2. In an experiment with a mass attached to a spring, a student measures the period of the oscillation t as follows: 1.94 s, 1.96 s, 2.01 s, 1.98 s, 2.02 s, 2.00 s, 1.99 s, 2.01 s, 1.98 s, 1.97 s. a) What is the average and b) standard deviation? Use the instructions for excel to find statistics on a set of similar numbers, round to the correct # sig figs. Zeroes to the right do. ALWAYS include units for a number that has units. These statistics have the same units all the way throug

Answers

Answer 1

Answer: a) The average of this set of numbers is 1.98 s. b) the standard deviation is 0.02 s.

Step-by-step explanation: The average is calculated by adding all numbers found in the experiment divided by the number of values. [tex]average = \frac{Sumxi}{n}[/tex]

The standard deviation is given by the square root of the squared sum of the difference between each number and the average divided by the total number of values. For instance, std deviation = [tex]\frac{sqrt{(1.94-1.98)^{2}+(1.96-1.98)^2+(2.01-1.98)^2...}}{11}[/tex] and that is done with all 11 terms of data minus the average to find the standard deviation.

In excel, the average of a certain set of numbers (displayed in cells A1 to A5) can be found by the commands =average(A1:A5). The standard deviation can be found by the commands =stdedv.p(A1:A5) in which p is the population. You can decrease decimals by clicking on the icon displayed in the Ribbon.


Related Questions

If AA and BB are countable sets, then so is A∪BA∪B.

Answers

Answer with Step-by-step explanation:

We are given that A and B are two countable sets

We have to show that if A and B are countable then [tex]A\cup B[/tex] is countable.

Countable means finite set or countably infinite.

Case 1: If A and B are two finite sets

Suppose A={1} and B={2}

[tex]A\cup B[/tex]={1,2}=Finite=Countable

Hence, [tex]A\cup B[/tex] is countable.

Case 2: If A finite and B is countably infinite

Suppose, A={1,2,3}

B=N={1,2,3,...}

Then, [tex]A\cup B[/tex]={1,2,3,....}=N

Hence,[tex]A\cup B[/tex] is countable.

Case 3:If A is countably infinite and B is finite set.

Suppose , A=Z={..,-2,-1,0,1,2,....}

B={-2,-3}

[tex]A\cup B[/tex]=Z=Countable

Hence, [tex]A\cup B[/tex] countable.

Case 4:If A and B are both countably infinite sets.

Suppose A=N and B=Z

Then,[tex]A\cup B[/tex]=[tex]N\cup Z[/tex]=Z

Hence,[tex]A\cup B[/tex] is countable.

Therefore, if A and B are countable sets, then [tex]A\cup B[/tex] is also countable.

Answer:

To remedy confusions like yours and to avoid the needless case analyses, I prefer to define X to be countable if there is a surjection from N to X.

This definition is equivalent to a few of the many definitions of countability, so we are not losing any generality.

It is a matter of convention whether we allow finite sets to be countable or not (though, amusingly, finite sets are the only ones whose elements you could ever finish counting).

So, if A and B be countable, let f:N→A and g:N→B be surjections. Then the two sequences (f(n):n⩾1)=(f(1),f(2),f(3),…) and (g(n):n⩾1)=(g(1),g(2),g(3),…) eventually cover all of A and B, respectively; we can interleave them to create a sequence that will surely cover A∪B:

(h(n):n⩾1):=(f(1),g(1),f(2),g(2),f(3),g(3),…).

An explicit formula for h is h(n)=f((n+1)/2) if n is odd, and h(n)=g(n/2) if n is even.

Hope it helps uh mate...✌

Use z scores to compare the given values.
Based on sample​ data, newborn males have weights with a mean of 3232.9 g and a standard deviation of 714.6 g. Newborn females have weights with a mean of 3094.9 g and a standard deviation of 586.3 g. Who has the weight that is more extreme relative to the group from which they​ came: a male who weighs 1600 g or a female who weighs 1600 ​g?
Since the z score for the male is z =__ and the z score for the female is z=__​, the
female or male(which one) has the weight that is more extreme.

Answers

Answer:

Female

Step-by-step explanation:

The formula for finding z-score is;

z=(x-μ)/δ

where x= weight given=1600g

μ=mean weight given

δ=standard deviation given

For male

x=1600g

μ=3232.9g

δ=714.6g

z=(1600-3232.9)/714.6

z= -2.3

The weight for the male is 2.3 standard deviations below the mean

For female

x=1600g

μ=3094.9g

δ=586.3g

z=(1600-3094.9)/586.3

z= -2.5

The weight for the female is 2.5 standard deviations below the mean

⇒The female has the weight that is more extreme

Answer:

z= 1

Step-by-step explanation:

got it on egde

"The diagram above represents a square garden. If each side of the garden is increased in length by 50%, by what percent is the area of the garden increased?" Whats the explanation for this problem, I cant seem to figure it out?

Answers

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%.

At a certain school, twenty-five percent of the students wear a watch and thirty percent wear a bracelet. Sixty percent of the students wear neither a watch nor a bracelet. (a) One of the students is chosen at random. What is the probability that this student is wearing a watch or a bracelet? (b) What is the probability that this student is wearing both a watch and a bracelet?

Answers

Answer:  a) 0.40    b) 0.15

Step-by-step explanation:

Let A denotes the event that students wear a watch and B denotes the event that students wear a bracelet.

Given : P(A)=0.25   ;   P(B)=0.30

[tex]P(A'\cup B')=0.60[/tex]

Since, [tex]P(A\cup B)=1-P(A'\cup B')=1-0.60=0.40[/tex]

Thus, the probability that this student is wearing a watch or a bracelet = 0.40

Also, [tex]P(A\cup B)=P(A)+P(B)-P(A\cap B)[/tex]

[tex]P(A\cap B)=P(A)+P(B)-P(A\cup B)[/tex]

[tex]P(A\cap B)=0.25+0.30-0.40\\\\\Rightarrow\ P(A\cap B)=0.15[/tex]

Thus,  the probability that this student is wearing both a watch and a bracelet= 0.15

Answer:

Step-by-step explanation:

Given that at a certain school, twenty-five percent of the students wear a watch and thirty percent wear a bracelet.

A- people who wear watch = 25%

B - people who wear bracelet = 30%

(AUB)' - People who wear neither a watch nor a bracelet=60%

[tex]A \bigcap B[/tex] - People who wear both =100%-60% = 40%

a) [tex]P(AUB) = P(A)+P(B)-P(AB) = 25%+30%-40%\\= 15%[/tex]

b) the probability that this student is wearing both a watch and a bracelet

= [tex]P(A \bigcap B) = 40%[/tex]

The Eco Pulse survey from the marketing communications firm Shelton Group asked individuals to indicate things they do that make them feel guilty (Los Angeles Times, August 15, 2012). Based on the survey results, there is a .39 probability that a randomly selected person will feel guilty about wasting food and a .27 probability that a randomly selected person will feel guilty about leaving lights on when not in a room. Moreover, there is a .12 probability that a randomly selected person will feel guilty for both of these reasons. a. What is the probability that a randomly selected person will feel guilty for either wasting food or leaving lights on when not in a room (to 2 decimals)? b. What is the probability that a randomly selected person will not feel guilty for either of these reasons (to 2 decimals)?

Answers

Answer:

a) There is a probability of 42% that the person will feel guilty for only one of those things.

b)There is a probability of 46% that a randomly selected person will not feel guilty for either of these reasons

Step-by-step explanation:

This probability problem can be solved by building a Venn like diagram for each probability.

I say that we have two sets:

-Set A, for those people that will feel guilty about wasting food.

-Set B, for those people that will feel guilty about leaving lights on when not in a room.

The most important information is that there is a .12 probability that a randomly selected person will feel guilty for both of these reasons. It means that [tex]P(A \cap B) = .12.[/tex]

The problem also states that there is a .39 probability that a randomly selected person will feel guilty about wasting food. It means that P(A) = 0.39. The probability of a person feeling guilty for only wasting food is PO(A) = .39-.12 = .27.

Also, there is a .27 probability that a randomly selected person will feel guilty about leaving lights on when not in a room. So, the probability of a person feeling guilty for only leaving the lights on is PO(B) = 0.27-0.12 = 0.15.

a) What is the probability that a randomly selected person will feel guilty for either wasting food or leaving lights on when not in a room?

This is the probability that the person feels guilt for only one of those things, so:

P = PO(A) + PO(B) = 0.27 + 0.15 = 0.42 = 42%

b) What is the probability that a randomly selected person will not feel guilty for either of these reasons

The sum of all the probabilities is always 1. In this problem, we have the following probabilies

- The person will not feel guilty for either of these reasons: P

- The person will feel guilty for only one of those things:  PO(A) + PO(B) = 0.42

- The person will feel guilty for both reasons: PB = 0.12

So

`P + 0.42 + 0.12 = 1

P = 1-0.54

P = 0.46

There is a probability of 46% that a randomly selected person will not feel guilty for either of these reasons

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. how many different meals are possible?

Answers

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

The Cabernet Cafe sold eight times as much Cabernet Sauvignon as Zinfandel. How many bottles of Cabernet Sauvignon were sold if the cafe sold 5 bottles of Zinfindel? This class is Food and Beverage Cost

Answers

Answer:

40 bottles

Step-by-step explanation:

In the problem it is said that there were sold eigth times as much Cabernet Sauvignon as Zinfandel, this menas that if for example the cafe sold 1 bottle of zinfandel it would be 8 bottles of Cabernet, so we have to multiply the number of bottles sold fon Zinfandel by 8, as there were 5 bottles of sauvingon sold we have to multiply

8x5=40 bottles.

Final answer:

The Cabernet Cafe sold 40 bottles of Cabernet Sauvignon, given that it sold 8 times as many as the 5 bottles of Zinfandel.

Explanation:

The question involves a simple proportion calculation, typical of mathematics problems. Given the information that the Cabernet Cafe sold eight times as much Cabernet Sauvignon as Zinfandel, we can set up a ratio and solve for the unknown. In this case, we know that the number of Cabernet Sauvignon sold is 8 times the amount of Zinfandel sold. If the cafe sold 5 bottles of Zinfandel, using the said proportion, we multiply 5 (number of Zinfandel bottles sold) by 8 (the given ratio) to find the number of Cabernet Sauvignon bottles sold - which is 40 bottles.

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4. Find the center and the radius of the circle which circumscribes the triangle with vertices ai, a, a3. Express the result in symmetric form.

Answers

Answer:

[tex]\left[\begin{array}{ccc}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{array}\right]=\left[\begin{array}{ccc}-a_{1}^{2}-b_{1}^{2}\\-a_{2}^{2}-b_{2}^{2}\\-a_{3}^{2}-b_{3}^{2}\end{array}\right][/tex]

Step-by-step explanation:

In the question,

We have to find out the circumcentre of the circle passing through the triangle with the vertices (a₁, b₁), (a₂, b₂) and (a₃, c₃).

So,

The circle is passing through these points the equation of the circle is given by,

[tex]x^{2}+y^{2}+ax+by+c=0[/tex]

On putting the points in the circle we get,

[tex]x^{2}+y^{2}+ax+by+c=0\\(a_{1})^{2}+(b_{1})^{2}+a(a_{1})+b(b_{1})+c=0\\and,\\(a_{2})^{2}+(b_{2})^{2}+a(a_{2})+b(b_{2})+c=0\\and,\\(a_{3})^{3}+(b_{3})^{3}+a(a_{3})+b(b_{3})+c=0\\[/tex]

So,

[tex](a_{1})^{2}+(b_{1})^{2}+a(a_{1})+b(b_{1})+c=0\\a(a_{1})+b(b_{1})+c=-(a_{1})^{2}-(b_{1})^{2}\,.........(1)\\and,\\a(a_{2})+b(b_{2})+c=-(a_{2})^{2}-(b_{2})^{2}\,.........(2)\\and,\\a(a_{3})+b(b_{3})+c=-(a_{3})^{3}-(b_{3})^{3}\,.........(3)\\[/tex]

On solving these equation using, Matrix method we can get the required equation of the circle,

[tex]\left[\begin{array}{ccc}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{array}\right]=\left[\begin{array}{ccc}-a_{1}^{2}-b_{1}^{2}\\-a_{2}^{2}-b_{2}^{2}\\-a_{3}^{2}-b_{3}^{2}\end{array}\right][/tex]

This is the required answer.

Final answer:

The center of the circumscribed circle of the symmetric isosceles triangle is at the origin, and the radius is equal to the length of the triangle's equal sides, denoted as r.

Explanation:

To find the center and the radius of the circle which circumscribes the triangle with vertices at ai, a, and a3, we must first understand the nature of the triangle. Given that the triangle is described to be symmetric with equal sides AB = BC = r, it is an isosceles triangle. The perpendicular bisector of the base a will pass through the midpoint of the base and the opposite vertex, given it is symmetric about this bisector. This will also be the diameter of the circumscribed circle. Consequently, as the triangle is isosceles and symmetric, we can use the properties of similar isosceles triangles to solve for the center and radius of the circumscribed circle.

Since the base a will also be the diameter of the circumscribed circle, and we know from geometry that the diameter is twice the radius (a = 2r), the radius of the circumscribing circle is r. The center of this circle is at the midpoint of the base a in the given symmetric form. Therefore, the center of the circle is at the origin due to the symmetric property and the radius remains r.

If we are performing a two-tailed test of whether mu = 100, the probability of detecting a shift of the mean to 105 will be ________ the probability of detecting a shift of the mean to 110.

Answers

Answer:

less than

Step-by-step explanation:

When we are performing a two-tailed test of whether μ = 100,  the probability of detecting a shift of the mean to 105 will be ___Less__than___ the probability of detecting a shift of the mean to 110.

this means that

mean of 105 < mean of 110.

Kim uses two ropes to help support a newly planted tree. The ropes are placed on both sides of the tree and pinned to the ground, as shown in the diagram above. The length of each rope from the tree to the ground is 14 feet. The distance from the base of the tree to where it is pinned to the ground is 8 feet.

Enter the acute angle each rope makes with the ground, rounded to the nearest degree.

Answers

Answer:

Angle = 35°

Step-by-step explanation:

The figure would form a triangle and the tree will divide it into two 90 degree triangles.

Height of the triangles will be 8 feet which is opposite to the angle you have to find.

Hypotenuse of both 90 degree triangles will be 14 feet.

To find the angle, use the sin formula

sin (angle) = opposite/hypotenuse

angle = ?

opposite = 8

hypotenuse = 14

sin (angle) = 8/14

angle = sin inverse (0.57)

angle = 34.8° rounded off to 35°

The angle will be the same for both triangles.

Therefore, the acute angle each triangle makes with the ground is 35°.

Population mean = 80
standard deviation = 20
sample = 60
What is the probability that the sample mean will be between84
and 88?

Answers

Answer:

The answer is : 0.0597

Step-by-step explanation:

Population mean = μ = 80

Standard deviation = σ = 20

Sample = N = 60

σ_mean = σ/√N

= (20)/√(60) = 2.582

Now we will find z1 and z2.

z1 = {(84) - μ}/σ_mean = [tex]{(84)-(80)}/(2.582)= 1.549[/tex]

z2 = {(88) - μ}/σ_mean = [tex]{(88)-(80)}/(2.582)= 3.098[/tex]

Now probability that the sample mean will be between 84 and 88 is given by:

Prob{ (1.549) ≤ Z≤ (3.098) } = (0.9990) -(0.9393) = 0.0597

You have an outdoor swimming pool that is 5.0 m wide and 12.0 m long. If weekly evaporation is 2.35 in, how many gallons of water must be added to the pool each week (if it doesn't rain)?

Answers

Answer:

Water should be greater than 946.11 gallons per week to prevent the pool from drying.

Step-by-step explanation:

You have an outdoor swimming pool that is 5.0 m wide and 12.0 m long.

The height of water evaporated is = 2.35 inch

We will convert this to m.

1 inch = 0.0254 meter

So, 2.35 inch = [tex]2.35\times0.0254=0.05969[/tex] meter

Now, volume of water in the pool = [tex]5\times12\times0.05969=3.5814[/tex] cubic meter per week.

1 cubic meter = 264.172 gallons

So, 3.5814 cubic meter = [tex]3.5814\times264.172=946.11[/tex]gallons

Hence, the volume of the water that should be poured in the swimming pool should be greater than 946.11 gallons per week to prevent the pool from drying.

Two angles of a triangle measure 30 and 45 degrees. If the side of the triangle opposite the 30-degree angle measures 6√2 units, what is the sum of the lengths of the two remaining sides? Express your answer as a decimal to the nearest tenth.

Answers

Answer:

  28.4 units

Step-by-step explanation:

If we call the given angles C and A, then the given side is c, and the other two sides can be found from the Law of Sines.

Angle B is the remaining angle of the triangle:

  180° -C -A = 180° -30° -45° = 105°

The remaining sides are ...

  b = sin(B)/sin(C)·c = sin(105°)/sin(30°)·6√2 ≈ 16.4

  a = sin(A)/sin(C)·c = sin(45°)/sin(30°)·6√2 = 12

Then the sum of the lengths of the remaining sides is ...

  a + b = 12 + 16.4 = 28.4 . . . units

Final answer:

By applying the principles of trigonometry and the Pythagorean theorem, we can determine that the sum of the lengths of the two remaining sides is approximately 25.5 units.

Explanation:

This question is about the application of trigonometric principles and the Pythagorean theorem. In a triangle, the sine of an angle is defined as the ratio of the side opposite that angle to the hypotenuse. Given a 30-degree angle and its opposite side of length 6√2, we can find the hypotenuse (h) using the fact that sin(30) = 1/2. So, 1/2 = 6√2 / h. Solving this, we get h = 2 * 6√2 = 12√2.

The angle 45 degrees helps determine the length of the remaining side. Using the fact that cos(45) = s/h, where s is the remaining side, we get cos(45) = s / 12√2. Solving for s, s = cos(45) * 12√2, s = √2/2 * 12√2 = 6√2.

So, the sum of the lengths of the two remaining sides is 12√2 + 6√2 = 18√2, which is approximately 25.5 when rounded to the nearest tenth.

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What is meant by the "complexity of an algorithm"?

Answers

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.

A. One day Annie weighed 24 ounces more than Benjie, and Benjie weighed 3 1/4 pounds less than Carmen. How did Annie’s and Carmen’s weights compare on that day?
B. Why can’t you tell how much each person weighed?

Draw diagrams to support your answer.

Answers

Answer:

A.

Let Annie's weight be = a

Let Benjie's weighs = b

Let Carmen's weight be = c

One day Annie weighed 24 ounces more than Benjie, equation forms:

[tex]a=b+24[/tex]        ......(1)

Benjie weighed 3 1/4 pounds less than Carmen.  

In ounces:

1 pound = 16 ounces

[tex]\frac{13}{4}[/tex] pounds = [tex]\frac{13}{4}\times16=52[/tex] ounces

[tex]b=c-52[/tex]  or

[tex]c=b+52[/tex]    ......(2)

Now adding (1) and (2), we get

a+b=b+24+c-52

=> [tex]a=c-28[/tex]

This gives  Annie weighs 28 ounces less than Carmen.

B.

We cannot know anyone's actual weight, as we only know their relative weights.

Medication Z is prescribed for a 50 kg woman and is to be given in enteric-coated pills of 125 mg per pill This medication is to be administered every 6 hours, but the total daily amount cannot exceed 8 mg.per kg of body weight per day. a) How many pills will this patient need? b) If the first dose is administered at 6 AM, when is the last dose given?

Answers

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.

For a certain event, 817 tickets were sold, for a total of $1919. If students paid $2 per ticket and nonstudents paid $3 per ticket, how many student tickets were sold?

Answers

Answer: 532

Step-by-step explanation:

Let x be the number of students and y be the number of non-students.

Then,  by considering the given information, we have

[tex]x+y=817-----(1)\\\\2x+3y=1919-----------(2)[/tex]

Multiply 2 on the both sides of equation (1), we get

[tex]2x+2y=1634--------(3)[/tex]

Subtract (3) from (2), we get

[tex]y=285[/tex]

Put the value of y in (1), we get

[tex]x+285=817\\\\\Rightarrow\ x=817-285=532[/tex]

Hence, 532 student tickets were sold .

Exercise 4.x2 Make a reasonable conjecture about the nth term in the sequence. 1 3 16 125 1296...

Answers

Answer:

The nth term in the sequence is given by the equation:

[tex]n_{th} =(n+1)^{n-1}[/tex]

Step-by-step explanation:

Arranging a table for n and nth:

[tex]\left[\begin{array}{ccc}n&nth\\1&1\\2&3\\2&16\\4&125\\5&1296\end{array}\right][/tex]

It is easier to notice that 16 and 125 result from the second power of 4 and the third of 5, respectively, which are one number below their respective position. That is why we can deduce that the base of the power is n+1.

For n=2, the base n+1 results in 3, which matches the nth term for n=2. Since 3 is the result of 3 to the power of 1, 16 is 4 to the power of 2, and 125 is 5 to the power of 3, all the powers are one number behind n, so the power is given by n-1, giving the equation:

[tex]n_{th} =(n+1)^{n-1}[/tex]

pls tell me the
domain
range
and if its a function or not thx

Answers

Answer:

  both domain and range are {-4, -3, -2, -1, 0, 1, 2, 3, 4}

  it IS a function

Step-by-step explanation:

The domain is the list of x-values of the plotted points. It is all the integers from -4 to +4, inclusive.

The range is the list of y-values of the plotted points. It is all the integers from -4 to +4, inclusive. (domain and range are the same for this function)

No x-value has more than one y-value associated with it, so this relation IS A FUNCTION.

Chose parameters h and k such that the system has a) a unique solution, b) many solutions, and c) no solution. X1 + 3x2 = 4 2x1 + kx2 = h

Answers

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:

If [tex]k\neq 6[/tex] then, the system has a unique solution for any value of [tex]k[/tex], meaning that the last row will transform back to the equation as:

[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.

if [tex]k=6[/tex] and [tex]h=8[/tex] the system has infinite solutions: this is very simple to see by substituting these values in the equation resulting from the last row:

[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

If you had carried out the algebra using variables before plugging numbers into your expressions, you would have found that (vf)α=−2qαΔVmα−−−−−−−√, where ΔV is measured in volts. To verify that this expression for (vf)α has the correct units of velocity, you need to perform some unit analysis. Begin by finding the equivalent of a volt in terms of basic SI units. What is a volt in terms of meters (m), seconds (s), kilograms (kg), and coulombs (C)? Express your answer using the symb

Answers

A volt in terms of meters, seconds, kilograms, and coulombs is

[tex]$\frac{\text{kilogram} \times \text{meter}^2}{\text{second}^2 \times \text{coulomb}}$[/tex]

We are given that;

(vf)α=−2qαΔVmα

Now,

To find the equivalent of a volt in terms of basic SI units, we can use the definition of a volt as the potential difference that causes one joule of energy to be transferred per coulomb of charge.

A joule is the unit of energy, which is defined as the work done by a force of one newton over a distance of one meter.

A newton is the unit of force, which is defined as the product of mass and acceleration. Therefore, we can write:

[tex]$1 \text{ volt} = \frac{1 \text{ joule}}{1 \text{ coulomb}} = \frac{1 \text{ newton} \times 1 \text{ meter}}{1 \text{ coulomb}} = \frac{1 \text{ kilogram} \times 1 \text{ meter} \times 1 \text{ meter}}{1 \text{ second}^2 \times 1 \text{ coulomb}}$[/tex]

Simplifying, we get:

[tex]$1 \text{ volt} = \frac{\text{kilogram} \times \text{meter}^2}{\text{second}^2 \times \text{coulomb}}$[/tex]

Therefore, by volt answer will be [tex]$\frac{\text{kilogram} \times \text{meter}^2}{\text{second}^2 \times \text{coulomb}}$[/tex].

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Final answer:

A volt is a unit of potential difference, which is measured in joules per coulomb (V). 1 volt is equal to 1 joule per coulomb (1 V = 1 J/C).

Explanation:

The potential difference between two points A and B, VB - VA, is defined as the change in potential energy divided by the charge q. This potential difference is measured in joules per coulomb, which is called a volt (V). To express a volt in terms of the basic SI units of meters (m), seconds (s), kilograms (kg), and coulombs (C), we need to use the equation J = V × C, where J represents the unit of energy, the joule.

We can rewrite the equation J = V × C as V = J / C. Since 1 V is equivalent to 1 J/C, we can express the volt in terms of meters (m), seconds (s), kilograms (kg), and coulombs (C) as:

1 V = 1 J/C

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Can someone help me solve this Statistics homeworl? Thanks!

Answers

Answer:

b. 5  

Step-by-step explanation:

In a stem and leaf plot,  each observation is split into a stem (the first digit or digits) and a leaf (usually the last digit).  

In this example, the entries in the 20s, 30s, and 40s are:

20, 20, 24, 25, 26, 35, 36, 43, 43

We omit 20, 20, 43, and 43 because they are not between 20 and 40.

That leaves the five observations

24, 25, 26, 35, 36

In the cost function below, C(x) is the cost of producing x items. Find the average cost per item when the required number of items is produced C(x)=7.6x + 10,800 a 200 items b. 2000 items c. 5000 items a. What is the average cost per item when 200 items are produced?

Answers

Answer:

The average cost per item when 200 items are produced is 61.6

Step-by-step explanation:

We start with the cost formula given by:

[tex]C(x)=7.6x+10,800[/tex]

Then we compute C(x) for x=200, 2000 and 5000 as follows:

[tex]C(200)=7.6*200+10,800=12,320\\C(2000)=7.6*2000+10,800=26,000\\C(5000)=7.6*5000+10,800=48,800[/tex]

Finally, to obtain the average cost per item when 200, 2,000 and 5,000 are produced (we will denote this by Av(200), Av(2000) and Av(5000) respectively) we just need to divide C(x) by the number of items produced. Then [tex]Av(x)=\frac{C(x)}{x}[/tex].

[tex]Av(200)=\frac{C(200)}{200}=\frac{12,320}{200}= 61.6\\Av(2000)=\frac{C(2000)}{2000}=\frac{26,000}{2000}= 13\\Av(5000)=\frac{C(5000)}{5000}=\frac{48,800}{5000}= 9.76\\[/tex]

A certain test preparation course is designed to help students improve their scores on the LSAT exam. A mock exam is given at the beginning and end of the course to determine the effectiveness of the course. The following measurements are the net change in 6 students' scores on the exam after completing the course:23,18,23,12,13,23Using these data, construct a 80% confidence interval for the average net change in a student's score after completing the course. Assume the population is approximately normal.

Answers

Final Answer:

We are 80% confident that the average net change in a student's score after completing the course lies between 7.88 and 21.12 points.

Explanation:

Here's how to construct the 80% confidence interval:

1. Calculate the sample mean and standard deviation:

Sample mean (xbar) = (23 + 18 + 23 + 12 + 13 + 23) / 6 = 18.33 points

Sample standard deviation (s) = √[(23 - 18.33)^2 + (18 - 18.33)^2 + ... + (23 - 18.33)^2] / (6 - 1) ≈ 5.38 points

2. Find the critical value for a 80% confidence interval:

For a two-tailed 80% confidence interval with 5 degrees of freedom (n-1), the critical value from the t-distribution table is approximately 1.942.

3. Calculate the margin of error:

Margin of error (ME) = critical value * standard deviation / √n

ME = 1.942 * 5.38 / √6 ≈ 4.49 points

4. Construct the confidence interval:

Lower limit = sample mean - margin of error = 18.33 - 4.49 ≈ 13.84 points

Upper limit = sample mean + margin of error = 18.33 + 4.49 ≈ 22.82 points

Therefore, we can be 80% confident that the true average net change in a student's score after completing the course falls within the range of 13.84 to 22.82 points, or rounded to the nearest whole number, 7.88 to 21.12 points.

Note: This calculation assumes the population is approximately normal. If there is reason to believe the population is not normal, a different method like bootstrapping may be more appropriate.

Question 4 options:

Given the following definitions:

U = {a, b, c, d, e, f, g}

A = {a, c, e, g}

B = {a, b, c, d}

Find A'

Answer in roster form, with a single space after each comma.

Answers

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}

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Solve. X1 – 3x2 + 4x3 = -4 3xı – 7x2 + 7x3 = -8 –4x1 + 6x2 – x3 = 7

Answers

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

1) Solve the word problem for the portion, rate, or base.


A quality control process finds 46.8 defects for every 7,800 units of production. What percent of the production is defective?


2) Solve the word problem for the portion, rate, or base.


A medical insurance policy requires Ana to pay the first $100 of her hospital expense. The insurance company will then pay 90% of the remaining expense. Ana is expecting a short surgical stay in the hospital, for which she estimates the total bill to be about $4,800.


How much (in $) of the total bill will Ana owe?

Answers

Answer:

1) 0.6% of the production is defective.

2) Ana will owe $570.

Step-by-step explanation:

Both questions here are percentage problems

Percentage problems can be explained 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. In this case, the rule of three is a cross multiplication.

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 case, the rule of three is a line multiplication.

Percentage problems have a direct relationship between the measures.

1) The problem states that a quality control process finds 46.8 defects for every 7,800 units of production. And asks what percent of the production is defective? We have to answers how many defects are there for 100 units of production. So:

46.8 defects - 7,800 units

x defects - 100 units

7,800x = 4680

[tex]x = \frac{4680}{7800}[/tex]

x = 0.6

0.6% of the production is defective.

2) The problem states that the medical insurance policy requires Ana to pay the first $100 of her hospital expense. The insurance company will then pay 90% of the remaining expense. The total bill is expected to be about $4,800.

Ana has to pay:

P = P1 + P2

-P1 :The first $100

-P2: 10% of the remaining expense. The remaining expense is $4,800-$100 = $4,700. Ana has to pay 10% of this. So

4700 - 100%

P2 - 10%

100P2 = 47000

[tex]P2 = \frac{47000}{100}[/tex]

P2 = $470

Ana will owe P = P1 + P2 = $100 + $470 = $570.

Final answer:

To find the percentage of defective units in the production, set up a ratio and calculate the percentage. Approximate 0.6% of the production is defective. To find how much Ana will owe of the total bill, subtract $100 from the total bill and calculate 10% of the remaining expense. Ana will owe $470 of the total bill.

Explanation:

To find the percent of the production that is defective, we need to find the ratio of the number of defective units to the total number of units produced. In this case, we have 46.8 defects for every 7,800 units of production. So, we can set up the ratio as:

Defective Units / Total Units = 46.8 / 7,800

To find the percentage, we can multiply the ratio by 100:

Percentage of Defective Units = (46.8 / 7,800) * 100

Now, calculate the value of the ratio and simplify to find the percentage:

Percentage of Defective Units = 0.006 * 100 = 0.6%

Therefore, approximately 0.6% of the production is defective.

For the second question, to find how much of the total bill Ana will owe, we need to calculate the 10% of the remaining expense after she pays the first $100. First, subtract $100 from the total bill:

Remaining Expense = $4,800 - $100 = $4,700

Next, calculate 10% of the remaining expense:

Amount Ana will owe = 10% of $4,700 = ($4,700 * 10) / 100 = $470

Therefore, Ana will owe $470 of the total bill.

Find the equation of the line going through the points (2,-1) and (5,2) 3x 2y

Answers

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]

Prove: If n is a positiveinteger and n2 is
divisible by 3, then n is divisible by3.

Answers

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.

Need Help Fast!!!!!!!!!!!!!!!!!!

Answers

7.2, 3, 8.09, 2.22, 5.06, 2.5
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