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spayn [35]
2 years ago
13

The diameter of a circle is 6 centimeters. What is the circle's area? Use 3.14 for pi

Mathematics
2 answers:
Lynna [10]2 years ago
7 0

Step-by-step explanation:

diameter=

2 \times\pi \times r = 6 \\ r =  \frac{6}{2 \times 3.14}  = 0.955 \\ area = \pi \times  {r}^{2}  = 3.14 \times  {0.955}^{2}  \\  =  2.87\:  {cm}^{2}

ELEN [110]2 years ago
5 0

Answer:

Diameter =>6cm

Radius =>6cm/2 =3cm

To find =>The area

Area =>πr²

Step-by-step explanation:

area =  > 3.14 \times 3 \times 3 \\ are a =  > 28.26cm {}^{2}

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Over the last week,the daily low temperatures in degrees Fahrenheit have been -4,6.2,18 1/2,-5.9,21,-1/4,1.75.List these numbers
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So L x W=A so I have a problem where one side is 9cm on the other sides there's nothing​
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Frank chooses a box at random. If boxes 1 and 3 are chosen, Frank extracts 2 balls without replacement. If box 2 is chosen Frank
mars1129 [50]

Complete question:

Consider 3 boxes, each of which contains 4 balls in particular, box 1 contains 4 white balls, box 2 contains 3 white balls and 1 red ball, box 3 contains 2 white balls and 2 red balls. Frank chooses a box at random. If boxes 1 and 3 are chosen, Frank extracts 2 balls without replacement. If box 2 is chosen Frank extracts 2 balls with replacement. USE ONLY THE CONDITIONAL PROBABILITY FORMALISM and derive an expression for the probability that the 2 extracted balls are red.

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11/288

Step-by-step explanation:

We are given:

Box 1: ( 4White, ORed)

Box 2: (3White, 1Red)

Box 3: (2White, 2 Red)

We are told that if Frank choose from Box 1 and Box 3, 2 balls are extracted without replacement.

Since there is no red ball in Box 1, there is no way 2 red balls will come out from Box1.

Our Event, E = getting 2 red balls.

Now Box 1 is ruled out, we have:

P[E(B1)]= 0

P[E/B3)] = (2 2) / (4 2)

= 1/6

If box 2 is chosen, 2 balls are extracted with replacement. Therefore for Box 2:

P(E/B2) = (1/4) *(1/4)

= 1/16

Therefore probability that 2 balls extracted are red, we have:

P(E)=P(E/B1) P(B1) + P(E/B2) P(B2)+P(E/B3) P(B3)

= 0 * \frac{1}{3} + \frac{1}{6}*\frac{1}{3} + \frac{1}{16}*\frac{1}{3}

= 11/288

3 0
3 years ago
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