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2 years ago

Find the exact area of a circle with a dimeter of 5 yards. Use 3.14 for pi

Mathematics

1 answer:

kenny6666 [7]2 years ago

7 0

Answer:

$A = 6.25(pi)$

$A$ ~ $19.63495$

Step-by-step explanation:

The formula for finding the area of a circle is;

$A = (pi)r^2$

Where ( $A$ ) represents the area of the circle, ( $pi$ ) represents the value ( $3.1415...$ ) and ( $r$ ) represents the radius of the circle. As per its definition, the diameter is the largest cord or line segment that can be drawn through a circle, the diameter always passes through the center of a circle. The radius ( a line segment that can be drawn from the center of the circle to any side of the circle) is always half of the diameter.

This means that the radius for the given circle is ( $2.5$ ), since ( $5$ ÷ $2=2.5$ )

Substitute these values into the equation and solve for the area of the circle.

$A = (pi)r^2\\\\A = (pi)(2.5)^2\\\\A = (pi)6.25\\$

$A$ ~ $19.63495$

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Find the center and radius of this circle: (x+2)^2 + (y-3)^2 =100

Flauer [41]

Answer:

center:-2,3

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Step-by-step explanation:Happy to help :)

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2 years ago

Use the function shown in the table to answer the questions below .

Lubov Fominskaja [6]

Answer:

a) -6

b) 3

c) Y = -6x + 3

Step-by-step explanation:

Pick two points and find rise/run = slope

rise y1 - y2 = 9 - 3 = 6

run x1 - x2 = -1 - 0 = -1

slope = 6 / -1 = -6

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y intercept = 3

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2 years ago

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aev [14]

This looks super duper hard wish I can solve it out

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2 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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3 years ago

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aev [14]

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1 year ago