7. When companies employ control charts to monitor the quality of their products, a series of small samples is typically used to determine if the process is "in control" during the period of time in which each sample is selected. Suppose a concrete-block manufacturer samples nine blocks per hour and tests the breaking strength of each. During one-hour’s test, the mean and standard deviation are 985.6 pounds per square inch (psi) and 22.9 psi, respectively. a. Construct a 99% confidence interval for the mean breaking strength of blocks produced

Answer:  300

Step-by-step explanation:

Given : Hugh and Janet are dining out at a cafe that has on its menu 6 entrees, 5 salads and 10 desserts.

They decide thy will each order a different entree, salad, and dessert, and share the portions.

Then, the number of different meals are possible :-

[tex]6\times5\times10=300[/tex]

Hence, the number of different meals are possible =300

Answer:

The solution is: [tex](x_{1}, x_{2}, x_{3}) = (1,0,-4)[/tex]

Step-by-step explanation:

The Gauss-Jordan elimination method is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.

We have the following system:

[tex]2x_{1} - x_{2} + 3x_{3} = -10[/tex]

[tex]x_{1} - 2x_{2} + x_{3} = -3[/tex]

[tex]x_{1} - 5x_{2} + 2x_{3} = -7[/tex]

This system has the following augmented matrix:

[tex]\left[\begin{array}{ccc}2&-1&3|-10\\1&-2&1|-3\\1&-5&2| -7\end{array}\right][/tex]

To make the reductions easier, i am going to swap the first two lines. So

[tex]L1 <-> L2[/tex]

Now the matrix is:

[tex]\left[\begin{array}{ccc}1&-2&1|-3\\2&-1&3|-10\\1&-5&2| -7\end{array}\right][/tex]

Now we reduce the first row, doing the following operations

[tex]L2 = L2 - 2L1[/tex]

[tex]L3 = L3 - L1[/tex]

So, the matrix is:

[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&3&1|-4\\0&-3&1| -4\end{array}\right][/tex]

Now we divide L2 by 3

[tex]L2 = \frac{L2}{3}[/tex]

So we have

[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&1&\frac{1}{3}|\frac{-4}{3}\\0&-3&1| -4\end{array}\right][/tex]

Now we have:

[tex]L3 = 3L2 + L3[/tex]

So, now we have our row reduced matrix:

[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&1&\frac{1}{3}|\frac{-4}{3}\\0&0&2| -8\end{array}\right][/tex]

We start from the bottom line, where we have:

[tex]2x_{3} = -8[/tex]

[tex]x_{3} = \frac{-8}{2}[/tex]

[tex]x_{3} = -4[/tex]

At second line:

[tex]x_{2} + \frac{x_{3}}{3} = \frac{-4}{3}[/tex]

[tex]x_{2} - \frac{4}{3} = -\frac{4}{3}[/tex]

[tex]x_{2} = 0[/tex]

At the first line

[tex]x_{1} -2x_{2} + x_{3} = -3[/tex]

[tex]x_{1} - 4 = -3[/tex]

[tex]x_{1} = 1[/tex]

The solution is: [tex](x_{1}, x_{2}, x_{3}) = (1,0,-4)[/tex]

Answer:

2 mL

Step-by-step explanation:

Given:

Volume of water for injection = 23 mL

Resulting concentration = 200,000 units per mL

Amount of drug in the vial = 5,000,000 units

Now,

Let the final volume of the solution be 'x' mL

Now, concentration = [tex]\frac{\textup{units of the powder}}{\textup{Total volume of the soltuion}}[/tex]

thus,

200,000 = [tex]\frac{\textup{5,000,000}}{\textup{x}}[/tex]

or

x = 25 mL

also,

Total volume 'x' = volume of water + volume of powder

or

25 mL = 23 + volume of powder

or

Volume of powder = 2 mL

Answer:

[tex]m\angle B=m\angle C=120^{\circ}[/tex]

[tex]m\angle A=m\angle D=60^{\circ}[/tex]

Step-by-step explanation:

Trapezoid ABCD is isosceles trapezoid, because AB = CD (given). In isosceles trapezoid, angles adjacent to the bases are congruent, then

Since BK ⊥ AD, the triangle ABK is right triangle. In this triangle,  AB = 8, AK = 4. Note that the hypotenuse AB is twice the leg AK:

[tex]AB=2AK.[/tex]

If in the right triangle the hypotenuse is twice the leg, then the angle opposite to this leg is 30°, so,

[tex]m\angle ABK=30^{\circ}[/tex]

Since BK ⊥ AD, then BK ⊥ BC and

[tex]m\angle KBC=90^{\circ}[/tex]

Thus,

[tex]m\angle B=30^{\circ}+90^{\circ}=120^{\circ}\\ \\m\angle B=m\angle C=120^{\circ}[/tex]

Now,

[tex]m\angle A=m\angle D=180^{\circ}-120^{\circ}=60^{\circ}[/tex]

Answer:

{ b, d, f }

Step-by-step explanation:

In the roster form we write the elements of a set by separating commas and enclose them within {} bracket.

We have give,

[tex]U = \{a, b, c, d, e, f, g\}[/tex],

[tex]A = \{a, c, e, g\}[/tex],

[tex]B = \{a, b, c, d\}[/tex],

[tex]\because A' = U - A[/tex]

[tex]\implies A' = \{a, b, c, d, e, f, g\} - \{a, c, e, g\}[/tex]

= { b, d, f }

Final answer:

A' is the complement of set A, consisting of elements in the universal set U that are not in A. Given U = {a, b, c, d, e, f, g} and A = {a, c, e, g}, A' is found to be {b, d, f}.

Explanation:

The complement of a set A, denoted by A', consists of all the elements in the universal set U that are not in A. Given the universal set U = {a, b, c, d, e, f, g} and set A = {a, c, e, g}, to find A' we need to identify all elements in U that are not in A.

To find A', we compare each element in U with the elements in A:

If the element is in A, we do not include it in A'.

If the element is not in A, we include it in A'.

After comparing, we find that A' consists of the elements {b, d, f} since these are the elements in U that are not present in set A.

Therefore, the roster form of A' with a single space after each comma is:

{b, d, f}

Learn more about Set Complement here:

https://brainly.com/question/30913259

#SPJ11

Answer: The relation is a function.

Step-by-step explanation: In this situation, we are given a relation in the form of a graph and we are asked if it represents a function. In this situation, we would you something called the vertical line test. In other words, if we can draw a vertical line that passes through more than one point on the graph, then the relation is not a function. Notice that in this problem, it's impossible to draw a vertical line that passes through more than one point on the graph so the relation is a function.

Therefore, this relation must be a function.

Answer:

Y = [tex]e^{t}[/tex] +  [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex] + t - 18

Step-by-step explanation:

y''' − 3y'' + 3y' − y = ex − x + 21

Homogeneous solution:

First  we propose a solution:

Yh = [tex]e^{r*t}[/tex]

Y'h = [tex]r*e^{r*t}[/tex]

Y''h = [tex]r^{2}*e^{r*t}[/tex]

Y'''h = [tex]r^{3}*e^{r*t}[/tex]

Now we solve the following equation:

Y'''h - 3*Y''h + 3*Y'h - Yh = 0

[tex]r^{3}*e^{r*t}[/tex] - 3*[tex]r^{2}*e^{r*t}[/tex] + 3*[tex]r*e^{r*t}[/tex] - [tex]e^{r*t}[/tex] = 0

[tex]r^{3} - 3r^{2} + 3r - 1 = 0[/tex]

To solve the equation we must propose a solution to the  polynomial :

r = 1

To find the other r we divide the polynomial by (r-1) as you can see  

attached:

solving the equation:

(r-1)([tex]r^{2} - 2r + 1[/tex]) = 0

[tex]r^{2} - 2r + 1[/tex] = 0

r = 1

So we have 3 solution [tex]r_{1} = r_{2} =r_{3}[/tex] = 1

replacing in the main solution

Yh =  [tex]e^{t}[/tex] +  [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex]

The t and [tex]t^{2}[/tex] is used because we must have 3 solution  linearly independent

Particular solution:

We must propose a Yp solution:

Yp = [tex]c_{1} (t^{3} + t^{2} + t + c_{4} )e^{t} + c_{2} t + c_{3}[/tex]

Y'p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + c_{1}( 3t^{2} + 2t + 1 )e^{t} + c_{2}[/tex]

Y''p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + c_{1}(6t + 2)e^{t}[/tex]

Y'''p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + 6c_{1}e^{t}[/tex]

Y'''p - 3*Y''p + 3*Y'p - Yp = [tex]e^{t} - t + 21[/tex]

[tex]6c_{1}e^{t} - 18c_{1} te^{t} - 6c_{1} e^{t} + 6c_{1} te^{t} + 9c_{1} t^{2} e^{t} + 3c_{1}e^{t} + 3c_{2} - c_{2} t -  c_{3}[/tex] = [tex]e^{t} - t + 21[/tex]

equalizing coefficients of the same function:

- 12c_{1} = 0

9c_{1} = 0

3c_{1} = 0

c_{1} = 0

3c_{2} - c_{3} = 21 => c_{5} = [tex]\frac{1}{3}[/tex]

-c_{2} = -1

c_{2} = 1

c_{3} = -18

Then we have:

Y = [tex]e^{t}[/tex] +  [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex] + t - 18

Answer:

a) The system has a unique solution for [tex]k\neq 6[/tex] and any value of [tex]h[/tex], and we say the system is consisted

b) The system has infinite solutions for [tex]k=6[/tex] and [tex]h=8[/tex]

c) The system has no solution for [tex]k=6[/tex] and [tex]h\neq 8[/tex]

Step-by-step explanation:

Since we need to base the solutions of the system on one of the independent terms ([tex]h[/tex]), the determinant method is not suitable and therefore we use the Gauss elimination method.

The first step is to write our system in the augmented matrix form:

[tex]\left[\begin{array}{cc|c}1&3&4\\2&k&h\end{array}\right][/tex]

The we can use the transformation [tex]r_0\rightarrow r_0 -2r_1[/tex], obtaining:

[tex]\left[\begin{array}{cc|c}1&3&4\\0&k-6&h-8\end{array}\right][/tex].

Now we can start the analysis:

[tex](k-6)x_2=h-8\\x_2=h-8/(k-6)[/tex]

from where we can see that only in the case of [tex]k=6[/tex] the value of [tex]x_2[/tex] can not be determined.

[tex](k-6)x_2=h-8\\0=0[/tex] which means that the second equation is a linear combination of the first one. Therefore, we can solve the first equation to get [tex]x_1[/tex] as a function of [tex]x_2[/tex] o viceversa. Thus,  [tex]x_2[/tex] ([tex]x_1[/tex]) is called a parameter since there are no constraints on what values they can take on.

if [tex]k=6[/tex] and [tex]h\neq 8[/tex] the system has no solution. Again by substituting in the equation resulting from the last row:

[tex](k-6)x_2=h-8\\0=h-8[/tex] which is false for all values of [tex]h\neq 8[/tex] and since we have something that is not possible [tex](0\neq h-8,\ \forall \ h\neq 8)[/tex] the system has no solution

Answer:

98w +115(18-w)

Step-by-step explanation:

The fees per credit hour at Westside are $98.

The fees per credit hour at Pinewood are $115.

The total amount of hours is 18, this means the hours at Westside plus the hours at Pinewood add up to 18.

The problem asks us to name "w" the credit hours at Westside, therefore, the credit hours at Pinewood would be "18 - w"

Combining this information, we have that the combined total amount  he paid would be:

(Fee per hour at Westside)(hours at Westside) + (Fee per hour at Pinewood)(hours at Pinewood) = Total

⇒98w + 115(18 - w) = Total

Final answer:

The expression for the combined total cost of Ravi's classes is 98w + 115(18 - w), where w represents the number of credit hours taken at Westside College, and (18 - w) represents the number of credit hours taken at Pinewood College.

Explanation:

To calculate the combined total cost of Ravi's classes at Westside Community College and Pinewood Community College, we can create an expression based on the number of credit hours he is taking at each school. If he is taking w credit hours at Westside, where the cost is $98 per credit hour, the total cost at Westside would be 98w. Since Ravi is taking a total of 18 credit hours at both schools combined, he would be taking (18 - w) credit hours at Pinewood, where the cost is $115 per credit hour. The total cost at Pinewood would therefore be 115(18 - w).

The expression for the combined total amount paid for Ravi's classes would be the sum of the costs for each school:

Total Paid = 98w + 115(18 - w)

Answer:

[tex]y(x) = e^{3x} [Acos x+Bsin x][/tex]

Step-by-step explanation:

Given is a differential equation

[tex]y" - 6y' +10y=0.[/tex]

We have characteristic equation as

[tex]m^2-6m+10 =0[/tex]

The above quadratic cannot be factorised hence use formula

[tex]m=\frac{6+/-\sqrt{36-40} }{2} \\=3+i, 3-i[/tex]

Hence general solution would be

[tex]y(x) = e^{3x} [Acos x+Bsin x][/tex]

Answer:

a) The patient weighs 75.15kg = 75.1kg, rounded to the nearest tenth.

b) The patient should 75.1mg a day of the drug.

c) The patient should receive 0.37mL per dose, rounded to the nearest hundreth.

Step-by-step explanation:

These problems can be solved by direct rule of three, in which we have cross multiplication.

a. How many kilograms does the patient weigh?

The problem states that patient weighs 167lb. Each lb has 0.45kg. So:

1 lb - 0.45kg

167 lb - xkg

[tex]x = 167*0.45[/tex]

[tex]x = 75.15[/tex]kg

The patient weighs 75.15kg = 75.1kg, rounded to the nearest tenth.

b. How many milligrams should the patient receive per day?

The drug has 1mg/kg. The patient weighs 75.1kg. So

1 mg - 1 kg

x mg - 75.1kg

[tex]x = 75.1[/tex]mg

The patient should 75.1mg a day of the drug.

c. How many milliliters should the patient receive per dose?

The drug is SubQ, q12h. This means that the drug is administered twice a day, so there are 2 doses. 75.1mg of the drug are administered a day. so:

2 doses - 75.1mg

1 dose - xmg

[tex]2x = 75.1[/tex]

[tex]x = \frac{75.1}{2}[/tex]

[tex]x = 37.5[/tex] SubQ, q12h

For each dose, the patient should receive 37.5mg. Each 40mg of the drug has 0.4mL. So:

40mg - 0.4ml

37.5mg - xmL

[tex]40x = 0.4*37.4[/tex]

[tex]x = \frac{0.4*37.4}{40}[/tex]

[tex]x = 0.374mL[/tex]

The patient should receive 0.37mL per dose, rounded to the nearest hundreth.

Answer:

MNP is a Congruent Triangle according to Side-Side-Side Congruence.

Step-by-step explanation:

Whenever we talk about Euclidean Geometry and Congruence of Triangles. We are taking into account that in any given plane, three given points, in this case, M, N, and P is and its segments between two points make up a Triangle.

In this case MNP and an identical one and PNM

To be called congruent, it's necessary to have the same length each side and when it comes to angles, congruent angles have the same measure.

A postulate, cannot be proven since it's self-evident. And there's one that fits for this case which says

"Every SSS (Side-Side-Side) correspondence is a congruence"

So this is why we can classify MNP as Side-Side-Side congruence since its segments are the same size MN, NP, and MP for both of them. The order of the letters does not matter.

Answer:

There is no solution for this system

Step-by-step explanation:

I am going to solve this system by the Gauss-Jordan elimination method.

The Gauss-Jordan elimination method is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.

We have the following system:

[tex]2x_{1} - x_{2} + 3x_{3} = -10[/tex]

[tex]x_{1} - 2x_{2} + x_{3} = -3[/tex]

[tex]x_{1} - 5x_{2} + 2x_{3} = -7[/tex]

This system has the following augmented matrix:

[tex]\left[\begin{array}{ccc}1&-3&4|-4\\3&-7&7|-8\\-4&6&-1|7\end{array}\right][/tex]

We start reducing the first row. So:

[tex]L2 = L2 - 3L1[/tex]

[tex]L3 = L3 + 4L1[/tex]

Now the matrix is:

[tex]\left[\begin{array}{ccc}1&-3&4|-4\\0&2&-5|4\\0&-6&15|-9\end{array}\right][/tex]

We divide the second line by 2:

[tex]L2 = \frac{L2}{2}[/tex]

And we have the following matrix:

[tex]\left[\begin{array}{ccc}1&-3&4|-4\\0&1&\frac{-5}{2}|2\\0&-6&15|-9\end{array}\right][/tex]

Now we do:

[tex]L3 = L3 + 6L2[/tex]

So we have

[tex]\left[\begin{array}{ccc}1&-3&4|-4\\0&1&\frac{-5}{2}|2\\0&0&0|3\end{array}\right][/tex]

This reduced matrix means that we have:

[tex]0x_{3} = 3[/tex]

Which is not possible

There is no solution for this system

Answer:

The dosage has 1.2ml

Step-by-step explanation:

This problem 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.

Unit conversion problems, like this one, is an example of a direct relationship between measures.

First step: The first step is the conversion from mcg to mg.

The bottle has 60 mcg. How much is this in ml? 1 mcg has 0.001mg. So:

1 mcg - 0.001mg

60mcg - x mg

x = 60*0.001

x = 0.06mg

Final step: The bottle of elixir contains 0.05 mg per 1 ml. Calculate the dosage in ml.

The dosage has 0.06 mg, so:

0.05mg - 1 ml

0.06mg - xml

0.05x = 0.06

[tex]x = \frac{0.06}{0.05}[/tex]

x = 1.2ml

The dosage has 1.2ml

Answer:

Step-by-step explanation:

Let's say that each side of the square is of length [tex]s[/tex]. The area of this square would then be:

[tex]A = s^{2}[/tex]

If we increase the length of each side by 50%, then the length becomes [tex]1.5s[/tex], which will result in the area being:

[tex]A = (1.5s)(1.5s)[/tex]

[tex]A = 2.25s^{2}[/tex]

This means the area has increased by [tex]1.25[/tex] the original amount, or 125%.

Answer:

[tex] (x-4)^2+(y+5)^2=90[/tex]

Step-by-step explanation:

The equation of a circle of radius r, centered at the point (a,b) is

[tex](x-a)^2+(y-b)^2=r^2[/tex]

We already know the center is at [tex](4,-5)[/tex], we are just missing the radius. To find the radius, we can use the fact that the circle passes through the point (7,4), and so the radius is just the distance from the center to this point (see attached image). So we find the distance by using distance formula between the points (7,4) and (4,-5):

radius[tex]=\sqrt{(7-4)^2+(4-(-5))^2}=\sqrt{3^2+9^2}=\sqrt{90}[/tex]

And now that we know the radius, we can write the equation of the circle:

[tex] (x-4)^2+(y-(-5))^2=\sqrt{90}^2[/tex]

[tex] (x-4)^2+(y+5)^2=90[/tex]

Answer:

number of 1 bedroom units are 96

number of 1 bedroom units are 64

number of 1 bedroom units are 32

Step-by-step explanation:

Let the number of 1 bedroom units be 'a'

number of 2 bedroom units be 'b'

and,

number of 3 bedroom units be 'c'

now,

according to the question

a + b + c = 192 ................. (1)

also,

b + c = a ..............(2)

and,

a = 3c ...................(3)

now,

substituting value of 'a' from 3  into 2, we get

b + c = 3c

or

b = 2c ...................(4)

also,

from 3 and 1

3c + b + c = 192

or

4c + b = 192  ................(5)

now from 4 and 5,

4c + 2c = 192

or

6c = 192

or

c = 32 units

now, substituting c in equation 4, we get

b = 2 × 32 = 64 units

and, substituting c in equation (3), we get

a = 3 × 32 = 96 units

Therefore,

number of 1 bedroom units are 96

number of 1 bedroom units are 64

number of 1 bedroom units are 32

Answer:

The equation of the line is:

[tex]y = x - 3[/tex]

Step-by-step explanation:

The general equation of a straight line is given by:

[tex]y = ax + b[/tex]

Being given two points, we can replace x and y, solve the system and find the values for a and b.

Solution:

The line goes through the point [tex](2,-1)[/tex]. It means that when [tex]x = 2, y = -1[/tex]. Replacing in the equation:

[tex]y = ax + b[/tex]

[tex]-1 = 2a + b[/tex]

[tex]2a + b = -1[/tex]

The line also goes through the point [tex](5,2)[/tex]. It means that when [tex]x = 5 y = 2[/tex]. Replacing in the equation:

[tex]y = ax + b[/tex]

[tex]2 = 5a + b[/tex]

[tex]5a + b = 2[/tex]

Now we have to solve the following system of equations:

[tex]1) 2a + b = -1[/tex]

[tex]2) 5a + b = 2[/tex]

From 1), we have:

[tex]b = -1 - 2a[/tex]

Replacing in 2)

[tex]5a - 1 - 2a = 2[/tex]

[tex]3a = 3[/tex]

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

[tex]a = 1[/tex]

[tex]b = -1 - 2a = -1 - 2 = -3[/tex]

The equation of the line is:

[tex]y = x - 3[/tex]

Answer:

3 : 8

Step-by-step explanation:

Let x quantity of 10% ethanol is mixed with y quantity of 65% ethanol to obtain 50% ethanol mixture,

Thus, the total quantity of resultant mixture = x + y

Also, ethanol in 10% ethanol mixture + ethanol in 65% ethanol mixture = ethanol in resultant mixture,

⇒ 10% of x + 65% of y = 50% of (x+y)

[tex]\implies \frac{10x}{100}+\frac{65y}{100}=\frac{50(x+y)}{100}[/tex]

⇒ 10x + 65y = 50(x+y)

⇒ 10x + 65y = 50x+50y

⇒ 10x - 50x = 50y - 65y

⇒ -40x = -15y

[tex]\implies \frac{x}{y}=\frac{15}{40}=\frac{3}{8}[/tex]

Answer:

a) The patient is going to need 3.2 pills. So four doses.

b) The last dose is at midnight, that i consider 12 PM

Step-by-step explanation:

This problem can be solved by a rule of three in which the measures are directly related, meaning that we have a cross multiplication.

a) How many pills will this patient need?

The first step is finding how many mg she is going to need.

She weighs 50kg, and the total daily amount cannot exceed 8 mg per kg of body. So:

1kg - 8 mg

50 kg - x mg

[tex]x = 50*8[/tex]

[tex]x = 400[/tex]mg.

Each dose has 125 mg, so she will need:

1 dose - 125 mg

x doses - 400mg

[tex]125x = 400[/tex]

[tex]x = \frac{400}{125}[/tex]

[tex]x = 3.2[/tex]

The patient is going to need 3.2 pills. So four doses.

b) If the first dose is administered at 6 AM, when is the last dose given?

There are four doses, administered every 6 hours.

6AM: First dose

12AM: Second dose

6PM: Third dose

12PM: Fourth Dose

The last dose is at midnight, that i consider 12 PM

Final answer:

To calculate the number of pills, we first determine the total daily allowable dosage based on the patient's weight and divide that by the dosage per pill, rounding up to provide whole pills. Then we schedule the medication every 6 hours starting from 6 AM, with the last dose at 12 AM.

Explanation:

The question involves calculating the appropriate dosage of medication for a patient based on weight and the maximum allowable daily dosage.

The medication prescribed is available as enteric-coated pills of 125mg each, and the patient requires a dosage that does not exceed 8mg per kg of their body weight per day.

The patient weighs 50 kg. To find the total daily dosage allowed, we multiply the patient's weight by the maximum mg per kg: 50 kg × 8 mg/kg = 400 mg.

Since the patient can have 400 mg of medication per day and the medication comes in 125 mg pills, we divide the total daily dosage by the dosage per pill: 400 mg ÷ 125 mg/pill = 3.2 pills.

Since we cannot give a patient a fraction of a pill, the patient will need 4 pills per day.

The medication must be administered every 6 hours. If the first dose is at 6 AM, the subsequent doses will be at 12 PM, 6 PM, and 12 AM, with the last dose given precisely 18 hours after the first dose.

Answer:

If [tex]n^2[/tex] is divisible by 3, the n is also divisible by 3.

Step-by-step explanation:

We will prove this with the help of contrapositive that is we prove that if n is not divisible by 3, then, [tex]n^2[/tex] is not divisible by 3.

Let n not be divisible by 3. Then [tex]\frac{n}{3}[/tex] can be written in the form of fraction [tex]\frac{x}{y}[/tex], where x and y are co-prime to each other or in other words the fraction is in lowest form.

Now, squaring

[tex]\frac{n^2}{9} = \frac{x^2}{y^2}[/tex]

Thus,

[tex]n^2 = \frac{9x^2}{y^2}[/tex]

[tex]\frac{n^2}{3} = \frac{3x^2}{y^2}[/tex]

It can be clearly seen that the fraction [tex]\frac{3x^2}{y^2}[/tex] is in lowest form.

Hence, [tex]n^2[/tex] is not divisible by 3.

Thus, by contrapositivity if [tex]n^2[/tex] is divisible by 3, the n is also divisible by 3.

Answer: The total purchase price is $ 29,032

Step-by-step explanation:

Hi, to solve this problem you have to solve the percentages of each taxes first.

So :

The next step is adding the taxes results to the speedboat cost.

so:

speedboat = $24,500

speedboat + taxes = $24,500 + $1,347.5 +$3,185 = $29,032

The total purchse price for the speedboat is $29,032.

Final answer:

The height of the pile after 3 weeks is 43 181/192 inches.

Explanation:

To find the height of the pile after 3 weeks, we need to add the increase in height for each week. The initial height of the pile is 17 3/4 inches. For each week, the height increases by 8 7/12 inches.

So after 1 week, the height is (17 3/4 + 8 7/12) inches.

After 2 weeks, the height is [(17 3/4 + 8 7/12) + 8 7/12] inches.

And after 3 weeks, the height is [((17 3/4 + 8 7/12) + 8 7/12) + 8 7/12] inches.

Let's calculate:

So, after 3 weeks, the height of the pile of newspapers is 43 181/192 inches.

Answer: Algorithm complexity or the complexity of an algorithm is known as a measure under which one evaluates degree of count of operations, that are specifically performed by an algorithm which is taken in consideration as a function of size of the data. In rudimentary terms, it is referred to as a rough approximation of number of stages required in order to enforce an algorithm.

Answer:

Angle A = 60º and Angle B = 30º

Step-by-step explanation:

We know that A and B are complementary angles, therefore ∠A + ∠B =90. (+)

On the other hand, ∠A is half as large as ∠B; this can be written algebraically as [tex]A=\frac{B}{2}[/tex]. (*)

If we substitute (*) in (+) we get:

[tex]\frac{B}{2}+B = 90\\\\ \frac{B+2B=180}{2} \\\\ B+2B=180\\\\ 3B=180\\\\ B=\frac{180}{3}\\\\ B=60[/tex]

And now we substitute the value of B in (+) and we get:

∠A+60 = 90

∠A = 90-60

∠A = 30

Answer:

Part 1.

In 12 hours 30 units of  Humulin R insulin in 300 mL is to be infused.

So, per hour = [tex]\frac{30}{12}= 2.5[/tex] units are to be infused.

Part 2.

Now 30 units Humulin R insulin in 300 mL of normal saline (NS).

So, 2.5 units will be in : [tex]\frac{300\times2.5}{30}= 25[/tex] mL

Hence. 2.5 units Humulin R insulin in 25 mL of normal saline (NS).

In this type of problems what we have to do is unit conversion. In order to do so we need all the equivalences which we will be mentioning during the explanation of the problem:

First of all the answer is asked to be in acre-feet and we can see the data we are getting from the rain is in [tex]in*Km^{2}[/tex] not even a volume unit.

To calculate the volume of poured rain we need to have both numbers in the same units, we will convert [tex]Km^{2}[/tex] to [tex]in^{2}[/tex] using the equivalence [tex]1 Km^{2}=1550001600in^{2}[/tex] like this:

[tex]101Km^{2}*\frac{1550001600in^{2} }{1Km^{2}}=156550161600in^{2}[/tex]

it is possible now to calculate the volume ([tex]Volume_{cuboid}=Area*Height[/tex]) like this:

[tex]Volume_{cuboid}=156550161600in^{2}*2.5in=391375404000in^{3}[/tex]

Now we just need to convert this volume to acre-feet and we will do so using the equivalence [tex]1acre-foot=751271680in^{3}[/tex] like this:

[tex]391375404000in^{3}*\frac{1acre-foot}{751271680in^{3}}=5199.50403658 acre-feet[/tex]

5199.50403658 acre-feet would be the answer to our problem

Answer: There were 469,532 American prison for drug related offenses in 2015.

Step-by-step explanation:

Since we have given that

Number of Americans in prison for drug related offenses in 1980 = 40,900

Rate of increment in Americans in prison from 1980 to 2015 = 1048%

So, Number of Americans who were in prison for drug related offenses in 2015 is given by

[tex]\dfrac{100+1048}{100}\times 40900\\\\\dfrac{1148}{100}\times 40900\\\\=1148\times 409\\\\=469,532[/tex]

Hence, there were 469,532 American prison for drug related offenses in 2015.