Dilation and reflection
-multiplication of - =reflection
example -(x)
-multiplication of any number other than 1=dilation
example 3x
-addition/subtraction inside the function=horizontal translation
Example (x-3)
-addition/subtraction outside the function=vertical translation
Example x-3
Answer:
Volume of room will be left after Michael stores the boxes 426 feet cube
Step-by-step explanation:
It is given that
Michael is putting 12 boxes in a storage shed.
Each box is in the shape of a rectangular prism with dimensions 1.5 feet by 1.5 feet by 2 feet.
If the storage shed has dimensions 6 feet by 10 feet by 8 feet,
<u>To find the volume of one box
</u>
Volume = 1.5 x 1.5 x 2= 4.5 feet cube
Volume of 12 such box = 4.5 x 12 =54 feet cube
<u>To find the volume of shed
</u>
Volume of Shed = 6 x 10 x 8 =480
<u>
To find remaining portion of shed</u>
Volume of room will be left after Michael stores the boxes =volume of shed - volume of box
=480 - 54 = 426 feet cube
Answer:
The pre-image of the quadrilateral is attached.
Step-by-step explanation:
Each point (x, y) is rotated with the rule (x, y) -> (-x, -y).
So, point W'(2, 3) was W(-2, -3) before rotation.
X'(6, 3) was X(-6, -3)
Y'(6, 4) was Y(-6, -4).
Z'(3, 5) was Z(-3, -5).
The pre-image of the quadrilateral is attached.
Answer:
a)0.099834
b) 0
Step-by-step explanation:
To solve for this question we would be using , z.score formula.
The formula for calculating a z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.
A candy maker produces mints that have a label weight of 20.4 grams. Assume that the distribution of the weights of these mints is normal with mean 21.37 and variance 0.16.
a) Find the probability that the weight of a single mint selected at random from the production line is less than 20.857 grams.
Standard Deviation = √variance
= √0.16 = 0.4
Standard deviation = 0.4
Mean = 21.37
x = 20.857
z = (x-μ)/σ
z = 20.857 - 21.37/0.4
z = -1.2825
P-value from Z-Table:
P(x<20.857) = 0.099834
b) During a shift, a sample of 100 mints is selected at random and weighed. Approximate the probability that in the selected sample there are at most 5 mints that weigh less than 20.857 grams.
z score formula used = (x-μ)/σ/√n
x = 20.857
Standard deviation = 0.4
Mean = 21.37
n = 100
z = 20.857 - 21.37/0.4/√100
= 20.857 - 21.37/ 0.4/10
= 20.857 - 21.37/ 0.04
= -12.825
P-value from Z-Table:
P(x<20.857) = 0
c) Find the approximate probability that the sample mean of the 100 mints selected is greater than 21.31 and less than 21.39.
Answer:
12
Step-by-step explanation:
Im going to divide this up to make it easier
first im going to do the triangle on top which we need the formula for that which would be
(4*2) / 2 which is( base * height )/2 then 4*2 = 8 /2 = 4
then ill deal with the rectangle just below that triangle
which is 3*2 which is 6
then last is triangle which is 2*2 which is 4 / 2 so 2
then add all together
4 + 6 + 2 = 12
