Answer:
Fraction of the garden used = 
Step-by-step explanation:
The fifth graders to plant a butterfly garden.
We are given that
Clover grows in ½ of the garden
Daisies grow in ¼ of the garden
Coneflowers grow in ⅛ of the garden
milkweed grows in ⅛ of the garden
The fifth graders notice that the butterflies land on the clover and milkweed.
We are asked to find out the fraction of the garden that the butterflies used.
Fraction of the garden used = clover + milkweed
Fraction of the garden used = 
The LCM of 2 and 8 is 8
Fraction of the garden used = 
Fraction of the garden used = 
Therefore, the butterflies used
fraction of the garden.
I have attached a screenshot with the images of the two popcorn bags used by the theater.
Part (a):Bag A:Volume of bag = area of base * height
Volume of bag = length * width * height
We have:
volume = 96 in³
length = 3 in
width = 4 in
Therefore:
96 = 3 * 4 * height
96 = 12 * height
height of bag A = 8 inBag B:Volume of bag = area of base * height
Volume of bag = length * width * height
We have:
volume = 96 in³
length = 4 in
width = 4 in
Therefore:
96 = 4 * 4 * height
96 = 16 * height
height of bag B = 6 inPart (b):To determine the amount of paper needed, we will need to calculate the surface of each bag. Excluding the top base, each bag will have 5 faces. Four side faces (each two opposite are equal) and the base.
Bag A:Surface area = area of base + 2*area of front face + 2*area of side face
Surface area = (3*4) + 2(8*4) + 2(8*3)
Surface area = 12 + 64 + 48
Surface area of bag A = 124 in²Bag B:Surface area = area of base + 2*area of front face + 2*area of side face
Surface area = (4*4) + 2(6*4) + 2(6*4)
Surface area = 16 + 48 + 48
Surface area of bag B = 112 in²From the above calculations, we can deduce that
the theater should choose bag B in order to reduce the amount of paper needed.
Hope this helps :)
Answer:
The company should take a sample of 148 boxes.
Step-by-step explanation:
Hello!
The cable TV company whats to know what sample size to take to estimate the proportion/percentage of cable boxes in use during an evening hour.
They estimated a "pilot" proportion of p'=0.20
And using a 90% confidence level the CI should have a margin of error of 2% (0.02).
The CI for the population proportion is made using an approximation of the standard normal distribution, and its structure is "point estimation" ± "margin of error"
[p' ±
]
Where
p' is the sample proportion/point estimator of the population proportion
is the margin of error (d) of the confidence interval.

So






n= 147.28 ≅ 148 boxes.
I hope it helps!
3/5= 18/30, 2/6= 10/30, so Tom jogged 8/30 more of a mile on Monday then tuesday