The mass of the eggs laid by a certain breed of hen is a normally distributed random variable with mean 0.06kg and standard deviation 0.009kg. If a random sample of 1800 eggs are sold, how many would you expect to have mass; (i)Less than 0.052 kg (standard size). (ii)Between 0.052 kg and 0.075kg (medium size).(iii)Greater than 0.075 kg (large size).

(b)If the cost of producing an egg in (a) is Gh¢0.45 and selling price of each standard, medium and large size eggs are Gh¢0.60, Gh¢0.80, and Gh¢0.95 respectively, calculate (i)The expected profit to be made from selling the standard size, medium size and large size eggs. (ii)The overall profit.​

Answers

Answer 1

a (i) The number of standard size is 288 eggs

a (ii) The number of medium size is 1,476 eggs

a (iii) The number of large size is 36 eggs

b (i) The expected profit made from selling the standard size is Gh¢43.2

b (ii) The expected profit made from selling the medium size is Gh¢516.6

b (iii) The expected profit made from selling the large size is Gh¢16.2

The given parameters:

the mean of the distribution, m = 0.06 kg

standard deviation (std) of the distribution, d = 0.009 kg

number of the samples, n = 1800 eggs

(i) Find the position of "less than 0.052 kg (standard size)":

1 standard deviation below the mean = m - d m - d = 0.06 kg  -  0.009 kg = 0.051 kg

(the standard size is 1 standard deviation below the mean)

less than 1 standard deviation below the mean in a normal distribution is equal to 16% of the data samplesNumber of standard size = 0.16 x 1800 = 288 eggs

(ii) Find the position of "Between 0.052 kg and 0.075kg (medium size)":

0.052 kg is 1 standard deviation below the mean2 standard deviation above the mean = m + 2dm + 2d = 0.06 + 2(0.009) = 0.078 kg

(the medium size is between 1 std below the mean and 2 std above the mean)

Between 1 std below the mean and 2 std above the mean in a normal distribution = (68 + 14)% = 82%Number of medium size = 0.82 x 1800 = 1,476 eggs

(iii) Find the position of "Greater than 0.075 kg (large size)":

0.078 kg is 2 standard deviation below the meangreater than 2 std above the mean in a normal distribution = 2 % of the data samples Number of large size = 0.02 x 1800 = 36 eggs

Check: 288 eggs + 1476 eggs + 36 eggs = 1,800 eggs

(b) the cost of production of an egg = Gh¢0.45

the selling price of the standard size = Gh¢0.60

the selling price of the medium size = Gh¢0.80

the selling price of the large size = Gh¢0.95

(i) The expected profit made from selling the standard size:

Profit = total revenue - cost of production

Profit = 288(0.6) - 288(0.45) = Gh¢43.2

(ii) The expected profit made from selling the medium size:

Profit = total revenue - cost of production

Profit = 1476(0.8) - 1476(0.45) = Gh¢516.6

(iii) The expected profit made from selling the large size:

Profit = total revenue - cost of production

Profit = 36(0.9) - 36(0.45) = Gh¢16.2

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The Mass Of The Eggs Laid By A Certain Breed Of Hen Is A Normally Distributed Random Variable With Mean

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7.5 -ln(0.25)

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We can first graph this function. Looking at the graph, we can separate this into two ranges -- from x=0.25 to x=1 and from x=1 to x=4.

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Step-by-step explanation:

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[tex]2=f(0)=\frac{3(0)+8}{2(0)-a}=-\frac8a[/tex]

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[tex]-\frac a8=\frac12[/tex]

Then multiplying by [tex]-8[/tex] gives

[tex]a=\frac12(-8)=-4[/tex]

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Answer:

[tex] \frac{3}{50} [/tex]

Step-by-step explanation:

[tex] \frac{1}{5} \times \frac{3}{10} [/tex]

➡️ [tex] \frac{1 \times 3}{5 \times 10} [/tex]

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Answer:

x = -2, y = 2

Step-by-step explanation:

-7x + 7y = 28

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y = 4 + x ----(1)

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sub (1) into (2)

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Answer

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Step by step explanation

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x=4

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[tex]\\ \rm\longmapsto 23458\times \dfrac{7}{100}[/tex]

[tex]\\ \rm\longmapsto \dfrac{164206}{100}[/tex]

[tex]\\ \rm\longmapsto 1642.06[/tex]

Vito Avido's commission on $23,458 in sales is $1466.06.

Here, we have to calculate Vito Avido's commission on $23,458 in sales, we'll break down the sales amount into different ranges and apply the respective commission rates for each range.

For the first $5,000 of sales: 5% commission

Commission = 5% of $5,000 = 0.05 * $5,000 = $250

For the next $15,000 of sales: 6.5% commission

Commission = 6.5% of $15,000 = 0.065 * $15,000 = $975

For the remaining amount over $20,000: 7% commission

Sales above $20,000 = $23,458 - $20,000 = $3,458

Commission = 7% of $3,458 = 0.07 * $3,458 = $241.06

Now, add up the commissions from each range:

Total Commission = $250 + $975 + $241.06 = $1466.06

Vito Avido's commission on $23,458 in sales is $1466.06.

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(A)

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Answer:

7 bags

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Step-by-step explanation:

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Answer:

z = 40 + 26.00x + 15.00y

Coefficients: 26.00, 15.00

Constant: 40

Variables: z, x, y

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Let x be the number of canoe rentals and y be the number of bicycle rentals.

(The family can hold only one annual pass, so we don't need a variable for that. Rather, we'll use a "constant." But the number of bicycle/canoe rentals can change!)

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Step-by-step explanation:

Let [tex]P_1(-1, -8)\:\text{and}\:P_2(-4, -9)[/tex]

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[tex]-8 = \frac{1}{3}(-1) + b \Rightarrow b = -\dfrac{23}{3}[/tex]

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90

Step-by-step explanation:

Answer:

90

Step-by-step explanation:

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A. 0
B. Nonexistent
C. 1
D. -1

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Answer:

[tex]\displaystyle \lim_{x\rightarrow 0^{+}} \frac{x\ln x}{\tan x}=-\infty\implies \text{B. Nonexistent (best answer)}[/tex]

Step-by-step explanation:

Recall L'Hopital's rule:

[tex]\displaystyle \lim_{x\rightarrow c}\frac{f(x)}{g(x)}=\lim_{x\rightarrow c}\frac{f'(x)}{g'(x)}[/tex]

First derivative of [tex]x\ln x[/tex]:

Recall the product rule:

[tex](f\cdot g)'=f'\cdot g+g'\cdot f[/tex]

[tex]\displaystyle \frac{d}{dx} (x\ln x)=\frac{d}{dx}(x)\cdot \ln (x)+\frac{d}{dx}(\ln x)\cdot x[/tex]

Note that:

[tex]\displaystyle \frac{d}{dx}(x)=1,\\\frac{d}{dx}(\ln (x))=\frac{1}{x}[/tex]

Simplifying, we get:

[tex]\displaystyle \frac{d}{dx} (x\ln x)=1\cdot \ln x+\frac{1}{x}\cdot x,\\\frac{d}{dx}(x\ln x)=\ln x+1[/tex]

First derivative of [tex]\tan x[/tex]:

[tex]\displaystyle \frac{d}{dx}(\tan x)=\sec^2 x[/tex]

Therefore, we have:

[tex]\displaystyle \lim_{x\rightarrow 0^{+}}\frac{x\ln x}{\tan x}=\lim_{x\rightarrow 0^{+}}\frac{\ln x+1}{\sec^2{x}}[/tex]

By definition, [tex]\cos x=\frac{1}{\sec x}[/tex]. Therefore,

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[tex]\displaystyle \lim_{x\rightarrow 0^{+}}\cos^2x=1,\\\lim_{x\rightarrow 0^{+}}\ln x+1=-\infty[/tex]

Substitute:

[tex]\displaystyle \lim_{x \rightarrow0^{+}} \cos^2x(\ln x+1)=1\cdot (-\infty)=-\infty[/tex]

Therefore, we have:

[tex]\displaystyle \lim _{x\rightarrow 0^{+}}\frac{x\ln x}{\tan x}=-\infty \text{}[/tex], which best corresponds with [tex]\boxed{\text{B. Nonexistent}}[/tex]

*Commentary:

Technically speaking, a limit exists only if it is equal to a real number. By proper definition, infinity is not a number. With that being said, you will see limits expressed as infinity or negative infinity.

Here's what I will say about this specific problem.

The problem is stipulating that we approach [tex]x[/tex] from the right side. Because of this condition, it may be unorthodox to say this limit doesn't exist. However, if the problem just asked for [tex]\displaystyle \lim_{x\rightarrow 0}\frac{x\ln x}{\tan x}[/tex], it is common and preferred to say this limit does not exist, since [tex]\displaystyle \lim_{x\rightarrow 0^{-}}\frac{x\ln x}{\tan x}\neq \displaystyle \lim_{x\rightarrow 0^{+}}\frac{x\ln x}{\tan x}[/tex].

For example, [tex]\displaystyle \lim_{x\rightarrow 0}\frac{1}{x}=\text{DNE}[/tex], because [tex]\displaystyle \lim_{x\rightarrow 0}\frac{1}{x}[/tex] diverges. In other words, [tex]\displaystyle \lim_{x\rightarrow 0^{-}}\frac{1}{x}=-\infty \neq \displaystyle \lim_{x\rightarrow 0^{+}}\frac{1}{x}=\infty[/tex].

But again, the problem is asking for the limit as [tex]x[/tex] approaches from the right, in which case [tex]\displaystyle \lim _{x\rightarrow 0^{+}}\frac{x\ln x}{\tan x}=-\infty }[/tex]. It's really a pedagogical choice whether to say a limit equal to infinity or negative infinity exists or not since infinity implies there is no limit, so saying the limit of something is infinity becomes an oxymoron. In this case, the person who wrote the answer choices chose to express a limit of infinity as nonexistent, but it is worth mentioning that someone else solving this problem might express [tex]-\infty[/tex] as the answer, and they would be just as, if not more, correct.

Without resorting to L'Hopital's rule, recall that

[tex]\displaystyle \lim_{x\to0}\frac{\sin(ax)}{ax} = 1[/tex]

for a ≠ 0. Then

[tex]\displaystyle \lim_{x\to0^+} \frac{x \ln(x)}{\tan(x)} = \lim_{x\to0^+}\frac x{\sin(x)} \times \lim_{x\to0^+}\cos(x) \times \lim_{x\to0^+}\ln(x)[/tex]

The first two limits exist and are equal to 1, but the last limit is -∞.

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Answers

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

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Answers

Answer:

[tex]thank \: you[/tex]

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What is an Equation?

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

It demonstrates the equality of the relationship between the expressions printed on the left and right sides.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

Given data ,

Let the equation be represented as n

Now , the value of A is

The number of candies which were eaten = 80 candies

The fraction of candies that were eaten = 2/3

So , ( 2/3 ) of the total = 80

( 2/3 )A = 80

Multiply by 3 on both sides , we get

2A = 240

Divide by 2 on both sides , we get

A = 120 candies

Now , the number of candies that were left n = total number of candies - number of candies that were eaten

On simplifying the equation , we get

The number of candies that were left n = 120 - 80

The number of candies that were left n = 40 candies

Hence , the number of candies that are left is 40 candies

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