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3 years ago

63 is 90% of what number

Mathematics

2 answers:

otez555 [7]3 years ago

8 0

The answer is 70.

You can just divide 63 by 0.9.

Hope this helped. :)

Mariana [72]3 years ago

7 0

X*90/100=63
90x=6300
x=6300/90
x=70
90% of 70 is 63

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QUESTION 3 [10 MARKS] A bakery finds that the price they can sell cakes is given by the function p = 580 − 10x where x is the nu

Harman [31]

Answer:

(a)Revenue function, $R(x)=580x-x^2$

Marginal Revenue function, R'(x)=580-2x

(b)Fixed cost =900 .

Marginal Cost Function=300+50x

(c)Profit, $P(x)=-35x^2+280x-900$

(d)x=4

Step-by-step explanation:

Part A

Price Function $= 580 - 10x$

The revenue function

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

The marginal revenue function

$\dfrac{dR}{dx}= \dfrac{d}{dx}(R(x))=\dfrac{d}{dx}(580x-x^2)=580-2x\\R'(x)=580-2x$

Part B

(Fixed Cost)

The total cost function of the company is given by $c=(30+5x)^2$

We expand the expression

$(30+5x)^2=(30+5x)(30+5x)=900+300x+25x^2$

Therefore, the fixed cost is 900 .

Marginal Cost Function

If $c=900+300x+25x^2$

Marginal Cost Function, $\frac{dc}{dx}= (900+300x+25x^2)'=300+50x$

Part C

Profit Function

Profit=Revenue -Total cost

$580x-10x^2-(900+300x+25x^2)\\580x-10x^2-900-300x-25x^2\\$Profit,P(x)=-35x^2+280x-900$

Part D

To maximize profit, we find the derivative of the profit function, equate it to zero and solve for x.

$P(x)=-35x^2+280x-900\\P'(x)=-70x+280\\-70x+280=0\\-70x=-280\\$Divide both sides by -70\\x=4$

The number of cakes that maximizes profit is 4.

6 0

3 years ago

A jar of peanut butter contains 454 g with a standard deviation of 10.2 g. Find the probability that a jar contains more than 46

Fofino [41]

Answer:

The probability that a jar contains more than 466 g is 0.119.

Step-by-step explanation:

We are given that a jar of peanut butter contains a mean of 454 g with a standard deviation of 10.2 g.

Let X = Amount of peanut butter in a jar

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 454 g

$\sigma$ = standard deviation = 10.2 g

So, X ~ Normal( $\mu=454 , \sigma^{2} = 10.2^{2}$ )

Now, the probability that a jar contains more than 466 g is given by = P(X > 466 g)

P(X > 466 g) = P( $\frac{X-\mu}{\sigma}$ > $\frac{466-454}{10.2}$ ) = P(Z > 1.18) = 1 - P(Z $\leq$ 1.18)

= 1 - 0.881 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

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just olya [345]

Answer: 31/35

Step-by-step explanation:

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4 years ago

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kobusy [5.1K]

Y = 4x - 1

// Plug this in for variable y in equation [1]

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3 years ago

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geniusboy [140]

Answer:

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3 years ago