MATHEMATICS COLLEGE

Evaluate each function.k(a) = |-2a + 3| - 1; Find k(3).

g(x) = 4^2x-1 + 7; Find g(1).

f(x) = |8x^2 - 5x + 3|; Find f(-2).

h(x) = -3x + 9; Find h(-1 + x).

f(n) = 5n - 1, Find f(- 3n).

Answers

Answer 1

Answer:

Step-by-step explanation:

k(a) = | -2a + 3 | - 1

k(3) = | -2(3) + 3 | - 1 replaced the a with 3

= | -6 + 3 | - 1

= | -3 | - 1

= 3 - 1

= 2

g(x) = 4²ˣ⁻¹ + 7

g(1) = 4²⁽¹⁾⁻¹ + 7 replaced the x with 1

= 4¹ + 7

= 4 + 7

= 11

f(x) = | 8x² - 5x + 3 |

f(-2) = | 8(-2)² - 5(-2) + 3 | replaced the x with -2

= | 8(4) - 10 + 3 |

= | 12 - 10 + 3 |

= | 5 |

= 5

h(x) = -3x + 9

h(-1 + x) = -3(-1 + x) + 9 replaced the x with -1 + x

= 3 - 3x + 9

= 12 - 3x

f(n) = 5n - 1

f(-3n) = 5(-3n) - 1 replaced the n with -3n

= -15n - 1

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Charlie has—$19 in his bank account. This is $28 more then Steve’s bank account. How much money does Steve have in his bank account?

Answers

Answer:

may be incorrect but probably not -9

Step-by-step explanation:

19 - 28 = -9

HELP PLEASE ASAP What is the length of the diagonal of the square shown below?

Answers

Answer:

C) diagonal = 5√2

Step-by-step explanation:

diagonal² = 5² + 5²

diagonal² = 50

diagonal = √50

diagonal = √25x2

diagonal = 5√2

Answer:

C) diagonal = 5√2

Step-by-step explanation:

diagonal² = 5² + 5²

diagonal² = 50

diagonal = √50

diagonal = √25x2

diagonal = 5√2

Solve each quadratic equation. Show your work. 14. (2x – 1)(x + 7) = 0 15. x^2 + 3x = 10 16. 4x2 = 100

Answers

14.)

2x(x + 7) -1(x + 7) = 2x² + 14x - x - 7 = 2x² + 13x - 7 ...//Ans ...

15.)

x² + 3x - 10 = x² + 5x - 2x - 10 = x(x + 5) -2(x + 5) = (x - 2) (x + 5) ...//Ans ...

16.)

4x² = 100
x² = 100 ÷ 4 = 25
x = √25 = 5 ...//Ans ...

A drawer contains 3 tan sweaters and two black sweaters. You randomly choose two sweaters. What is the probability that both sweaters are black?

Answers

A = event that you select a black sweater
P(A) = 2/5 since there are 2 black out of 2+3 = 5 total

After you make a selection, we have the event
B = event that you select another black sweater assuming event A has happened already

P(B) = 1/4 because there's 1 black sweater left out of 5-1 = 4 left over

Multiply the probabilities
P(A and B) = P(A)*P(B)
P(A and B) = (2/5)*(1/4)
P(A and B) = 2/20
P(A and B) = 1/10

The answer as a fraction is 1/10
In decimal form, it is 0.1
As a percent, the answer is 10%

A reservation service employs six information operators who receive requests for information independently of one another, each according to a Poisson process with rate ???? = 2 per minute. a. What is the probability that during a given 1 min period, the first operator receives no requests? (Round your answer to three decimal places.) b. What is the probability that during a given 1 min period, exactly three of the six operators receive no requests? (Round your answer to five decimal places.)

Answers

Answer:

a) 0.135 = 13.5% probability that during a given 1 min period, the first operator receives no requests.

b) 0.03185 = 3.185% probability that during a given 1 min period, exactly three of the six operators receive no requests

Step-by-step explanation:

To solve this question, we need to understand the Poisson distribution and the binomial distribution.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)$

In which $C_(n,x)$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_(n,x) = (n!)/(x!(n-x)!)$

And p is the probability of X happening.

Poisson process with rate 2 per minute

This means that $\mu = 2$

a. What is the probability that during a given 1 min period, the first operator receives no requests?

Single operator, so we use the Poisson distribution.

This is P(X = 0).

$P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)$

$P(X = 0) = (e^(-2)*2^(0))/((0)!) = 0.135$

0.135 = 13.5% probability that during a given 1 min period, the first operator receives no requests.

b. What is the probability that during a given 1 min period, exactly three of the six operators receive no requests?

6 operators, so we use the binomial distribution with $n = 6$

Each operator has a 13.5% probability of receiving no requests during a minute, so $p = 0.135$

This is P(X = 3).

$P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)$

$P(X = 3) = C_(6,3).(0.135)^(3).(0.865)^(3) = 0.03185$

0.03185 = 3.185% probability that during a given 1 min period, exactly three of the six operators receive no requests

farmer ed has 3000 meters of fencing and wants to enclose a rectangle plot that borders on a river. if farmer ed does not fence the side along the river what is the largest area that can be enclosed

Answers

Answer:

area = 1500× 750 = $1125000 m^2$

Step-by-step explanation:

we know area of rectangle

for length = l m

and width = b m

$A = lb$

and perimeter

$Perimeter = 2 (length + width)$

but one side length measures is not required because of the river so

He does not use the fence along the side of the river

so we use this formula

Perimeter = P = L + 2 b

Perimeter is 3000 m

$so \ \ 3000 = l +2b$

$l = 3000 - 2b$

so area will be

$A = (3000-2b)b$

it is a quadratic function whose max or min will

occur at the average of the Solutions.

on Solving (3000 - 2b)b = 0

3000 - 2b = 0 or b=0

2b =3000

$b =(3000)/(2) \nb = 1500 m$

or $b = 0 m$

The average of the values are $((0+1500))/(2) = 750$

so for max area we use b= $750 m$

The Length is then L=3000 - 2(750) = 3000 - 1500 = 1500

for max area

length = 1500 m

bredth = 750 m

area = 1500× 750 = $1125000 m^2$

Final answer:

The largest area that can be enclosed by Farmer Ed with 3000 meters of fencing along a river (with only three sides fenced) equals 1,125,000 square meters by using principles of mathematical optimization.

Explanation:

In this question, Farmer Ed wants to maximize the area of a rectangle with only three sides fenced, since one side borders on a river. We can use the principles of optimization in mathematics to solve this problem.

With 3000 meters of fencing for three sides, if we denote one side perpendicular to the river as X and the side parallel to the river (which forms the base of the rectangle) as Y, then, the perimeter would be Y+2X which is equal to 3000 meters. So, Y = 3000-2X.

The area A of a rectangle is length times width, or, in this case, A = XY. Substituting Y from the equation above: A = X(3000-2X) = 3000X - 2X^2. To maximize this area, we need to find values of X for which this equation has its maximum value.

The maximum or minimum of a function can be found at points where its derivative is zero. So, we take the derivative of A with respect to X, set it equal to zero, and solve for X.

The derivative, dA/dX is 3000 - 4X. Setting this equal to 0 gives X = 3000/4 = 750. So, the maximum area that Farmer Ed can enclose is when X is 750, and Y is 3000 - 2X = 1500, so the maximum area is 750 * 1500 = 1,125,000 square meters.