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10 months ago

10

How is a circumference related to an arc

1 answer:

koban [17]10 months ago

6 0

Answer:

An arc of a circumference or of a circle mostly is a portion of the circumference. The length of an arc for all intents and purposes is simply the length of this portion of the circumference in a definitely major way. The circumference itself can particularly be considered an arc that goes around the circle in a fairly major way.

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A circle has an area of 144 cm2.Find the circumference

lara [203]

Circumference of Circle = 2πrArea of circle = πr²144 = 3.14 * r²r² = 144/3.14r² = 45.85r = √45.85r = 6.77 Circumference = 2πr = 2*3.14*6.77Circumference = 42.51 m²
So, your answer is 42.51 m²

5 0

2 years ago

In baseball, Shintaro had 4 hits in his last 12 times at bat.

swat32

Answer:

The probability of shintaro is 4/12 simplified which is 1/3

Step-by-step explanation:

It's like saying.. Someone had won 4 races every 12 race

You divide both side of 4/12 till you can't divide anymore

4/12

3/9

2/6

1/3

7 0

2 years ago

Find 1/4+13/20<br> . Write your answer as a fraction in simplest form.

jek_recluse [69]

The LCD here is 20. Thus, 1/4 = 5/20. The sum of 5/20 and 13/20 is 18/20, which can be reduced to 9/10.

4 0

2 years ago

Estimate the length. Then use cubes to measure

NemiM [27]

Answer:

10 cubes

Step-by-step explanation:

7 0

2 years ago

A new security system needs to be evaluated in the airport. The probability of a person being a security hazard is 4%. At the ch

ivolga24 [154]

Answer:

(a) 0.9412

(b) 0.9996 ≈ 1

Step-by-step explanation:

Denote the events a follows:

$P$ = a person passes the security system

$H$ = a person is a security hazard

Given:

$P (H) = 0.04,\ P(P^{c}|H^{c})=0.02\ and\ P(P|H)=0.01$

Then,

$P(H^{c})=1-P(H)=1-0.04=0.96\\P(P|H^{c})=1-P(P|H)=1-0.02=0.98\\$

(a)

Compute the probability that a person passes the security system using the total probability rule as follows:

The total probability rule states that: $P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})$

The value of P (P) is:

$P(P)=P(P|H)P(H)+P(P|H^{c})P(H^{c})\\=(0.01\times0.04)+(0.98\times0.96)\\=0.9412$

Thus, the probability that a person passes the security system is 0.9412.

(b)

Compute the probability that a person who passes through the system is without any security problems as follows:

$P(H^{c}|P)=\frac{P(P|H^{c})P(H^{c})}{P(P)} \\=\frac{0.98\times0.96}{0.9412} \\=0.9996\\\approx1$

Thus, the probability that a person who passes through the system is without any security problems is approximately 1.

7 0

2 years ago