What is the probability that the spinner lands on an even number or on the unshaded section? one-fifth two-fifths three-fifths f
1 answer:
The probability that the spinner lands on an even number or on the unshaded section is; 3/5
<h3>How to find the probability?</h3>
From the spinner attached, we see that the number of sections on the spinner is 5 sections.
Now, we can also see that;
Number of even numbers section = 2
Number of shaded sections = 1
Thus, probability that the spinner lands on an even number or on the unshaded section is;
P(even number or unshaded section) = (2 + 1)/5 = 3/5
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Answer:
The exact value of its surface area = 144π m²
The exact value of its volume = 288π m³
Step-by-step explanation:
∵ The diameter of the sphere is 12m
∴ The radius of the sphere = 12 ÷ 2 = 6m
∵ The surface area of the sphere = 4πr²
∴ The surface area = 4 × π × 6² = 144π m²
∵ The volume of the sphere = 4/3 πr³
∴ The volume = = 288π m³