given:
bull's eye radius= x
width of surrounding rings=y

solution:
Radius of the circle=x+4y
Area of the outermost ring=Area of the circle-Area of the penultimate ring =Ď€(x+4y)^2-Ď€(x+3y)^2
=Ď€(x^2+8xy+16y^2-x^2-9y^2-6xy)
=Ď€(2xy+7y^2)
hence the area of the outermost ring in terms of x and y is Ď€(2xy+7y^2).

4 0

3 years ago

A small business owner estimates his mean daily profit as $970 with a standard deviation of $129. His shop is open 102 days a ye

Katena32 [7]

Answer:

The probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we select appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.

Then, the mean of the distribution of the sum of values of X is given by,

$\mu_{x}=n\mu$

And the standard deviation of the distribution of the sum of values of X is given by,

$\sigma_{x}=\sqrt{n}\sigma$

The information provided is:

μ = $970

σ = $129

n = 102

Since the sample size is quite large, i.e. n = 102 > 30, the Central Limit Theorem can be used to approximate the distribution of the shopkeeper's annual profit.

Then,

$\sum X\sim N(\mu_{x}=98940,\ \sigma_{x}=1302.84)$

Compute the probability that the shopkeeper's annual profit will not exceed $100,000 as follows:

$P (\sum X \leq 100,000) =P(\frac{\sum X-\mu_{x}}{\sigma_{x}}$

$=P(Z$

*Use a z-table for the probability.

Thus, the probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

6 0

3 years ago

Which is the simplified form of the expression ((p squared) (q Superscript 5 Baseline)) Superscript negative 4 Baseline times ((

Alexus [3.1K]

Answer:

$q^{-30}$

Step-by-step explanation:

Given the expression:

$((p^2) (q ^5)) ^{-4} X ((p ^{-4}) (q^5)) ^{-2}$

First, we open the outer brackets

$=(p^{2X-4}) (q ^{5X-4})X (p ^{-4X-2}) (q^{5X-2})\\\\=(p^{-8}) (q ^{-20})X (p ^{8}) (q^{-10})\\\\$Next, we can collect like terms\\\\=p^{-8}Xp ^{8}Xq ^{-20}Xq ^{-10}\\\\$Applying addition law of indices\\\\=p^{-8+8}Xq^{-20+(-10)}\\\\=p^0 X q^{-30}\\\\=1 X q^{-30}\\\\=q^{-30}$

Therefore:

$((p^2) (q ^5)) ^{-4} X ((p ^{-4}) (q^5)) ^{-2}=q^{-30$

8 0

4 years ago

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