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3 years ago

Why is y =ax+b called a linear equation

Mathematics

1 answer:

grigory [225]3 years ago

3 0

It's called a linear equation because the linear equation formula is y= mx+b, so they just changed the m to an a

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If the circle below is cut from the square of plywood below, how many square inches of plywood would be left over? Use π = 3.14,

Elis [28]

Answer:

86in²

Step-by-step explanation:

Find the complete question attached

Area of the circle = πr²

Area of the square = L²

Length of the side of the square

r is the radius of the circle

r = L/2 = 20/2 =10in

Given

Length of the side of the circle = 20in

Area of the square = 20²

Area of the square = 400in²

Area of the circle =π(10)²

Area of the circle = 3.14(100)

Area of the circle= 314in²

Amount of plywood left over = 400in²-314in² = 86in²

8 0

3 years ago

A bakery finds that the price they can sell cakes is given by the function p = 580 − 10x where x is the number of cakes sold per

dexar [7]

Answer:

The total revenue is $TR=580x-10x^2$ .

The marginal revenue is $MR=580-20x$ .

The fixed cost is $900.

The marginal cost function is $MC=50x+300$ .

Step-by-step explanation:

The Total Revenue ( $TR$ ) received from the sale of $x$ goods at price $p$ is given by

$TR=p\cdot x$

The Marginal Revenue ( $MR$ ) is the derivative of total revenue with respect to demand and is given by

$MR=\frac{d(TR)}{dx}$

From the information given we know that the price they can sell cakes is given by the function $p=580-10x$ , where $x$ is the number of cakes sold per day.

So, the total revenue is

$TR=(580-10x)\cdot x\\TR=580x-10x^2$

And the marginal revenue is

$MR=\frac{d}{dx}(580x-10x^2) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\MR=\frac{d}{dx}\left(580x\right)-\frac{d}{dx}\left(10x^2\right)\\\\MR=580-20x$

The Fixed Cost ( $FC$ ) is the amount of money you have to spend regardless of how many items you produce.

The Marginal Cost ( $MC$ ) function is the derivative of the cost function and is given by

$MC=\frac{d(TC)}{dx}$

We know that the total cost function of the company is given by $C=(30+5x)^2$ , which it is equal to

$\mathrm{Apply\:Perfect\:Square\:Formula}:\quad \left(a+b\right)^2=a^2+2ab+b^2\\a=30,\:\:b=5x\\\\\left(30+5x\right)^2=30^2+2\cdot \:30\cdot \:5x+\left(5x\right)^2=25x^2+300x+900\\\\C=25x^2+300x+900$

From the total cost function and applying the definition of fixed cost, the fixed cost is $900.

And the marginal cost function is

$MC=\frac{d}{\:dx}\left(25x^2+300x+900\right)\\\\MC=\frac{d}{dx}\left(25x^2\right)+\frac{d}{dx}\left(300x\right)+\frac{d}{dx}\left(900\right)\\\\MC=50x+300+0=50x+300$

4 0

3 years ago

Suppose the number of children in a household has a binomial distribution with parameters n=12n=12 and p=50p=50%. Find the proba

nadya68 [22]

Answer:

a) 20.95% probability of a household having 2 or 5 children.

b) 7.29% probability of a household having 3 or fewer children.

c) 19.37% probability of a household having 8 or more children.

d) 19.37% probability of a household having fewer than 5 children.

e) 92.71% probability of a household having more than 3 children.

Step-by-step explanation:

Binomial probability 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

$n = 12, p = 0.5$

(a) 2 or 5 children

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

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.1934$

$p = P(X = 2) + P(X = 5) = 0.0161 + 0.1934 = 0.2095$

20.95% probability of a household having 2 or 5 children.

(b) 3 or fewer children

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0002 + 0.0029 + 0.0161 + 0.0537 = 0.0729$

7.29% probability of a household having 3 or fewer children.

(c) 8 or more children

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

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

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.1208$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.0537$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.0161$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.0029$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.0002$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.1208 + 0.0537 + 0.0161 + 0.0029 + 0.0002 = 0.1937$

19.37% probability of a household having 8 or more children.

(d) fewer than 5 children

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

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

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.1208$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0029 + 0.0161 + 0.0537 + 0.1208 = 0.1937$

19.37% probability of a household having fewer than 5 children.

(e) more than 3 children

Either a household has 3 or fewer children, or it has more than 3. The sum of these probabilities is 100%.

From b)

7.29% probability of a household having 3 or fewer children.

p + 7.29 = 100

p = 92.71

92.71% probability of a household having more than 3 children.

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