b=2.35 + 0.25x
c=1.75+0.40x
In the equations above,b and c represent the price per pound,in dollars, of beef and chicken,respectively,x weeks after July 1 of last summer.What was the price per pound of beef when it was equal to the price per pound of chicken?

A.2.60
B.2.85
C.2.95
D.3.35

The answer & explanation for this question is given in the attachment below.

The number of aces in the first game and the number of spades in the 5th game follow a Hypergeometric Distribution while the number of games receiving at least one ace can be modeled by a Binomial distribution. The event of all cards being from the same suit can be thought of as a Uniform distribution.

a) The number of aces you get in the first game follows a Hypergeometric Distribution. In such a distribution, you are drawing cards without replacement. The parameters are N=52 (the population size), K=4 (the number of success states in the population i.e., the number of aces in a deck), and n=13 (the number of draws).

b) The number of games in which you receive at least one ace can be modeled by a Binomial distribution. Each game you play (out of 50) is a single trial, with the probability of success (getting at least one ace) being the same for every trial. The parameters are n=50 (the number of trials/games) and p (the probability of getting at least one ace).

c) The likelihood of all your cards being from the same suit in a game is heavily reliant on chance, can be modeled as a Uniform distribution given its rare occurrence. Essentially, the parameters would be minimum = 0 and maximum = 1. However, determining the parameters would require calculation of the specific probabilities, which is complex due to the nature of the game.

d) The number of spades you receive in the 5th game also follows a Hypergeometric distribution, similar to the situation in the first game. The parameters in this case are N=52, K=13 (number of spades in a deck), and n=13 (the number of drawn cards).

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Answer:

[tex] 8.637 \leq \sigma^2 \leq 28.93[/tex]

[tex] 2.939 \leq \sigma \leq 5.379[/tex]

Step-by-step explanation:

Data given and notation

s represent the sample standard deviation

[tex]\bar x[/tex] represent the sample mean

n=23 the sample size

Confidence=95% or 0.95

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]

The sample standard deviation for this case was s = 3.8

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

[tex]df=n-1=23-1 =22[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a tabele to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.025,22)" "=CHISQ.INV(0.975,22)". so for this case the critical values are:

[tex]\chi^2_{\alpha/2}=36.78[/tex]

[tex]\chi^2_{1- \alpha/2}=10.98[/tex]

And replacing into the formula for the interval we got:

[tex]\frac{(22)(3.8)^2}{36.78} \leq \sigma^2 \leq \frac{(22)(3.8)^2}{10.98}[/tex]

[tex] 8.637 \leq \sigma^2 \leq 28.93[/tex]

Now we just take square root on both sides of the interval and we got:

[tex] 2.939 \leq \sigma \leq 5.379[/tex]

Answer:

Part 1) [tex]AC=40\ units[/tex]

Part 2) [tex]DC=29\ units[/tex]

Part 3) [tex]m\angle ABE=39^o[/tex]

Part 4) [tex]m\angle BCE=51^o[/tex]

Step-by-step explanation:

we know that

A kite has two pairs of consecutive, congruent sides the diagonals are perpendicular and the non-vertex angles are congruent

Part 1) Find AC

we know that

BD is the axis of symmetry, bisects the diagonal AC

so

[tex]AE=EC[/tex]

we have

[tex]EC=20\ units[/tex]

[tex]AC=AE+EC[/tex] ----> by segment addition postulate

therefore

[tex]AC=20+20=40\ units[/tex]

Part 2) Find CD

we know that

CDE is a right triangle (the diagonals are perpendicular)

so

Applying Pythagorean Theorem

[tex]DC^2=EC^2+ED^2[/tex]

substitute the values

[tex]DC^2=20^2+21^2[/tex]

[tex]DC^2=841\\DC=29\ units[/tex]

Part 3) Find m∠ABE

we know that

In the right triangle ABE

[tex]51^o+m\angle ABE=90^o[/tex] ----> by complementary angles

[tex]m\angle ABE=90^o-51^o=39^o[/tex]

Part 4) Find m∠BCE

we know that

[tex]m\angle BCE=m\angle BAE[/tex] ----> diagonal BD is the axis of symmetry

we have

[tex]m\angle BAE=51^o[/tex]

therefore

[tex]m\angle BCE=51^o[/tex]

The measure of AC is 40 units.

The measure of the length DC is 29 units.

The measure of the angle ABE is 39 degrees.

The measure of the angle BCE is 51 degrees.

The missing lengths and angle measures in kite ABCD.

According to the question

The rhombus is a four-sided quadrilateral with all its four sides equal in length.

Rhombus is a kite with all its four sides congruent.

A kite is a special quadrilateral with two pairs of equal adjacent sides.

1. The measure of the length of AC is;

In the figure, BD is the axis of symmetry, bisects the diagonal AC.

Then,

[tex]\rm AE = EC[/tex]

And the measure of AC is,

[tex]\rm AC = AE+EC \\ \\ AC = 20+20 \\ \\ AC = 40 \ units[/tex]

The measure of AC is 40 units.

2. In the figure, CDE is a right triangle (the diagonals are perpendicular)

Then,

By applying the Pythagoras Theorem

[tex]\rm DC^2=EC^2+ED^2\\ \\ DC^2=20^2+21^2\\\\ DC^2 = 400+441 \\ \\ DC^2 = 841\\ \\ DC = 29 \ \rm units[/tex]

The measure of the length DC is 29 units.

3. In the right triangle ABE by the complementary angles;

[tex]\rm 51+ \angle ABE = 90\\ \\ \angle ABE=90-51\\ \\ \angle ABE=39 \ degrees[/tex]

The measure of the angle ABE is 39 degrees.

4. By the axis of symmetry the diagonal BD is;

[tex]\rm m \angle BCE = m \angle BAE\\\\ m \angle BCE = m \angle BAE = 51 \ degrees[/tex]

The measure of the angle BCE is 51 degrees.

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Answer:

The controllable canonical form of the system =

Y = (5 0 0 0) (x1)

                     (x2)

                     (x3)

                     (x4)

Step-by-step explanation:

The detailed explanation is as shown in the attached file.

By definition, the half life is a quantity is the time it takes to lose half of that quantity.

So, if the half life is 60 days, it means that after 60 days you have lost half the initial amount, i.e. you're left with 100 grams.

After another 60 days, you've lost, again, half of that amount, so you're left with 50 grams.

In other words, every time a half life period passes, you're left with half the quantity you had at the beginning of that period.

So, after two half-life periods (i.e. 120 days), you'll have half of the half of the initial quantity, i.e. one quarter.

Answer:

Step-by-step explanation:

The sum of the angles in a triangle is 180 degrees. This means that in triangle ABC,

Angle A + angle B + angle C = 180

Therefore,

6x - 1 + 20 + x + 14 = 180

6x + x + 20 + 14 - 1 = 180

7x + 33 = 180

Subtracting 33 from the left hand side and the right hand side of the equation, it becomes

7x + 33 - 33 = 180 - 33

7x = 147

Dividing the left hand side and the right hand side of the equation by 7, it becomes

7x/7 = 147/7

x = 21

Therefore

Angle A = 6x - 1 = 6 × 21 - 1

Angle A = 125 degrees

Angle C = x + 14 = 21 + 14

Angle C = 35 degrees.

Answer:

[tex]n \geq 36[/tex]

Step-by-step explanation:

For this case we know the mean and the deviation:

[tex] \mu = 1.12 kg , \sigma= 0.0009[/tex]

The mean is given by this:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

If we find the expected value for the sample mean we have:

[tex] E(\bar X) = E(\frac{\sum_{i=1}^n X_i}{n}) =\frac{1}{n} \sum_{i=1}^n E(X_i)[/tex]

Since each observation in the sample [tex] X_1, X_2,...., X_{25}[/tex] have an expectation [tex] E(X_i) = \mu , i =1,2,...,25[/tex] so then we have that:

[tex] E(\bar X) = \frac{1}{n} n\mu = \mu[/tex]

Now for the variance of the sample mean we have this:

[tex]Var (\bar X) = Var(\frac{\sum_{i=1}^n X_i}{n})= \frac{1}{n^2} \sum_{i=1}^n Var(X_i)[/tex]

And again each observation have a variance [tex] \sigma^2_i = 0.0009 , i =1,2,...,25[/tex] then we have:

[tex]Var (\bar X) =\frac{1}{n^2} n(\sigma^2) =\frac{\sigma^2}{n}[/tex]

And then the standard deviation would be:

[tex] Sd(\bar X) =\sqrt{\frac{\sigma^2}{n}}= \frac{\sigma}{\sqrt{n}}[/tex]

And we want that the standard deviation for the sample no more than 0.005 so we have this condition:

[tex]\frac{\sigma}{\sqrt{n}} \leq 0.005[/tex]

And since we know the value of [tex] \sigma= \sqrt{0.0009}=0.03[/tex] we can solve for the value of n like this:

[tex] \frac{0.03}{0.005} \leq \sqrt{n}[/tex]

[tex] 6 \leq \sqrt{n}[/tex]

And if we square both sides we got:

[tex] n\geq 36[/tex]

Final answer:

To achieve an average weight standard deviation of no more than 0.005 kilograms for a certain type of brick with a variance of 0.0009 kilograms², 36 bricks need to be selected.

Explanation:

The question concerns the calculation of the number of bricks needed so that the average weight of these bricks has a standard deviation of no more than 0.005 kilograms. Given that the weight of a certain type of brick has an expectation (mean) of 1.12 kilograms and a variance of 0.0009 kilograms2, the first step is to understand that the variance of the average weight of n bricks is equal to the variance of a single brick divided by n.

To achieve a standard deviation of the average weight of no more than 0.005 kilograms, we use the formula for the standard deviation of the mean, which is √(variance/n). The variance given is 0.0009 kilograms2. Setting the equation √(0.0009/n) <= 0.005 and solving for n gives:

0.0009/n = 0.0052

0.0009/n = 0.000025

n = 0.0009 / 0.000025 = 36

Therefore, to ensure the standard deviation of the average weight is no more than 0.005 kilograms, 36 bricks would need to be selected.

Answer:

There is an 88% probability that a course has a final exam or a research paper.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

E is the probability that a course has final exam.

P is the probability that a course requires research paper.

We have that:

[tex]E = e + (E \cap P)[/tex]

In which e is the probability that a course has final exam but does not require research paper and [tex]E \cap P[/tex] is the probability that a course has both of these things.

By the same logic, we have that:

[tex]P = p + (E \cap P)[/tex]

(a) Find the probability that a course has a final exam or a research paper.

This is

[tex]Pr = e + p + (E \cap P)[/tex]

Suppose that 26% of courses have a research paper and a final exam.

This means that

[tex]E \cap P = 0.26[/tex]

43% of courses require research papers.

So [tex]P = 0.43[/tex]

[tex]P = p + (E \cap P)[/tex]

[tex]0.43 = p + 0.26[/tex]

[tex]p = 0.17[/tex]

71% of courses have final exams

So [tex]E = 0.71[/tex]

[tex]E = e + (E \cap P)[/tex]

[tex]0.71 = e + 0.26[/tex]

[tex]e = 0.45[/tex]

The probability is

[tex]Pr = e + p + (E \cap P) = 0.45 + 0.17 + 0.26 = 0.88[/tex]

There is an 88% probability that a course has a final exam or a research paper.

Answer:

The probability that Umbrella returns any shipment is 0.0062.

Step-by-step explanation:

Let X = the impurity level in the material per drum

Then it is provided that,

[tex]X\sim N(\mu = 0.03,\ \sigma=0.004)[/tex]

Also if the impurity level is more than or equal to 4% or 0.04 in any shipment, then Umbrella returns the entire drum to the supplier.

Compute the probability that Umbrella returns any shipment as:

[tex]P(X\geq 0.04)=P(\frac{X-\mu}{\sigma}\geq \frac{0.04-0.03}{0.004}) \\=P(Z\geq 2.5)\\=1-P(Z<2.5)\\=1-0.99379\\=0.00621\\\approx0.0062[/tex]

Use the z-table fro left z-scores to determine the probability.

Thus, the probability that Umbrella returns any shipment is 0.0062.

Answer:

The information above is just a part of a long question.The main question is  found in the attached as well as the answer following a step by step methodical approach.

Step-by-step explanation:

Take a good at the full question before venturing into the statements prepared so as to have a thorough understanding of the requirements and how the answers provided have touched upon the requirements.

The answer is a) [tex]X=X_1 +X_2 +X_3+..X_n[/tex]  b) The expression  is

[tex]E[X] = \dfrac{1}{n+m} +\dfrac{1}{n+m-1} +....+ \dfrac{1}{m+1}[/tex].

Given:

An urn contains n+m balls of which n is red and m is black

a)

Expressing X in terms of the defined random variables:

Let be the indicator random variable for the event that red ball i is taken before any black ball is chosen. It takes the value 1 if this event occurs and 0 otherwise.

Now, the value of X is the number of red balls removed before the first black ball is chosen. This can be expressed as the sum of the individual indicator random variables for each red ball:

[tex]X=X_1 +X_2 +X_3+..X_n[/tex]

b)

Find E[X] (expected value of X):

The expected value of a sum of random variables is the sum of the expected values of those random variables. To find E[X],

Find the expected value of each indicator random variable and then sum them up.

Therefore, the expected value   [tex]X_i[/tex] is:

[tex]E[X_i] = 1. \dfrac{1}{n+m-i+1} + 0\cdot \dfrac{m}{n+m-i+1}[/tex]

[tex]= \dfrac{1}{n+m-i+1}[/tex]

Now, to find  E[X], we sum up the expected values of the indicator random variables for all red balls:

[tex]E[X] = E[X_1]+E[X_2]+E[X_3]+.....+E[X_n][/tex]

Substitute the values of [tex]E[X_i][/tex] into the sum and simplify:

[tex]E[X] = \dfrac{1}{n+m} +\dfrac{1}{n+m-1} +....+ \dfrac{1}{m+1}[/tex]

a) [tex]X=X_1 +X_2 +X_3+..X_n[/tex]  b) [tex]E[X] = \dfrac{1}{n+m} +\dfrac{1}{n+m-1} +....+ \dfrac{1}{m+1}[/tex]

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X can be expressed as X = R_1 + 2R_2 + 3R_3 + ... + nR_n. To find E[X], use the linearity of expectation and the probabilities of R_i.

To determine E[X], we need to express X in terms of the random variables and then find the expected value. Let R_i denote the event that red ball i is taken before any black ball is chosen. The expression for X is:

To find the expected value, we use the linearity of expectation. Since the probabilities for each R_i are the same, we have:

Now, we need to determine the probabilities of R_i. Since the balls are drawn without replacement, the probability that red ball i is drawn before any black ball is chosen is:

Substituting this probability into the expression for E[X], we get:

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Answer:

[tex]P(k) = \frac{(n C k) [(N-n) C (m-k)]}{(NCm)}[/tex]

Step-by-step explanation:

Step 1: Number of possible combination of selecting ‘m’ deer in second sample

Total number of deer are N and therefore the combinations can be calculated as (N С m).  

Step 2: Number of possible combination of marked deer ‘k’ in second sample

Total number of marked deer in total population is ‘n’. Therefore, the possible number of selecting marked deer is (n C k).

Step 3: Number of possible combination unmarked deer in second sample

Since we have already calculated the total combinations of selecting marked deer in the second sample. Hence, we have to calculate the total unmarked deer in total population which is N-n and number of unmarked deer in the second sample which is m-k.

Therefore, total possible combination of unmarked deer in second sample is [(N-n) C (m-k)].

Step 4: Probability of selecting unmarked deer in the second sample is

Let the probability of selecting unmarked deer in the second sample be P(k)

Therefore,

                                 [tex]P(k) = \frac{(n C k) [(N-n) C (m-k)]}{(NCm)}[/tex]

Answer:

Step-by-step explanation:

Fractions are used in telling time.

Fractions are used to determine discounts when there is sales going on.

Fraction is used to represent part of whole .

Fractions are used in baking to tell how much of an ingredient to use.

Answer:

2683

Step-by-step explanation:

Using the linear regression equation that predict the relationship between the weight of the luggage and the total number of passenger y = 103 + 30x, we can plug in the number of passenger x = 86 to predict the weight of the luggage on a flight:

y = 103 + 30*86 =  2683

Answer:

36% probability that she will find an acceptable house in the first two weeks that she looks.

Step-by-step explanation:

For each house, there is a 20% probability of it being acceptable. And a 100-20 = 80% of not being acceptable

What is the probability that she will find an acceptable house in the first two weeks that she looks (round off to second decimal place)?

She only looks one house a week.

So this is the same as the probability of taking two or less weeks to find a house.

The are two outcomes

Finding an acceptable house in the first week, with 20% probability

Not finding an acceptable house in the first week, with 80% probability, and then finding an acceptable house in the second week, with 20% probability.

Probability:

[tex]P = 0.2 + 0.8*0.2 = 0.36[/tex]

36% probability that she will find an acceptable house in the first two weeks that she looks.

The correct answer is 0.36 or 36%.

To solve this problem, we need to calculate the probability that the house hunter does not find an acceptable house in the first two weeks and then subtract this probability from 1 to find the probability that she does find an acceptable house within that time frame.

 Let's denote the probability of finding an acceptable house in a given week as[tex]\( p \)[/tex] and the probability of not finding an acceptable house in a given week as [tex]\( q \)[/tex]. Since 20% of the houses are in acceptable condition, [tex]\( p = 0.20 \) and \( q = 1 - p = 1 - 0.20 = 0.80 \).[/tex]

The probability that the house hunter does not find an acceptable house in the first week is[tex]\( q \).[/tex] The probability that she does not find an acceptable house in the first two weeks is the product of the probabilities that she does not find one in each of the two weeks separately, which is [tex]\( q \times q \) or \( q^2 \).[/tex]

Now, we calculate [tex]\( q^2 \):[/tex]

[tex]\[ q^2 = (0.80)^2 = 0.64 \][/tex]

This is the probability that she will not find an acceptable house in the first two weeks. To find the probability that she will find an acceptable house in the first two weeks, we subtract this value from 1:

[tex]\[ \text{Probability of finding an acceptable house in two weeks} = 1 - q^2 \][/tex]

[tex]\[ \text{Probability of finding an acceptable house in two weeks} = 1 - 0.64 \][/tex]

[tex]\[ \text{Probability of finding an acceptable house in two weeks} = 0.36 \][/tex]

The resulting probability is approximately 100%.

To find the probability that the plywood is less than 24mm thick, we need to calculate the z-score for 24mm using the given mean and standard deviation. The z-score formula is:

z = (x - mean) / standard deviation

Substituting the values, we get:

z = (24 - 5) / 0.2 = 95

We can then look up the z-score in a standard normal distribution table to find the corresponding probability.

In this case, the probability is practically 1 (or 100%) since 24mm is significantly greater than the mean.

Therefore, the probability that the plywood is less than 24mm thick is approximately 100%.

Answer:

a) y = 7.74*x + 7.5

b)  y = 1.148*x + 6.036

Step-by-step explanation:

Given:

                                  f(x) = 6 - 10*x^2

                                  g(x) = 8 - (x-2)^2

Find:

(a) The line simultaneously tangent to both graphs having the LARGEST slope has equation

(b) The other line simultaneously tangent to both graphs has equation,

Solution:

- Find the derivatives of the two functions given:

                                f'(x) = -20*x

                                g'(x) = -2*(x-2)

- Since, the derivative of both function depends on the x coordinate. We will choose a point x_o which is common for both the functions f(x) and g(x). Point: ( x_o , g(x_o)) Hence,

                                g'(x_o) = -2*(x_o -2)

- Now compute the gradient of a line tangent to both graphs at point (x_o , g(x_o) ) on g(x) graph and point ( x , f(x) ) on function f(x):

                                m = (g(x_o) - f(x)) / (x_o - x)

                                m = (8 - (x_o-2)^2 - 6 + 10*x^2) / (x_o - x)

                                m = (8 - (x_o^2 - 4*x_o + 4) - 6 + 10*x^2)/(x_o - x)

                                m = ( 8 - x_o^2 + 4*x_o -4 -6 +10*x^2) /(x_o - x)

                                m = ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x)

- Now the gradient of the line computed from a point on each graph m must be equal to the derivatives computed earlier for each function:

                                m = f'(x) = g'(x_o)

- We will develop the first expression:

                                m = f'(x)

                                ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

Eq 1.                          (-2 - x_o^2 + 4*x_o + 10*x^2) = -20*x*x_o + 20*x^2

And,

                              m = g'(x_o)

                              ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

                              -2 - x_o^2 + 4*x_o + 10*x^2 = -2(x_o - 2)(x_o - x)

Eq 2                       -2 - x_o^2 + 4*x_o+ 10*x^2 = -2(x_o^2 - x_o*(x + 2) + 2*x)

- Now subtract the two equations (Eq 1 - Eq 2):

                              -20*x*x_o + 20*x^2 + 2*x_o^2 - 2*x_o*(x + 2) + 4*x = 0

                              -22*x*x_o + 20*x^2 + 2*x_o^2 - 4*x_o + 4*x = 0

- Form factors:       20*x^2 - 20*x*x_o - 2*x*x_o + 2*x_o^2 - 4*x_o + 4*x = 0

                              20*x*(x - x_o) - 2*x_o*(x - x_o) + 4*(x - x_o) = 0

                               (x - x_o)(20*x - 2*x_o + 4) = 0  

                               x = x_o   ,     x_o = 10x + 2    

- For x_o = 10x + 2  ,

                               (g(10*x + 2) - f(x))/(10*x + 2 - x) = -20*x

                                (8 - 100*x^2 - 6 + 10*x^2)/(9*x + 2) = -20*x

                                (-90*x^2 + 2) = -180*x^2 - 40*x

                                90*x^2 + 40*x + 2 = 0  

- Solve the quadratic equation above:

                                 x = -0.0574, -0.387      

- Largest slope is at x = -0.387 where equation of line is:

                                  y - 4.502 = -20*(-0.387)*(x + 0.387)

                                  y = 7.74*x + 7.5          

- Other tangent line:

                                  y - 5.97 = 1.148*(x + 0.0574)

                                  y = 1.148*x + 6.036

The ball travels a distance of 52 ft during the time interval [3,3.5]. The average velocity over this interval is 104 ft/sec. When we calculate over increasingly smaller time intervals, it appears the instantaneous velocity at t=3 is 96 ft/sec.

(a) We need to find the displacement Δs over the time interval [3,3.5]. For that, we subtract the position at time 3 from the position at time 3.5:

Δs = s(3.5) - s(3) = 16(3.5^2) - 16(3^2) = 196 - 144 = 52 ft.

(b) We calculate the average velocity by dividing Δs by Δt. That is Δs/Δt = 52/0.5 = 104 ft/sec.

(c) To estimate the instantaneous velocity at t=3, we have to compute the average velocities over smaller and smaller intervals centered at t=3. Let's calculate the average velocities for [3,3.01],[2.999,3]:

For [3,3.01]: Δs/Δt = [16*(3.01)^2 - 16*3^2]/0.01 = 96.12 ft/sec. For [2.999,3]: Δs/Δt = [16*3^2 - 16*(2.999)^2]/0.001 = 96 ft/sec.

Instantaneous velocity at t=3, v(3) can be estimated as the limit of these average velocities, which seems to be approaching about 96 ft/sec.

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(a) The ball travel Δs= 52 ft during the time interval [3,3.5].

(b) The average velocity over [3,3.5] is Δs/Δt= 104 ft/sec

(c) The object's instantaneous velocity at t=3 .V(3)= 96 ft/sec

Let's solve the problem step-by-step.

Given:
The distance traveled by the ball after t seconds is given by the function: [tex]s(t) = 16t^2[/tex]

(a) To find the distance traveled, we need to calculate the change in the distance function, which is
[tex]\Delta s = s(3.5) - s(3)[/tex]

First, compute [tex]s(3)[/tex]:
[tex]s(3) = 16(3)^2 = 16 \times 9 = 144[/tex] ft

Next, compute [tex]s(3.5)[/tex]:
[tex]s(3.5) = 16(3.5)^2 = 16 \times 12.25 = 196[/tex] ft

So,
[tex]\Delta s = 196 - 144 = 52[/tex] ft

(b) Average velocity is calculated by dividing the change in distance by the change in time, which is
[tex]\frac{\Delta s}{\Delta t} = \frac{52}{0.5} = 104[/tex] ft/sec

(c) For the time intervals, we compute each average velocity:

[3, 3.01]:
[tex]\Delta t = 0.01[/tex]
[tex]s(3.01) = 16(3.01)^2 = 16 \times 9.0601 = 144.9616[/tex] ft
[tex]\Delta s = 144.9616 - 144 = 0.9616[/tex] ft
[tex]\text{Average velocity} = \frac{0.9616}{0.01} = 96.16[/tex] ft/sec

[3, 3.001]:
[tex]\Delta t = 0.001[/tex]
[tex]s(3.001) = 16(3.001)^2 = 16 \times 9.006001 = 144.096016[/tex] ft
[tex]\Delta s = 144.096016 - 144 = 0.096016[/tex] ft
[tex]\text{Average velocity} = \frac{0.096016}{0.001} = 96.016[/tex] ft/sec

[3, 3.0001]:
[tex]\Delta t = 0.0001[/tex]
[tex]s(3.0001) = 16(3.0001)^2 = 16 \times 9.00060001 = 144.00960016[/tex] ft
[tex]\Delta s = 144.00960016 - 144 = 0.00960016[/tex] ft
[tex]\text{Average velocity} = \frac{0.00960016}{0.0001} = 96.0016[/tex] ft/sec

[2.9999, 3]:
[tex]\Delta t = 0.0001[/tex]
[tex]s(2.9999) = 16(2.9999)^2 = 16 \times 8.99940001 = 143.99040016[/tex] ft
[tex]\Delta s = 144 - 143.99040016 = 0.00959984[/tex] ft
[tex]\text{Average velocity} = \frac{0.00959984}{0.0001} = 95.9984[/tex] ft/sec

[2.999, 3]:
[tex]\Delta t = 0.001[/tex]
[tex]s(2.999) = 16(2.999)^2 = 16 \times 8.994001 = 143.904016[/tex] ft
[tex]\Delta s = 144 - 143.904016 = 0.095984[/tex] ft
[tex]\text{Average velocity} = \frac{0.095984}{0.001} = 95.984[/tex] ft/sec

[2.99, 3]:
[tex]\Delta t = 0.01[/tex]
[tex]s(2.99) = 16(2.99)^2 = 16 \times 8.9401 = 143.0416[/tex] ft
[tex]\Delta s = 144 - 143.0416 = 0.9584[/tex] ft
[tex]\text{Average velocity} = \frac{0.9584}{0.01} = 95.84[/tex] ft/sec

From these values, we can see that as the interval gets smaller, the average velocities are approaching a specific value. This helps us estimate the instantaneous velocity at [tex]t = 3[/tex].

The volume of a box is the amount of space in it.

The expression that represents volume is: [tex]\mathbf{V = (12 -2x) (9 - 2x)x}[/tex]

The dimension of the cardboard is given as:

[tex]\mathbf{Length = 12}[/tex]

[tex]\mathbf{Width = 9}[/tex]

Assume the cut-out is x.

So, the dimension of the box is:

[tex]\mathbf{Length =12-2x}[/tex]

[tex]\mathbf{Width =9 - 2x}[/tex]

[tex]\mathbf{Height = x}[/tex]

The volume of the box is:

[tex]\mathbf{V = (12 -2x) (9 - 2x)x}[/tex]

Hence, the expression that represents volume is:

[tex]\mathbf{V = (12 -2x) (9 - 2x)x}[/tex]

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The volume of a box formed by cutting a square of side x from a 9" by 12" sheet and folding the sides is given by the formula V = x(12 - 2x)(9 - 2x).

The volume of a box is formed by multiplying its length, width, and height. In this case, if you cut a square of side x from each corner of the sheet, the new dimensions of your box would be:

Therefore, the formula to express the volume V in terms of x is:

V = x(12 - 2x)(9 - 2x)

.

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To find the probability of selling more than 4 properties in one week, use the binomial probability formula for each number of sales above 4 and sum them, or subtract the sum of probabilities of 4 or fewer sales from 1.

To compute the probability of selling more than 4 properties in one week, given that the chance of selling any one property is 50% and is independent of selling another property, we can use the binomial probability formula. In our scenario, we have a total of 14 properties, and the random variable X represents the number of properties sold in a week.

The binomial probability formula is:

P(X = k) = (n choose k) * p^k * (1-p)^(n-k)

Where:

However, instead of calculating the probability of selling exactly k properties, we are interested in selling more than 4 properties. This means we need to calculate the probabilities for selling 5, 6, ..., up to 14 properties and sum them up. Alternately, we can calculate 1 minus the probability of selling 4 or fewer properties to save time, since the probabilities must sum to 1.

The probability that exactly three out of six individuals have group A blood, with a 40 percent chance for each individual, is found using the binomial probability formula. The calculation yields approximately 27.648% as the probability for exactly three individuals having group A blood.

The question asks for the probability that exactly three out of six individuals have group A blood, given that 40 percent of all individuals have group A blood. This is a binomial probability problem because there are two outcomes (having group A blood or not) and a fixed number of trials (six individuals).

Let's denote X as the random variable representing the number of individuals with group A blood. The probability of success on any given trial is p = 0.40 (having group A blood). The probability of failure is q = 1 - p = 0.60 (not having group A blood).

The binomial probability formula is:

[tex]P(X = k) = C(n, k) * p^k * q^(n-k)[/tex]

where:

For our problem:

Plugging these values into the binomial formula gives:

First, calculate the combination:

C(6, 3) = 6! / (3! * (6-3)!) = 20

Then, calculate the probability:

[tex]P(X = 3) = 20 * (0.40)^3 * (0.60)^3 = 20 * 0.064 * 0.216 = 0.27648[/tex]

Hence, the probability that exactly three of the individuals have group A blood is approximately 0.27648 or 27.648%.

Answer: False

Step-by-step explanation:

To know if the circle passes through the given point, we simply insert the coordinates of the given point (5,6) into the general equation Of the Circle. If left hand side of the equation balances with the right hand side, then the circle passed through the point.

The equation of the circle is (x-2)² + (y-6)² = 4

By inserting the given coordinates (5,6) we have.

(5-2)² + (6-6)² = 3² =9

Since the answer gotten is not 4, then the circle does not pass through the point (5,6)

Answer:

Step-by-step explanation:

Given that in a high school graduating class of 202 students, 95 are on the honor roll. Of these, 71 are going on to college, and of the other 107 students, 53 are going on to college.

Total students = 202

Honor roll = 95

College going out of honor = 71

Non college going out of honor = 24

Not in honor roll = 107

Not going to college from not in honour = 53

The probabilities that the person chosen is

(a) going to college,  = [tex]\frac{71+53}{202} \\=0.614[/tex]

(b) not going to college, =[tex]\frac{202-124}{202} \\=0.485[/tex]

(c) on the honor roll, but not going to college

=[tex]\frac{24}{202} \\=0.119[/tex]

Final answer:

The probabilities for a student selected at random from the class are: going to college, not going to college, and being on the honor roll but not going to college, calculated based on provided numbers for a graduating class of 202 students.

Explanation:

To solve this problem, we first need to understand the given information about the high school graduating class consisting of 202 students, with 95 on the honor roll and 107 not on the honor roll. Out of those on the honor roll, 71 are going to college, and of those not on the honor roll, 53 are going to college. Let's calculate the probabilities for each scenario.

Probability of Going to College

Total students going to college = Students on the honor roll going to college + Students not on the honor roll going to college = 71 + 53 = 124

Probability = (Total students going to college) / (Total students) = 124 / 202

Probability of Not Going to College

Total students not going to college = Total students - Total students going to college = 202 - 124 = 78

Probability = 78 / 202

Probability of Being on the Honor Roll but Not Going to College

Total students on the honor roll but not going to college = Total students on the honor roll - Students on the honor roll going to college = 95 - 71 = 24

Probability = 24 / 202

Answer: 2.4

Step-by-step explanation:

Given : A large disaster cleaning company estimates that 30 percent of the jobs it bids on are finished within the bid time.

i..e The proportion of  the jobs it bids on are finished within the bid time: p = 30%= 0.30

Sample size of jobs that it has contracted : n= 8

Then ,  the mean number of jobs completed within the bid time : [tex]\mu=np[/tex]

[tex]= 8\times0.30=2.4[/tex]

Hence, the mean number of jobs completed within the bid time is 2.4 .

Answer:

0.25 or 25%

Step-by-step explanation:

3, 5 and 7 are prime numbers.

There are two possible outcomes for which the experimenter ends up with exactly twenty-five dollars:

A) Choosing urn 5 (5 x 5 = 25).

[tex]P(A) = \frac{1}{5}[/tex]

B) Choosing urn 4 and then urn 3 ([4 x 4] + [3 x 3] = 25).

[tex]P(B)= \frac{1}{5} *\frac{1}{4}=\frac{1}{20}[/tex]

The probability that the experimenter ends up with exactly $25 is:

[tex]P(x=\$25)=P(A)+P(B)= \frac{1}{5}+\frac{1}{20}\\P(x=\$25)=0.25=25\%[/tex]

Answer:

The answer to your question is below

Step-by-step explanation:

In Mayan notation there are only 3 symbols

   dot   .    means one

   line   ---   means five

  snail          means zero

And the are three levels,     the lower means     x 1

                                              the first means       x 20

                                              the second means  x 400

                                              the higher means    x 8000

Then, 135 is lower than 400, so we must start in the first level and need to divide 135 by 20.                         6

                                        20    135

                                                   15

Finally place 6 in the fist level and 15 in the lower lever, like this

And 120 + 15 = 135.          

Final answer:

To convert 135 to Mayan notation, one must work with a base-20 system. In this system, 135 is broken down to (6 x 20) + 15, which is written in Mayan as a bar and three dots on top, and below that, six dots.

Explanation:

The Maya used a vigesimal (base-20) numeral system for their calculations, which is different from our familiar base-10 system. To convert the number 135 to Mayan notation, you need to express it in base-20.

135 in base-20 is calculated as:

In Mayan notation, numbers are written vertically with the largest value at the top, so 135 would be represented with a bar and three dots on top (for 15), and under that, six dots (for 6).

Answer:

A. X-intercepts

Step-by-step explanation:

The zeros of a function are the values of x for which y is 0.

For example:

y = 2x - 4

Has zeros

2x - 4 = 0

2x = 4

x = 2

Which means that when x = 2, y = 0. Looking at the graphic of this function, it crosses the x line when x = 2. So the zeros are also called x intercepts.

So the correct answer is

A. X-intercepts

Answer:

X-intercepts

Step-by-step explanation:

TRUST ME

Answer:

Option A The linear model on tar content accounts for​ 92.4% of the variability in nicotine content.

Step-by-step explanation:

R-square also known as coefficient of determination measures the variability in dependent variable explained by the linear relationship with independent variable.

The given scenario demonstrates that nicotine content is a dependent variable while tar content is an independent variable. So, the given R-square value 92.4% describes that 92.4% of variability in nicotine content is explained by the linear relationship with tar content. We can also write this as "The linear model on tar content accounts for​ 92.4% of the variability in nicotine content".

The Upper R squared or the coefficient of determination here represents the percentage of the variability in the nicotine content that can be explained by the tar content in the regression model, which in this case is 92.4%.

In this context, the meaning of Upper R squared is represented by option A. The linear model on tar content accounts for 92.4% of the variability in nicotine content. This indicates that 92.4% of the change in nicotine content can be explained by the amount of tar content based on the linear regression model used. This measure is also known as the coefficient of determination. Meanwhile, options B, C, and D are not correct interpretations of the R squared in this context. Both B and D wrongly relate the percentage to the predictability of the data points and option C incorrectly associates this percentage with the residual magnitude.

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Answer:

x  =  7

y  =  3

z (max)  =  4950/3  =  1650

Step-by-step explanation:

Let call  

x numbers of church goup and

y numbers of Union Local

Then  

First contraint

2*x + 2*y ≤ 20

Second one

1*x + 3*y ≤ 16

Objective Function

z = 150*x + 200*y

Then the system is

z = 150*x + 200*y To maximize

Subject to:

2*x + 2*y ≤ 20

1*x + 3*y ≤ 16

x ≥ 0 y ≥ 0

We will solve by using the Simplex method

z - 150 *x - 200*y = 0

2*x + 2*y + s₁ = 20  

1*x + 3*y + 0s₁ + s₂ = 16

First Table

z           x          y          s₁           s₂          Cte

1        -150    -200       0            0     =      0

0          2         2           1            0     =    20

0          1          3          0            1      =     16

First iteration:

Column pivot   ( y column ) row pivot (third row) pivot 3

Second table

z           x                y          s₁           s₂                Cte

1       -250/3           0          0         200/3    =    3200/3

0      - 4/3               0          -1           2/3       =    -20/3

0         1/3               1            0           1/3       =    -20/3

Second iteration:

Column pivot  ( x column ) row pivot  (second row)  pivot  -4/3

Third table

z           x                y          s₁           s₂                Cte

1            0               0      750/12    700/6    =   4950/3

0            1                0        3/4          -1/2     =    7

0            0                1         -1/4          1/2     =  9/3

Donna should contact four church groups and four labor unions to maximize her fundraising, which would yield $1400. The maximum money she could 'lose' (or not earn) is $0 if she does not contact any group.

We need to set up linear inequalities to represent the constraints described by Donna's situation. We will let x represent the number of church groups and y represent the number of labor unions she contacts.

The time required for letter writing can be described by the inequality: 2x + 2y <= 20, and the time for follow-ups is x + 3y <= 16.

We want to maximize the sum z = 150x + 200y which represents the total money she can raise.

To find the maximum, one method is to graph the feasible region defined by the constraints and find the vertices. Then evaluate the objective function at the vertices. In this case, the vertices are (0,0), (0,5), (8,0) and (4,4).

Upon calculation, contacting four church groups and four labor unions yields the highest amount of $1400.

As for the loss, given the nature of the task, the most money she could 'lose' in a month is just not raising any money at all, which would be $0 if she doesn't contact any group.

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Answer = 321.7

Explanation:

radius = y

height = [tex]4y^{2} - y^{3}[/tex]

area of cylinder = 2π*r*h

Integrate the area to calculate the volume:

[tex]= \int\limits^0_4 {2\pi(4y^{3} -y^{4}) } \, dy[/tex]

[tex]= 2\pi (\int\limits^0_4 {4y^{3}dy -\int\limits^0_4 y^{4} dy )} \,[/tex]

[tex]= 2\pi (256-\frac{1024}{5} )[/tex]

[tex]=\frac{512\pi }{5}[/tex]

[tex]= 321.7[/tex]

Final answer:

To compute the volume of the solid formed by rotating the region bounded by the curves x = 4y² − y³ and x = 0 around the x-axis, utilize the cylindrical shells method with the relevant volume formula for cylindrical shells and integrate over the intersecting interval of y-values.

Explanation:

Finding the Volume of a Solid of Revolution

To find the volume of the solid obtained by rotating the region bounded by the curves x = 4y2 − y3 and x = 0 about the x-axis using the method of cylindrical shells, we follow these steps:

The relevant formulas for the volume and surface area of a sphere are (4/3)πr3 and 4πr2, respectively. It's important to note that the volume depends on the cube of the radius R3, while the surface area is a function of the square of the radius R2.