7. A magnifying glass shows you an image of an object that is 7 times the object's actual size. An ant is 0.2 cm in length. What
is the length of its image in the magnifying glass?
1 point
1.4 cm
14 cm
.03 cm
3 cm
8. Which isometry is equivalent to two reflections across parallel lines?
1 point
glide reflection
translation
rotation
dilation
9. Find the glide reflection rule that maps Triangle NOP to Triangle N'O'P'.
1 point
Image below
(x, y) --> (x - 1, y + 4) and reflect across x = -1
(x, y) --> (x + 4, y -1) and reflect across x = -1
(x, y) --> (x - 1, y + 4) and reflect across y = -1
(x, y) --> (x + 4, y - 1) and reflect across y = -1
10. Which transformations result in a preimage and image with opposite orientations?
1 point
Dilations only
Translations and rotations
Rotations and reflections
Reflections and glide reflections
1 point
1.4 cm
14 cm
.03 cm
3 cm
8. Which isometry is equivalent to two reflections across parallel lines?
1 point
glide reflection
translation
rotation
dilation
9. Find the glide reflection rule that maps Triangle NOP to Triangle N'O'P'.
1 point
Image below
(x, y) --> (x - 1, y + 4) and reflect across x = -1
(x, y) --> (x + 4, y -1) and reflect across x = -1
(x, y) --> (x - 1, y + 4) and reflect across y = -1
(x, y) --> (x + 4, y - 1) and reflect across y = -1
10. Which transformations result in a preimage and image with opposite orientations?
1 point
Dilations only
Translations and rotations
Rotations and reflections
Reflections and glide reflections
1 answer:
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Answer:
(a) P(X 20) = 0.9319
(b) Expected number of defective light bulbs = 15
(c) Standard deviation of defective light bulbs = 3.67
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.
Standard deviation of defective light bulbs is given by = S.D. = =
= 3.67
