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3 years ago

I = Prt (solve for P)

Mathematics

2 answers:

bearhunter [10]3 years ago

6 0

Answer:

$p=\frac{i}{rt}$

Step-by-step explanation:

Phantasy [73]3 years ago

5 0

Answer:

Step-by-step explanation:

Divide both sides by rt

$P=\frac{I}{rt}$

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Write a polynomial function of minimum degree in standard form with real coefficients whose zeros and their multiplicities inclu

Softa [21]

Answer-

The polynomial function is,

$y=x^4-10x^3+38x^2-64x+40$

Solution-

The zeros of the polynomial are 2 and (3+i). Root 2 has multiplicity of 2 and (3+i) has multiplicity of 1

The general form of the equation will be,

$\Rightarrow y=(x-(2))^2(x-(3+i))(x-(3-i))$ ( ∵ (3-i) is the conjugate of (3+i) )

$\Rightarrow y=(x-2)^2(x-3-i)(x-3+i)$

$\Rightarrow y=(x^2-4x+4)((x-3)-i)((x-3)+i)$

$\Rightarrow y=(x^2-4x+4)((x-3)^2-i^2)$

$\Rightarrow y=(x^2-4x+4)((x^2-6x+9)+1)$

$\Rightarrow y=(x^2-4x+4)(x^2-6x+10)$

$\Rightarrow y=x^2x^2-6x^2x+10x^2-4x^2x+4\cdot \:6xx-4\cdot \:10x+4x^2-4\cdot \:6x+4\cdot \:10$

$\Rightarrow y=x^4-10x^3+14x^2+24x^2-40x-24x+40$

$\Rightarrow y=x^4-10x^3+38x^2-64x+40$

Therefore, this is the required polynomial function.

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,

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