PLEASE HELP THANKS


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Answer:

Step-by-step explanation:

the n number of value of x

[tex] \displaystyle x_{1},x _{2} \dots x_{n}[/tex]

let it be

[tex] \displaystyle x_{1} = x _{2} = x_{3}{\dots }= x_{n} = a[/tex]

now, the A.M of x is

[tex] \rm \displaystyle \: A.M = \frac{ x_{1} + x_{2} + \dots \dots \: + x_{n} }{n} [/tex]

since every value equal to a

substitute:

[tex] \rm \displaystyle \: A.M = \frac{ a + a + \dots \dots \: + a}{n} [/tex]

[tex] \rm \displaystyle \: A.M = \frac{ na}{n} [/tex]

reduce fraction:

[tex] \rm \displaystyle \: A.M = a[/tex]

the G.M of x is

[tex] \rm\displaystyle \: G.M =( x_{1} \times x _{2} {\dots }\times x_{n} {)}^{ {1}^{}/ {n}^{} } [/tex]

since every value equal to a

substitute:

[tex] \rm\displaystyle \: G.M =( a \times a{\dots }\times a{)}^{ {1}^{}/ {n}^{} } [/tex]

recall law of exponent:

[tex] \rm\displaystyle \: G.M =( {a}^{n} {)}^{ {1}^{}/ {n}^{} } [/tex]

recall law of exponent:

[tex] \rm\displaystyle \: G.M = a[/tex]

the H.M of x is

[tex] \displaystyle \: H.M = \frac{n}{ \frac{1}{ x_{1}} + \frac{1}{ x_{2} } {\dots } \: { \dots}\frac{1}{x _{n} } } [/tex]

since every value equal to a

substitute:

[tex] \displaystyle \: H.M = \frac{n}{ \frac{1}{ a} + \frac{1}{ a } {\dots } \: { \dots}\frac{1}{a } } [/tex]

[tex] \displaystyle \: H.M = \frac{n}{ \dfrac{n}{a} } [/tex]

simplify complex fraction:

[tex] \displaystyle \: H.M = n \times \frac{a}{n} [/tex]

[tex] \displaystyle \: H.M = a \: [/tex]

so

[tex] \displaystyle \: A.M = G.M = H.M = a[/tex]

hence,

[tex]\text{Proven}[/tex]

Answer:

Answer:

Aceleracion = 0.8 m/s²

Step-by-step explanation:

Dados los siguientes datos;

Velocidad inicial = 12 m/s

Velocidad final = 20 m/s

Tiempo, t = 10 segundos

Para encontrar la aceleración;

Aceleración se puede definir como la tasa de cambio de la velocidad de un objeto con respecto al tiempo.

Esto simplemente significa que la aceleración viene dada por la resta de la velocidad inicial de la velocidad final a lo largo del tiempo.

Por lo tanto, si restamos la velocidad inicial de la velocidad final y la dividimos por el tiempo, podemos calcular la aceleración de un objeto. Matemáticamente, la aceleración viene dada por la fórmula;

[tex] Aceleracion = \frac{final \; velocidad - inicial \; velocidad}{tiempo}[/tex]

Sustituyendo en la fórmula, tenemos;

[tex] Aceleracion = \frac{20 - 12}{10}[/tex]

[tex] Aceleracion = \frac{8}{10}[/tex]

Aceleracion = 0.8 m/s²

Answer:

D

i thank

Step-by-step explanation:

The answer to question is given below.

(i) To find the mean, E(N), use the following formula:

mean = E(N) = ∑[n · P(N = n)] for all values of n.

The distribution given above is geometric: P(N = n) = p^ n (1-p)

where n = 1,2,3,...

Therefore, E(N) = ∑[n · P(N = n)] = ∑ [n · p^ n (1-p)] for n = 1,2,3,...

Since this sum is infinite, we have to truncate the summation and compute the mean for a finite number of terms. We can use 40 since there are 40 independent observations. Therefore, we have:

E(N) ≈ ∑[n · P(N = n)] for n = 1 to 40

= 1 · p(1-p) + 2 · p^2(1-p) + 3 · p^3(1-p) + ... + 40 · p^40(1-p)

= (1-p) ∑[n · p^n] for n = 1 to 40

= (1-p) [p + 2p^2 + 3p^3 + ... + 40p^40]

= (1-p) p ∑[n · p^(n-1)] for n = 1 to 40

= (1-p) p [1 + 2p + 3p^2 + ... + 40p^39]

= p(1-p) [1p^(1-1) + 2p^(2-1) + 3p^(3-1) + ... + 40p^(40-1)]

= p(1-p) ∑[n · p^(n-1)] for n = 1 to 40

= p(1-p) ∑[(n-1+1) · p^(n-1)] for n = 1 to 40

= p(1-p) [∑[(n-1) · p^(n-1)] + ∑[1 · p^(n-1)]] for n = 1 to 40

= p(1-p) [∑[n · p^(n-1)] - ∑[p^(n-1)]] + p(1-p) ∑[p^(n-1)] for n = 1 to 40

= p(1-p) [d/dp ∑[p^n]] - p(1-p) (1/(1-p)) + p(1-p) (1/(1-p))

= p(1-p) [d/dp (1/(1-p)) ∑[(1-p)p^n]] + 1

= p(1-p) [d/dp (1/(1-p)) (1-p)/(1-p)^(40+1)] + 1

= p(1-p) [d/dp (1-p)^(-40)) + 1

= p(1-p) (40(1-p)^(-41)) + 1

= 40p/(1-p) - 40

(ii) Since this is a geometric distribution, the maximum likelihood estimator (MLE) for the unknown parameter p is given by:

MLE: ê = x/n

where x is the number of successes (species of insect caught in a trap) and n is the sample size (40). To find an equation that determines the MLE ê, differentiate the log-likelihood and equate it to zero to find the maximum of the likelihood function. The log-likelihood for a geometric distribution is given by:

L = ∑ [log(P(N = n))] for n = 1 to 40

= ∑ [log(p^n (1-p))] for n = 1 to 40

= ∑ [n · log(p) + log(1-p)] for n = 1 to 40

= (40 · log(p) + ∑ [n · log(p)] + ∑ [log(1-p)])

Use the formula MLE = x/n to replace p by x/n. This gives:

L = (40 · log(x/n) + ∑ [n · log(x/n)] + ∑ [log(1-x/n)])

Differentiate L with respect to x/n and equate to zero to obtain the MLE ê:0 = dL/d(x/n) = (40/x) - (n/x) - ∑ [1/(1 - x/n)]

Solving this equation will give ê in terms of x and n.

(iii) The Fisher information, I(p), is defined as:

I(p) = -E[d^2/dp^2 L] = E[d/dp (d/dp L)]

where L is the log-likelihood function.

From part (ii), we have:

L = (40 · log(x/n) + ∑ [n · log(x/n)] + ∑ [log(1-x/n)])

Therefore,

∂L/∂p = (40/x) - (n/x) and∂^2L/∂p^2 = -40/x^2.The Fisher information is therefore:

I(p) = -E[d^2/dp^2 L] = E[40/x^2] = 40E[x/n]^(-2) = 40/p^2.

Using the asymptotic normality of the MLE, the 95% confidence interval for p is approximately given by:

p ± 1.96 · sqrt(Var(p))

where Var(p) = 1/I(p) = p^2/40.

Using Ô = 0.75, we have ê = x/n = (N1 + N2 + ... + N40)/40 = (28 + 30 + ... + 22)/40 = 0.625.

Therefore, p ± 1.96 · sqrt(Var(p))= 0.625 ± 1.96 · sqrt(0.625^2/40)= (0.478, 0.772).

(iv) When N1 = 100, the maximum likelihood estimate, ê(1), can be found iteratively as follows:

ê(1) = 0.70MLE = ê(1) = x/n

where x is the number of species of insect caught in a trap and n = 40. Therefore, ê(1) can be computed from the data. For example, if x = 25, then ê(1) = 25/40 = 0.625. To obtain ê(2), we need to solve the equation obtained in part (ii) for n = 100:0 = (40/x) - (100/x) - ∑ [1/(1 - x/100)]

We can use Newton's method to solve this equation numerically. Let ê(2) be the root obtained after one iteration of Newton's method. Then, we have:

ê(2) = ê(1) - f(ê(1))/f'(ê(1))where f(p) = (40/x) - (100/x) - ∑ [1/(1 - x/100)] and f'(p) = -∑ [x/100(x - 100)^2].

For example, if x = 25 and ê(1) = 0.625, then:

ê(2) = 0.625 - f(0.625)/f'(0.625)= 0.625 - (-0.1155)/(0.1869)= 0.625 + 0.6171= 1.242.

This value is not valid since the MLE must lie between 0 and 1. Therefore, we need to use a different starting value of ê. Let ê(1) = 0.80. Then, we have:

f(0.80) = (40/x) - (100/x) - ∑ [1/(1 - x/100)] = 0.0142f'(0.80) = -∑ [x/100(x - 100)^2] = -0.1026

Using Newton's method, we have:

ê(2) = 0.80 - f(0.80)/f'(0.80)= 0.80 - (0.0142)/(-0.1026)= 0.9367

ê(3) can be obtained in a similar manner by solving the equation obtained in part (ii) for n = 100 using ê(2) as the starting value.

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Answer:

Convenience sampling

Step-by-step explanation:

From the question, we understand that 40 of all students are to be surveyed. This implies that from the 41st till the last student will not be a part of the survey.

When sampling is done, taking a part of the whole population that is near or readily available; such sampling is referred to as convenience sampling.

The whole population, in this case are all students while the convenience being selected are the first 40 students.

Hi there!

[tex]\large\boxed{x < 8}[/tex]

3(x - 2) < 18

Begin by dividing both sides by 3:

3(x - 2)/3 < 18/3

x - 2 < 6

Add 2 to both sides:

x - 2 + 2 < 6 + 2

x < 8

x< 8

Hope this is the answer you are looing for.

Answer:

The probability that Scott will wash is 2.5

Step-by-step explanation:

Given

Let the events be: P = Purple and G = Green

[tex]P = 2[/tex]

[tex]G = 3[/tex]

Required

The probability of Scott washing the dishes

If Scott washes the dishes, then it means he picks two spoons of the same color handle.

So, we have to calculate the probability of picking the same handle. i.e.

[tex]P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)[/tex]

This gives:

[tex]P(G_1\ and\ G_2) = P(G_1) * P(G_2)[/tex]

[tex]P(G_1\ and\ G_2) = \frac{n(G)}{Total} * \frac{n(G)-1}{Total - 1}[/tex]

[tex]P(G_1\ and\ G_2) = \frac{3}{5} * \frac{3-1}{5- 1}[/tex]

[tex]P(G_1\ and\ G_2) = \frac{3}{5} * \frac{2}{4}[/tex]

[tex]P(G_1\ and\ G_2) = \frac{3}{10}[/tex]

[tex]P(P_1\ and\ P_2) = P(P_1) * P(P_2)[/tex]

[tex]P(P_1\ and\ P_2) = \frac{n(P)}{Total} * \frac{n(P)-1}{Total - 1}[/tex]

[tex]P(P_1\ and\ P_2) = \frac{2}{5} * \frac{2-1}{5- 1}[/tex]

[tex]P(P_1\ and\ P_2) = \frac{2}{5} * \frac{1}{4}[/tex]

[tex]P(P_1\ and\ P_2) = \frac{1}{10}[/tex]

Note that: 1 is subtracted because it is a probability without replacement

So, we have:

[tex]P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)[/tex]

[tex]P(Same) = \frac{3}{10} + \frac{1}{10}[/tex]

[tex]P(Same) = \frac{3+1}{10}[/tex]

[tex]P(Same) = \frac{4}{10}[/tex]

[tex]P(Same) = \frac{2}{5}[/tex]

is it a science question

Inglis please. Thank you.

Answer:

It’s a: The area of the floor would be a rational number

Step-by-step explanation:

Answer:

I believe it is 60

Step-by-step explanation:

It's going to be COS and since you don't have and angle you are going to click [tex]cos^{-1}[/tex] and then type in 9/18 so it should look like this

[tex]cos^{-1}(9/18) = 60[/tex]

Answer:

it represents a polynomial

If the probabilities of their outcomes are 20%, 60%, and 20% respectively, the expected value of these outcomes is $346.

To calculate the expected value of the outcomes, we multiply each outcome by its corresponding probability and then sum up the results.

In this case, the projected outcomes are $240, $310, and $560, with probabilities of 20%, 60%, and 20% respectively.

To calculate the expected value, we use the formula:

Expected value = (Outcome 1 * Probability 1) + (Outcome 2 * Probability 2) + (Outcome 3 * Probability 3) + ...

Expected value = ($240 * 0.20) + ($310 * 0.60) + ($560 * 0.20)

Expected value = $48 + $186 + $112

Expected value = $346

The expected value represents the average value or the long-term average outcome we can expect from the given probabilities and outcomes. It provides a summary measure that helps in understanding the central tendency of the distribution of outcomes.

In this case, the expected value indicates that, on average, we can expect the project's outcome to be around $346.

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a. The initial amount of money deposited into the account is $1575.

b. The annual interest rate being paid on the account is 4.5%.

c. The amount of money in the account after 15 years is approximately $2946.27.

d. It will take approximately 20 years for the account to reach a balance of at least $4000.

To answer these questions, let's analyze the given equation:

A(1) = 1575 * (1.045)^t

a. The initial amount of money deposited into the account is $1575. This is evident from the equation, where A(1) represents the amount of money after 1 year.

b. To determine the annual interest rate, we can compare the given equation with the general formula for compound interest:

A = P * (1 + r)^t

Comparing the two equations, we can see that the interest rate in the given equation is 4.5% (0.045) since (1 + r) is equal to 1.045.

c. To find the amount of money in the account after 15 years, we can substitute t = 15 into the equation and calculate the result:

A(15) = 1575 * (1.045)^15 ≈ $2946.27 (rounded to the nearest dollar)

Therefore, after 15 years, the amount of money in the account will be approximately $2946.

d. To find the number of years it will take for the account to have at least $4000, we need to solve the equation for t. Let's set up the equation and solve for t:

4000 = 1575 * (1.045)^t

To solve this equation, we can take the logarithm of both sides (with base 1.045):

log(4000) = log(1575 * (1.045)^t)

Using logarithm properties, we can simplify the equation:

log(4000) = log(1575) + log((1.045)^t)

log(4000) = log(1575) + t * log(1.045)

Now, we can isolate t by subtracting log(1575) from both sides and then dividing by log(1.045):

t = (log(4000) - log(1575)) / log(1.045)

Calculating this expression, we find:

t ≈ 19.56 (rounded to two decimal places)

Therefore, it will take approximately 20 years (rounded to the nearest whole year) for the account to have at least $4000.

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Answer:

V = 50 m³

Step-by-step explanation:

The formula for the volume of a pyramid of base area b and height h is V = (1/3)(b)(h).

Here the volume is V = (1/3)(5 m)²(6 m), or V = 50 m³

The best-fit quadratic model is given by the equation y = -0.19378x² - 0.15265x + 31.0148, where x represents the year after 1900.

According to the best-fit quadratic model obtained from the data, approximately 14,000 families will live in Sunnyvale in 2015.

To estimate the number of families in 2015, we substitute x = 115 (2015 - 1900) into the quadratic model:

y = -0.19378(115)² - 0.15265(115) + 31.0148

≈ 14,000

Therefore, based on the best-fit quadratic model, it is estimated that approximately 14,000 families will live in Sunnyvale in 2015. The model is obtained by performing a quadratic regression using the given data points, and it provides a reasonable estimate for the number of families in 2015 based on the trend observed in the data.

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Answer:

158

Step-by-step explanation:

The matrix A for the linear transformation T relative to the bases B = {1, x, x2} and B' = {1, x, x2, x3, x4} is:
[4 0 0]  
[0 4 0]
[0 0 4]
[0 0 0]
[0 0 0]

The given linear transformation is T: P2 → P4 and T(p) = 4x2p.

We are to find the matrix A for T relative to the bases B = {1, x, x2} and B' = {1, x, x2, x3, x4}.

Consider the linear transformation of each element of the first basis B.

We have; T(1) = 4x2(1) = 4x2(1) + 0x3 + 0x4 = 4x2T(x) = 4x2(x) = 0x2 + 4x3 + 0x4 = 0x2 + 4x3T(x^2) = 4x2(x^2) = 0x2 + 0x3 + 4x4 = 0x2 + 0x3 + 4x4

Thus, the matrix of T relative to B is: [4 0 0] [0 4 0] [0 0 4]

Next, we will find the coordinates of each element of the basis B' under the basis B.

Using the relations;x3 = x3x^3 = x2.x   [x3]B = [0 0 1]T(x^3) = 4x2(x^3) = 0x2 + 0x3 + 0x4 = 0x2 + 0x3 + 0x4

Thus, the coordinate vector of x3 relative to B is [0 0 1].

Using the relation; x4 = x4 - x3x^4 = x^4 - x2.x   [x4]B = [0 -1 0]T(x^4) = 4x2(x^4) = 0x2 + 0x3 + 4x4 = 0x2 + 0x3 + 4(x3 + x2.x) = 0x2 + 0x3 + 4x3 + 0x2 = 0x2 + 4x3

Thus, the coordinate vector of x4 relative to B is [0 -1 0].

Thus, the matrix of T relative to B' is [4 0 0 0 0] [0 4 0 0 0] [0 0 4 0 0] [0 0 0 0 0] [0 0 0 0 0]

Therefore, the matrix A for T relative to the bases B = {1, x, x2} and B' = {1, x, x2, x3, x4} is: [4 0 0] [0 4 0] [0 0 4] [0 0 0] [0 0 0].

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Answer:

s = 4.5 cm

l= 3.75 cm

Step-by-step explanation:

Mathematically, when two shapes are similar, the ratio of their corresponding sides are equal

According to this rule, we have it that;

4/6 = 3/s

4 * s = 3 * 6

4s = 18

s = 18/4

s = 4.5 cm

Similarly;

4/5 = 3/l

4 * l = 5 * 3

4l = 15

l = 15/4

l = 3.75 cm

Answer:

C

20 character minimum

The probability that they shoot free throws in alphabetical order depends on the number of players participating. If there are only two players, the probability would be 1/2 or 0.5.

Since each player has a different name, there are only two possible orders in which they can shoot free throws: alphabetical order or reverse alphabetical order. Out of these two possibilities, only one is the desired outcome (alphabetical order). Therefore, the probability of shooting free throws in alphabetical order is 1 out of 2, which can be expressed as 1/2 or 0.5

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Answer:

D<28

Step-by-step explanation:

The required inequality for the given situation can be, d > 28 or d < 28

A relationship between two expressions or values that are not equal to each other is called inequality.

Given that, write an inequality for the situation.

Situation :-

A number of days, d, of sunshine is not 28.

Here, the number of days is said to be not equal to 28, we are not given if is less or more,

It can be less or more than 28 but not 28

So, here we get two situations,

Either, the number of days, d, of sunshine is less than 28, or the number of days, d, of sunshine is more than 28.

If we write these situations, mathematically, the inequalities we will have, are :-

1) The number of days, d, of sunshine is less than 28 :-

d < 28

2) The number of days, d, of sunshine is more than 28 :-

d > 28

Therefore, the two inequalities, are :- d < 28 or d > 28

Hence, the required inequality for the given situation can be, d > 28 or d < 28

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The complete question is :-

Write an inequality for the situation.

A number of days, d, of sunshine is not 28 .

Probability of selecting two defective cylinders ≈ 0.1071. Probability of selecting at least one defective cylinder ≈ 0.6429

To calculate the probability of selecting two defective cylinders when two cylinders are chosen at random without replacement, we need to consider the total number of cylinders and the number of defective cylinders. Given: Total number of cylinders: 8, Number of defective cylinders: 3. Probability of selecting two defective cylinders: To calculate this probability, we first need to determine the total number of ways to choose two cylinders out of the eight available. This can be calculated using the combination formula (nCr). Total ways to choose two cylinders out of eight: C(8, 2) = 8! / (2! * (8-2)!) = 28.

Next, we need to determine the number of ways to choose two defective cylinders out of the three available. Number of ways to choose two defective cylinders out of three: C(3, 2) = 3! / (2! * (3-2)!) = 3. Therefore, the probability of selecting two defective cylinders is: P(Two defective cylinders) = Number of ways to choose two defective cylinders / Total ways to choose two cylinders = 3/28 ≈ 0.1071 (rounded to four decimal places). Probability of selecting at least one defective cylinder: To calculate this probability, we can consider the complementary event, which is the probability of selecting no defective cylinders. Then, we subtract this probability from 1 to obtain the probability of selecting at least one defective cylinder.

Number of ways to choose two non-defective cylinders out of five remaining non-defective cylinders: C(5, 2) = 5! / (2! * (5-2)!) = 10. Total ways to choose two cylinders out of eight: C(8, 2) = 28 (as calculated earlier). Number of ways to choose at least one defective cylinder = Total ways to choose two cylinders - Number of ways to choose two non-defective cylinders= 28 - 10 = 18. Therefore, the probability of selecting at least one defective cylinder is: P(At least one defective cylinder) = Number of ways to choose at least one defective cylinder / Total ways to choose two cylinders= 18/28 ≈ 0.6429 (rounded to four decimal places).

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Answer:

D

Step-by-step explanation:

Answer:

First row of blanks: -3 -3

Second row of blanks: 2 2 3

Third row blank: 6

Step-by-step explanation:

x/2 + 3 = 6

The objective is to isolate the variable:

Subtract both sides by 3

x/2 = 3

Multiply both sides by 2

x = 6

Answer:

Step-by-step explanation:

this is sorta a thought process to figure this out.   so think it thur with me.  b/c they are saying that the line is the same length with the lines thur it... see them?  so then that means that two sides of the right triangle are of a certain length so that means we know the base and length of the sides of a box, if we were to add the other right triangle to it.. see that?

so we can say that  6*7.5 = box   and we'll have two total boxes, see that?  the top portion of the parallelogram and the bottom are each a box .. but cut down the middle and put as two opposite parts of a triangle.

so

2*(6*7.5) = 90 [tex]in^{2}[/tex]

Answer:

2(x+2) = x+4

Step-by-step explanation:

The pair that shows equivalent expressions is:

2(x+2) = x+4

This is because when we distribute the 2 to the terms inside the parentheses, we get:

2x + 4 = x + 4

By subtracting x from both sides of the equation, we get:

2x - x + 4 = 4

Simplifying further, we have:

x + 4 = 4

Therefore, the expression 2(x+2) is equivalent to x+4.

Hope this helps!

Answer:

13.8

Step-by-step explanation:

Hello There!

Basic concept: Geometry

We can solve for x using trigonometric ratios

Here are the Trigonometric Ratios

Remember SOHCAHTOA

Sin = Opposite over Hypotenuse

Cos = Adjacent over Hypotenuse

Tan = Opposite over Adjacent

For the angle that has a measure of 37 degrees

We are given its adjacent side length (11) and we need to find the hypotenuse

Basic concept: Alegbra 1

Adjacent and Hypotenuse corresponds with Cos so we are going to use cosine to create an equation and solve for x

Remember cos = adjacent over hypotenuse

So

[tex]cos37=\frac{11}{x}[/tex]

now we solve for x

what we want to do is get rid of the 11, to do so we divide each side by 11

[tex]\frac{11}{cos37} =x\\cos37=.79863551\\\frac{11}{0.79863551} =13.773749224[/tex]

we're left with x = 13.773749224

finally we want to round to the nearest tenth

The answer would be 13.8

For function f:(a,b) → R (continuous), and c ∈ (a,b), then

(i) If derivative is negative before c and positive after c, then f has an absolute minimum at c.

(ii) If derivative is positive before c and negative after c, then f has an absolute maximum at c.

Part (i) : If derivative of a function f(x) is negative for values of x between a and c, and positive for values of x between c and b, then the function has an absolute minimum at c.

This means that at point c, function reaches its lowest-value compared to all other points in the interval (a, b). The negative derivative before c indicates a decreasing trend, while the positive derivative after c indicates an increasing trend.

The change from decreasing to increasing at c suggests a minimum point. By the continuity of the function, we can conclude that the minimum value is achieved at c.

Part (ii) : Conversely, if  derivative of a function f(x) is positive for values of x between a and c, and negative for values of x between c and b, then the function has an absolute maximum at c.

This means that at point c, the function reaches its highest-value compared to all other points in the interval (a, b). The positive derivative before c indicates an increasing trend, while the negative derivative after c indicates a decreasing trend.

The change from increasing to decreasing at c suggests a maximum point. By the continuity of the function, we can conclude that the maximum value is achieved at c.

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Prove the theorems below: Let f:(a,b) → R be continuous. Let c ∈ (a,b) and suppose f is differentiable on (a, c) and (c, b).

(i) if f'(x) < 0 for x ∈ (a, c) and f'(x) > 0 for x ∈ (c, b), then f has an absolute minimum at c.

(ii) if f'(x) > 0 for x € (a, c) and f'(x) < 0 for x ∈ (c, b), then f has an absolute maximum at c.

Answer:

283 1/3 pizzas

Step-by-step explanation:

I did 3,400 div 12 = 1,420