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3 years ago

Solve for A if is 3.14 and R = 3

Mathematics

1 answer:

gavmur [86]3 years ago

5 0

Answer:

A = 28.26

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Step-by-step explanation:

Step 1: Define

Equation: A = πr²

Given: π = 3.14, r = 3

Step 2: Evaluate

Substitute: A = 3.14(3)²
Exponents: A = 3.14(9)
Multiply: A = 28.26

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3 years ago

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Solve y = -7(-13) I'm giving 30 points!

VARVARA [1.3K]

y = -7(-13)

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3 years ago

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The physical plant at the main campus of a large state university recieves daily requests to replace fluorescent lightbulbs. The

faltersainse [42]

Answer:

34%

Step-by-step explanation:

Given that the distribution of daily light bulb request replacement is approximately bell shaped with ;

Mean , μ = 45 ; standard deviation, σ = 3

Using the empirical formula where ;

68% of the distribution is within 1 standard deviation from the mean ;

95% of the distribution is within 2 standard deviation from the mean

Lightbulb replacement numbering between ;

42 and 45

Number of standard deviations from the mean /

Z = (x - μ) / σ

(x - μ) / σ < Z < (x - μ) / σ

(42 - 45) / 3 = -1

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Hence, the approximate percentage is : 68% / 2 = 34%

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3 years ago

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

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