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2 years ago

10

What is the area of a circle with a radius of 4 inches?

2 answers:

Ostrovityanka [42]2 years ago

8 0

Question: what is the area of a circle with a radius of 4 inches
Answer: A≈50.27in²

klasskru [66]2 years ago

7 0

The area would be 50.27 inches

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the figure is made of up of a cone and a cylinder, to the nearest whole number what is the approximate volume of this figure. PL

Deffense [45]

The total volume is the sum of volume of cone and volume of cylinder.

Volume of cylinder = πr²h
π = 3.14
r = 4/2 = 2 mm
h = 2 mm

So, volume of cylinder equals:

$3.14 *(2)^{2}*2 =25.12$ mm³

Volume of cone = $\frac{1}{3} \pi r^{2}h$
π = 3.14
r = 2 mm
h = 3 mm

So, volume of cylinder equals:

$\frac{1}{3}*3.14* (2)^{2}*3=18.84$ mm³

Thus, total volume of the figure = 25.12 + 18.84 = 43.96 mm³

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3 years ago

What are the factors of 36?

Rashid [163]

1 2 3 4 6 9 12 18 36

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3 years ago

Tony bought a video game for $15.13. He gave the clerk a $20 bill. Which of the following could be the change he received?

Tanya [424]

Answer:

$4.87

Step-by-step explanation:

20 - 15.13 = 4.87

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2 years ago

Help!!i will give BRAINLIEST to whoever answers

Gnom [1K]

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3 years ago

A city hosted a music festival that included three concerts. According to the sales database, 28% of the audience attended the f

Vitek1552 [10]

Answer:

$P(X\cap Y\cap Z)=0.05$

Step-by-step explanation:

From the question we are told that:

Percentage of audience in first concert $P(X)=0.28$

Percentage of audience in second concert $P(Y)=0.42$

Percentage of audience in third concert $P(Z)=0.30$

Audience Percentage at at-least one concert $P(X \cup Y \cup Z)=0.80$

Percentage of audience at first & second concert $P(X \cap Y)=0.10$

Percentage of audience in first & third concert $P(X \cap Z)=0.08$

Percentage of audience in second & third concert $P(Y\cap Z)=0.07$

Generally the equation for probability of attending all concerts $P(X\cap Y\cap Z)$ is mathematically given by

$P(X \cup Y \cup Z)=P(X)+P(Y)+P(Z)-P(X \cap Y)-P(X \cap Z)-P(Y\cap Z)+P(X\cap Y\cap Z)$

$P(X\cap Y\cap Z)=P(X \cup Y \cup Z)-P(X)-P(Y)-P(Z)+P(X \cap Y)+P(X \cap Z)+P(Y\cap Z)$

$P(X\cap Y\cap Z)=0.80-0.28-0.42-0.30+0.10+0.80+0.70$

$P(X\cap Y\cap Z)=0.05$

Therefore the probability that a randomly selected audient attended all the concerts

$P(X\cap Y\cap Z)=0.05$

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3 years ago