Han and Tyler both started walking toward each other from opposite ends of a 18-mile path. Han walks at a pace of 4 miles per ho
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A semicircle is a part of a circle, and it is referred to as half of a given circle. Thud the area of the <em>semi-circle</em> window is 982 .
A circle is a shape that is <u>bounded</u> by a <em>curved</em> path which is referred to as the <em>circumference</em>. Some <u>parts</u> of a circle are radius, diameter, sector, arc, semi-circle, circumference, etc.
A <em>semicircle</em> is a <u>part</u> of a <u>circle</u>, and it is referred to as <em>half </em>of a given <em>circle</em>.
such that:
<em>Area</em> of a <u>circle</u> =
and
area of a <u>semicircle</u> =
where: r is the <u>radius </u>of the <u>circle</u>, and is a <u>constant </u>with a value of
.
Thus from the given question, it can be inferred that;
r =
= 25
r = 25 in
Thus, the area of the<em> semi-circle</em> can be determined as;
area of the <em>semi-circle</em> = *
*
= 982.1429
area of the semi-circle = 982.14
The area of the <em>semi-circle</em> window is approximately 982 .
for more clarifications on the area of a semi-circle, visit: brainly.com/question/15937849
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Answer:
(a) P(X 20) = 0.9319
(b) Expected number of defective light bulbs = 15
Step-by-step explanation:
We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.
Firstly, the above situation can be represented through binomial distribution, i.e.;
where, n = number of samples taken = 150
r = number of success
p = probability of success which in our question is % of bulbs that
are defective, i.e. 10%
<em>Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.</em>
So, Let X = No. of defective bulbs in a box
<u>Mean of X</u>, =
=
= 15
<u>Standard deviation of X</u>, =
=
= 3.7
So, X ~ N(
Now, the z score probability distribution is given by;
Z = ~ N(0,1)
(a) Probability that this box will contain at most 20 defective light bulbs is given by = P(X 20) = P(X < 20.5) ---- using continuity correction
P(X < 20.5) = P( <
) = P(Z < 1.49) = 0.9319
(b) Expected number of defective light bulbs found in such boxes, on average is given by = E(X) = =
= 15.