20 units Humulin R insulin in 100 mL of normal saline (NS) to infuse for 20 hours. (Round to the nearest tenth if applicable) a. How many units per hour will be infused? ________ b. How many milliliters per hour will be infused? ________

Answer:

0.04

Step-by-step explanation:

X~N(μ=14.2; σ=0.9), a=7

[tex]P(X>7)=1-P(X\leq 7)[/tex]

[tex]P(0\leq X\leq a)=P(\frac{0-\mu}{\sigma} \leq \frac{X-\mu}{\sigma}\leq \frac{a-\mu}{\sigma}), Z= \frac{X-\mu}{\sigma}=P(0\leq X\leq a)=P(\frac{0-\mu}{\sigma} \leq \frac{X-\mu}{\sigma}\leq \frac{a-\mu}{\sigma}), Z= \frac{X-\mu}{\sigma}[/tex]

[tex]P(0\leq X\leq 7) = P(\frac{0-5.5}{1.5} \leq Z\leq \frac{7-5.5}{1.5}) = \Phi(1.66)-\Phi(-3.66) = \Phi(1.66) -(1-\Phi(3.66)) = 0.95-(1-0.99)=0.96[/tex]

P(X>7)=1-0.96=0.04

[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:

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.

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:

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:

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].

Answer:

John’s weekly revenue from hot dogs = [tex]\$705[/tex]

Step-by-step explanation:

John  sells hot dogs from a cart outside an office building 5 days a week .

Price of a hot dog = [tex]\$3[/tex]

Number of hot dogs sold each day = 47

So, total number of hot dogs sold on 5 days = 5 × Number of hot dogs sold each day = 5 × 47 = [tex]235[/tex]

We need to find John’s weekly revenue from hot dogs if he sells 47 each day.

John’s weekly revenue from hot dogs = Price of a hot dog × total number of hot dogs sold on 5 days = 3 × 235 = [tex]\$705[/tex]

Answer:

0.5mL of the injection should be administered to obtain 0.02 g of the drug.

Step-by-step explanation:

First step: The first step of this problem is the conversion of 0.02g to mg.

Each gram has 1000 miligrams. So:

1g - 1000mg

0.02g - xmg

x = 1000*0.02

x = 20mg

Final step:

A vial contains 80 mg of drug in 2 mL of injection. How many mL should be administered to obtain 0.02 g = 20mg of the drug.

This can be solved as a rule of three problem.

In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.

When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too.

When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease.

In this step, as the dose of the injection increases, so the quantity of the drug. So the relationship between the measures is direct. So:

80 mg - 2mL

20 mg - xmL

80x = 40

[tex]x = \frac{40}{80}[/tex]

x = 0.5mL

0.5mL of the injection should be administered to obtain 0.02 g of the drug.

To obtain 0.02 grams of the drug, the required volume to administer from the vial containing 80 mg in 2 mL is 0.5 mL.

To solve this problem, we need to determine how many milliliters contain 0.02 grams of the drug. First, convert the amount we want from grams to milligrams since our given concentration is in milligrams: 0.02 grams is equal to 20 milligrams.

Given that the vial contains 80 mg of drug in 2 mL, we can calculate the volume required for 20 mg. The formula we will use is:

(desired dose / concentration of vial) × volume of vial = required volume

(20 mg / 80 mg) × 2 mL = 0.5 mL

Therefore, to obtain 0.02 grams (20 mg) of the drug, the required volume to administer would be 0.5 mL.

Answer:

The sum of two odd integers is even

Step-by-step explanation:

Proof by contradiction:

We are going to assume that the sum of two odd integers is odd.

An odd integer is written as 2p+1 where p is an integer and an even integer is written as 2p where p is an integer

So, if the sum of two odd integers is odd we would have

[tex](2k+1) + (2p+1) = 2r+1\\2k+1+2p+1=2r+1\\2k+2p+2=2r+1\\2(k+p+1)=2r+1[/tex]

The left side of the equation is clearly an even number while the right side of the equation is odd. Therefore, our hypothesis is wrong and we can conclude that the sum of two odd integers is even.

Answer and Explanation:

Given : The quadratic function [tex]f(x)=-x^2+x+12[/tex]

To find : Determine the following ?

Solution :

The x -intercept are where f(x)=0,

So, [tex]-x^2+x+12=0[/tex]

Applying middle term split,

[tex]-x^2+4x-3x+12=0[/tex]

[tex]-x(x-4)-3(x-4)=0[/tex]

[tex](x-4)(-x-3)=0[/tex]

[tex]x=4,-3[/tex]

The x-intercepts are (4,0) and (-3,0).

The smallest x-intercept is x=-3

The largest x-intercept is x=4

The y -intercept are where x=0,

So, [tex]f(0)=-(0)^2+0+12[/tex]

[tex]f(0)=12[/tex]

The y-intercept is y=12.

The quadratic function is in the form, [tex]y=ax^2+bx+c[/tex]

On comparing, a=-1 , b=1 and c=1 2

The vertex of the graph is denote by (h,k) and the formula to find the vertex is

For h, The x-coordinate of the vertex is given by,

[tex]h=-\frac{b}{2a}[/tex]

[tex]h=-\frac{1}{2(-1)}[/tex]

[tex]h=\frac{1}{2}[/tex]

For k, The y-coordinate of the vertex is given by,

[tex]k=f(h)[/tex]

[tex]k=-h^2+h+12[/tex]

[tex]k=-(\frac{1}{2})^2+\frac{1}{2}+12[/tex]

[tex]k=-\frac{1}{4}+\frac{1}{2}+12[/tex]

[tex]k=\frac{-1+2+48}{4}[/tex]

[tex]k=\frac{49}{4}[/tex]

The vertex of the function is [tex](h,k)=(\frac{1}{2},\frac{49}{4})[/tex]

The x-coordinate of the vertex i.e. [tex]x=-\frac{b}{2a}[/tex] is the axis of symmetry,

So, [tex]x=-\frac{b}{2a}=\frac{1}{2}[/tex] (solved above)

The axis of symmetry is [tex]x=\frac{1}{2}[/tex].

Answer:

$180 go into savings

Step-by-step explanation:

Given : Savings are represented by 45° on a circle graph showing all expenses.

Total expenses are $1440

To Find : How much go into savings?

Solution:

Total angle of expense = 360°

Savings = [tex]\frac{\text{Saving angle}}{\text{Total angle}} \times 1440[/tex]

            = [tex]\frac{45}{360} \times 1440[/tex]

            = [tex]180[/tex]

Hence $180 go into savings

Answer:

It is not true

Step-by-step explanation:

Suppose your domain is the integer numbers. Define

P(x)="x is even"

Q(x)="x is odd"

So we have that the predicate [tex]\forall x(P(x) \vee Q(x))[/tex] is always true because the integers are always even or odd. But the predicate [tex]\forall x P(x) \vee \forall x Q(x)[/tex] means that all the integer numbers are even or all the integer numbers are odd, which is false. So we can't deduce [tex]\forall x P(x) \vee \forall x Q(x)[/tex] from [tex]\forall x(P(x) \vee Q(x))[/tex].

Answer:

36.3255814 or 36 (when rounded)

Step-by-step explanation:

Calculator

Answer:

  24

Step-by-step explanation:

4P3 = 4!/(4-3)! = 4·3·2 = 24

Answer:

3600 mg of drug per day.

[tex]\text{The amount of drug per dosage}=\text{1200 mg}[/tex]

Step-by-step explanation:

We have been given that a patient is ordered 60 mg/kg/day of an antibiotic.

Since the weight of the patient is 60 kg, so the dosage of drug per day would be 60 mg times 60.

[tex]\text{The dosage of drug per day}=\frac{\text{60 mg}}{\text{ kg}}\times 60\text{ kg}[/tex]

[tex]\text{The dosage of drug per day}=\text{60 mg}\times 60[/tex]

[tex]\text{The dosage of drug per day}=\text{3600 mg}[/tex]

Therefore, the patient should receive 3600 mg of drug per day.

Since the patient gets 3 doses per day, so dosage of drug per dosage would be amount of drug per day divided by 3.

[tex]\text{The amount of drug per dosage}=\frac{\text{3600 mg}}{3}[/tex]

[tex]\text{The amount of drug per dosage}=\text{1200 mg}[/tex]

Therefore, the amount of drug per dosage would be 1200 mg.

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]

Answer:

1.538 g of streptomycin sulfate

Step-by-step explanation:

As we know, we have 650 μg of streptomycin base in 1 milligram of streptomycin sulfate.

If we convert everithing to grams:

650 μg= 0.00065 g of Streptomycin base for every 0.001 grams of Streptomycin Sulfate so we have :0.001 grs Streptomycin Sulfate/0.00065 gr Streptomycin base=1.538 gr Streptomycin Sulfate/Streptomycin base

Now if we want 1 gram of Streptomycin base we will need:

1 g of Streptomycin base*1.538 gr Streptomycin Sulfate/Streptomycin base= 1.538 gr Streptomycin Sulfate

Answer:

0,00027 grams!

Step-by-step explanation:

This is a scientific notation problem.

When a number is followed by a [tex]10^{-1}[/tex], it means that said number has one zero at the beginning.

In this case, the number is followed by [tex]10^{-4}[/tex], so that means that 2.7 has four zeros at the beginning. So, 2.7x[tex]10^{-4}[/tex] grams is equal to 0,00027 grams! (always the comma goes after the first zero).

The expression "2.7x10^-4 grams" is already given in grams, using scientific notation, and does not require a unit conversion since it is already in the unit we are interested in (grams). To understand what this quantity represents in standard decimal form, let's break down the scientific notation:

Scientific notation is a way of expressing very large or very small numbers in a compact form. The notation "2.7x10^-4" means that you take the number 2.7 and multiply it by 10 raised to the power of -4.

The term "10^-4" means "1 divided by 10 to the 4th power," which is the same as 0.0001 (1 followed by 4 zeros in the denominator).

To convert "2.7x10^-4" to its decimal form, you would perform the multiplication:

2.7 × 0.0001 = 0.00027

So, "2.7x10^-4 grams" is equal to 0.00027 grams in standard decimal notation.

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:

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:

91 people take Russian

26 people take French and Russian but not German

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of this set.

I am going to say that:

-The set A represents the students that take French.

-The set B represents the students that take German

-The set C represents the students that take Russian.

We have that:

[tex]A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)[/tex]

In which a is the number of students that take only Franch, A \cap B is the number of students that take both French and German , A \cap C is the number of students that take both French and Russian and A \cap B \cap C is the number of students that take French, German and Russian.

By the same logic, we have:

[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]

[tex]C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)[/tex]

This diagram has the following subsets:

[tex]a,b,c,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)[/tex]

There are 155 people in my school. This means that:

[tex]a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 155[/tex]

The problem states that:

90 take Franch, so:

[tex]A = 90[/tex]

83 take German, so:

[tex]B = 83[/tex]

22 take French, Russian, and German, so:

[tex]A \cap B \cap C = 22[/tex]

42 take French and German, so:

[tex]A \cap B = 42 - (A \cap B \cap C) = 42 - 22 = 20[/tex]

41 take German and Russian, so:

[tex]B \cap C = 41 - (A \cap B \cap C) = 41 - 22 = 19[/tex]

22 take French as their only foreign language, so:

[tex]a = 22[/tex]

Solution:

(1) How many take Russian?

[tex]C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)[/tex]

[tex]C = c + (A \cap C) + 19 + 22[/tex]

[tex]C = c + (A \cap C) + 41[/tex]

First we need to find [tex]A \cap C[/tex], that is the number of students that take French and Russian but not German. For this, we have to go to the following equation:

[tex]A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)[/tex]

[tex]90 = 22 + 20 + (A \cap C) + 22[/tex]

[tex](A \cap C) + 64 = 90[/tex].

[tex](A \cap C) = 26[/tex]

----------------------------

The number of students that take Russian is:

[tex]C = c + 26 + 41[/tex]

[tex]C = c + 67[/tex]

------------------------------

Now we have to find c, that we can find in the equation that sums all the subsets:

[tex]a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 155[/tex]

[tex]22 + b + c + 20 + 26 + 19 + 22 = 155[/tex]

[tex]b + c + 109= 155[/tex]

[tex]b + c = 46[/tex]

For this, we have to find b, that is the number of students that take only German. Then we go to this eqaution:

[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]

[tex]B = b + 19 + 20 + 22[/tex]

[tex]B = b + 61[/tex]

[tex]b + 61 = 83[/tex]

[tex]b = 22[/tex]

-------

[tex]b + c = 46[/tex]

[tex]c = 46 - b[/tex]

[tex]c = 24[/tex]

The number of people that take Russian is:

[tex]C = c + 67[/tex]

[tex]C = 24 + 67[/tex]

[tex]C = 91[/tex]

91 people take Russian

(2) How many take French and Russian but not German?

[tex](A \cap C) = 26[/tex]

26 people take French and Russian but not German

Answer:

There is a 57.68% probability that this last marble is red.

There is a 20.78% probability that we actually drew the same marble all four times.

Step-by-step explanation:

Initially, there are 50 marbles, of which:

30 are red

20 are blue

Any time a red marble is drawn:

The marble is placed back, and another three red marbles are added

Any time a blue marble is drawn

The marble is placed back, and another five blue marbles are added.

The first three marbles can have the following combinations:

R - R - R

R - R - B

R - B - R

R - B - B

B - R - R

B - R - B

B - B - R

B - B - B

Now, for each case, we have to find the probability that the last marble is red. So

[tex]P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8}[/tex]

[tex]P_{1}[/tex] is the probability that we go R - R - R - R

There are 50 marbles, of which 30 are red. So, the probability of the first marble sorted being red is [tex]\frac{30}{50} = \frac{3}{5}[/tex].

Now the red marble is returned to the bag, and another 3 red marbles are added.

Now there are 53 marbles, of which 33 are red. So, when the first marble sorted is red, the probability that the second is also red is [tex]\frac{33}{53}[/tex]

Again, the red marble is returned to the bag, and another 3 red marbles are added

Now there are 56 marbles, of which 36 are red. So, in this sequence, the probability of the third marble sorted being red is [tex]\frac{36}{56}[/tex]

Again, the red marble sorted is returned, and another 3 are added.

Now there are 59 marbles, of which 39 are red. So, in this sequence, the probability of the fourth marble sorted being red is [tex]\frac{39}{59}[/tex]. So

[tex]P_{1} = \frac{3}{5}*\frac{33}{53}*\frac{36}{56}*\frac{39}{59} = \frac{138996}{875560} = 0.1588[/tex]

[tex]P_{2}[/tex] is the probability that we go R - R - B - R

[tex]P_{2} = \frac{3}{5}*\frac{33}{53}*\frac{20}{56}*\frac{36}{61} = \frac{71280}{905240} = 0.0788[/tex]

[tex]P_{3}[/tex] is the probability that we go R - B - R - R

[tex]P_{3} = \frac{3}{5}*\frac{20}{53}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{937570} = 0.076[/tex]

[tex]P_{4}[/tex] is the probability that we go R - B - B - R

[tex]P_{4} = \frac{3}{5}*\frac{20}{53}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{968310} = 0.0511[/tex]

[tex]P_{5}[/tex] is the probability that we go B - R - R - R

[tex]P_{5} = \frac{2}{5}*\frac{30}{55}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{972950} = 0.0733[/tex]

[tex]P_{6}[/tex] is the probability that we go B - R - B - R

[tex]P_{6} = \frac{2}{5}*\frac{30}{55}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{1004850} = 0.0493[/tex]

[tex]P_{7}[/tex] is the probability that we go B - B - R - R

[tex]P_{7} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{33}{63} = \frac{825}{17325} = 0.0476[/tex]

[tex]P_{8}[/tex] is the probability that we go B - B - B - R

[tex]P_{8} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{30}{65} = \frac{750}{17875} = 0.0419[/tex]

So, the probability that this last marble is red is:

[tex]P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8} = 0.1588 + 0.0788 + 0.076 + 0.0511 + 0.0733 + 0.0493 + 0.0476 + 0.0419 = 0.5768[/tex]

There is a 57.68% probability that this last marble is red.

What's the probability that we actually drew the same marble all four times?

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

[tex]P_{1}[/tex] is the probability that we go R-R-R-R. It is the same [tex]P_{1}[/tex] from the previous item(the last marble being red). So [tex]P_{1} = 0.1588[/tex]

[tex]P_{2}[/tex] is the probability that we go B-B-B-B. It is almost the same as [tex]P_{8}[/tex] in the previous exercise. The lone difference is that for the last marble we want it to be blue. There are 65 marbles, 35 of which are blue.

[tex]P_{2} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{35}{65} = \frac{875}{17875} = 0.0490[/tex]

[tex]P = P_{1} + P_{2} = 0.1588 + 0.0490 = 0.2078[/tex]

There is a 20.78% probability that we actually drew the same marble all four times

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.

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:

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:

a)

x  |  y

==========

16  |  3

12  |  6

8   |  9

4   |  12

b) A large basket contains 23 eggs

Step-by-step explanation:

a) Isolating 3x, we get

3x = 60-4y

This means that 60-4y is a multiple of 3.

Besides, as x is a positive integer, then x = 60-4y has to be greater than zero

60-4y>0

Solving the inequality for y

60>4y

60/4 > y

15>y (or y<15, which is the same)

The multiples of 3 which are smaller than 15 are 3,6,9 and 12

Now we draw a table for these values of y and found the value of x by replacing in the equation 3x = 60-4y---> x = (60-4y)/3

x  |  y

=======

16  |  3

12  |  6

8   |  9

4   |  12

And these are all the solutions in positive integers.

b) This problem is pretty much like problem a)

Let's call x the number of eggs contained in the large baskets and y the number of eggs contained in the small baskets.

Then

Number of eggs bought - number of eggs sold = number of eggs left over.

In this case

11x-39y = 19

But we also know that 0<y<12 because the small baskets hols fewer than a dozen.

Isolating x in the equation 11x-39y + 19, we get

x=(19+39y)/11

Now we make a table with these values of x and y remembering that y is a positive integer smaller than 12

x   | y

=======

5.2  | 1

8.8  | 2

12.4 | 3

15.9 | 4

19.5 | 5

23   | 6

26.6 | 7

30.1 | 8

33.6 | 9

37.2 |10

40.7 |11

From this table we see that the only integer solution for x is 23, so a large basket contains 23 eggs.

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 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]

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:

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.