tex]  = We have to find and known student is late and traveled by car.

[tex]P(\frac{A}{L})[/tex] = [tex]\frac{P(A)\times P(A_1)}{P(A)\times P(A_1)+(P(C)\times P(C_1)}[/tex]

= [tex]\frac{\frac{1}{2}\times \frac{1}{10}}{\frac{1}{2}\times \frac{1}{10}+\frac{1}{100}\times \frac{9}{10}}[/tex]

= [tex]\frac{\frac{1}{20} }{\frac{1}{20}+\frac{9}{1000}}[/tex]

= [tex]\frac{\frac{1}{20}}{\frac{50+9}{1000}}[/tex]

= [tex]\frac{1}{20}\times \frac{1000}{59}[/tex]

= [tex](\frac{50}{59})[/tex]

= 0.8475

Answer:

Step-by-step explanation:

Given that the three orders are independent of each other.

X - no of ships not shipped on time

X is binomial with p = 0.05 and q = 0.95, n = 3

a)  the probability that all are shipped on time=[tex]0.95^3 =0.854[/tex]

b) the probability that exactly one is not shipped ontime

=P(X=1) =[tex]3C1(0.05)(0.95)^2 = 0.135[/tex]

c) the probability that two or more orders are notshipped on time

[tex]= P(X=2)+P(x=3)\\= 1-P(x=0)-P(x=1)\\=1-0.854-0.135\\=0.011[/tex]

Final answer:

The probabilities for the shipping orders, assuming independence, are calculated: all on time is 0.857, exactly one not on time is 0.135, and two or more not on time is 0.008.

Explanation:

The probability that a customer's order is not shipped on time is 0.05. Since the orders are independent events, we can calculate the following:

Probability that all are shipped on time: Since the orders are independent, the probability of all orders shipped on time is the product of their individual probabilities: (1-0.05)3 = (0.95)3 = 0.857.

Probability that exactly one is not shipped on time: There are three scenarios where exactly one order can be not shipped on time (NSO for not shipped, SO for shipped): NSO-SO-SO, SO-NSO-SO, SO-SO-NSO. The probability for each scenario is 0.05 * 0.95 * 0.95. Since there are three such scenarios, multiply by 3: 3 * (0.05 * 0.95 * 0.95) = 0.135.

Probability that two or more orders are not shipped on time: The probability of at least one order being shipped on time is 1 minus the probability of none being shipped on time, which is 1 - (0.05)3. From this, we subtract the probability of all being shipped on time and the probability of exactly one not being shipped on time to get our answer: 1 - (0.95)3 - (3 * 0.05 * 0.95 * 0.95) = 0.008.

Answer:

The level of significance observed is 0.99154

Step-by-step explanation:

Assuming that in a sample of size 50 people stated that they do not like the snack (p = 17/50), you have:

Proportion in the null hypothesis [tex]\pi_0=0.5[/tex]

Sample size [tex]n=50[/tex]

Sample proportion [tex]p=17/50=0.34[/tex]

The expression for the calculated statistic is:

[tex] = \frac{(p - \pi_0)\sqrt{n}}{\sqrt{\pi_0(1-\pi_0)}}[/tex]

[tex]= \frac{(0.34 - 0.5)\sqrt{50}}{\sqrt{0.34(0.66)}} = -2,38833[/tex]

The level of significance observed is obtained from the value of the statistic calculated:

[tex]P(Z>Z_{calculated}) = 0.99154[/tex]

Answer:

The required ratio is 9 : 61

Step-by-step explanation:

Given,

$72 for 488 photos,

That is, the price of 488 photos = 72 dollars,

So, the ratio of price of photos and number of photos = [tex]\frac{72}{488}[/tex]

∵ HCF(72, 488) = 8,

Thus, the ratio of price of photos and number of photos = [tex]\frac{72\div 8}{488\div 8}[/tex]

= [tex]\frac{9}{61}[/tex]

Answer:

474.84 ft of fencing is needed

Step-by-step explanation:

We know that the angles of a triangle sum up 180º. We already know 2 of the triangle's angles (55º and 63º). Therefore the third angle measures:

180 - 55 - 63 = 62.

To know how much fencing is needed, we need the perimeter of the triangle, so we need to find out how much the other sides of the lot measure.

We will use law of sins to solve this problem.

First we solve for y:

[tex]\frac{150}{sin55}= \frac{y}{sin63}  \\y=150 (sin63)/(sin55)\\y=133.65/.8191\\y=163.16[/tex]

Now we solve for the other side of the lot, x:

[tex]\frac{150}{sin55}=\frac{x}{sin62}\\  x=150(sin62)/(sin55)\\x=150(.8829)/.8191\\x=132.435/.8191\\x=161.68[/tex]

Now that we have the measures of all the sides we sum them up

total fencing needed= 150 + 163.16 + 161.68 = 474.84

Answer:

Option C.

Step-by-step explanation:

Negative 5 represents (-5)

Square root of negative 44 is [tex]\sqrt{(-44)}[/tex]

By the statement "negative 5 minus the square root of negative 44." will be

(-5) - [tex]\sqrt{(-44)}[/tex]

= -5 -[tex]\sqrt{(-1)(44)}[/tex]

= [tex]-5-\sqrt{(-1)(4\times11)}[/tex]

= [tex]-5-2\sqrt{11(-1)}[/tex]

= [tex]-5-2(\sqrt{11} )[\sqrt{(-1)} ][/tex]

= [tex]-5-2i\sqrt{11}[since\sqrt{(-1)=i}][/tex]

Option c will be the answer.

Answer:

R is reflexive

R is not symmetric

R is transitive

Step-by-step explanation:

R is reflexive.

To show this, we have to verify that for any pair of integers [tex](x_1,x_2)[/tex]

[tex](x_1,x_2)R(x_1,x_2)[/tex].

But this is obvious because

[tex]x_1=x_1[/tex] and [tex]x_2\leq x_2[/tex].

R is not symmetric.

To show it, we need to find two pairs [tex](x_1,x_2)[/tex] and [tex](y_1,y_2)[/tex] such that

[tex](x_1,x_2)R(y_1,y_2)[/tex]

but [tex](y_1,y_2) \not \mathrel{R} (x_1,x_2)[/tex]

For example (1,1) and (1,2).

[tex](1,1)R(1,2)[/tex] for 1=1 and [tex]1\leq 2[/tex] but  

[tex](1,2) \not \mathrel{R} (1,1)[/tex] because [tex]2\not \leq 1[/tex]

Finally, R is transitive.

If we take 3 pairs of integers [tex](x_1,x_2), (y_1,y_2)[/tex] and [tex](z_1,z_2)[/tex]

Such that

[tex](x_1,x_2)R(y_1,y_2)[/tex] and [tex](y_1,y_2)R(z_1,z_2)[/tex] then

[tex]x_1=y_1[/tex] and [tex]x_2\leq y_2[/tex]

[tex]y_1=z_1[/tex] and [tex]y_2\leq z_2[/tex]

But then,

[tex]x_1=z_1[/tex] and [tex]x_2\leq z_2[/tex]

So  

[tex](x_1,x_2)R(z_1,z_2)[/tex].