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Luba_88 [7]
1 year ago
8

A candle manufacturer sells cylindrical candles in sets of three. Each candle in the set is a different size. The smallest candl

e has a radius of 0.5 inches and a height of 3 inches. The other two candles are scaled versions of the smallest, with scale factors of 2 and 3. How much wax is needed to create one set of candles?
Mathematics
2 answers:
mafiozo [28]1 year ago
4 0
<span> it is 27 π cubic inches </span>
Margaret [11]1 year ago
3 0

Answer:

The amount of  wax that is needed to create one set of candles is:

                              27π cubic inches.

Step-by-step explanation:

  • The radius(r) of the smallest candle=0.5 inches.

Height(h) of smallest candle= 3 inches.

      Hence, Volume of smallest candle V_1 is:

                 V_1=\pi r^2h

  • Now the medium candle is one which is obtained by taking a scale factor of 2.

  Hence, radius of medium candle=2r

 Height of medium candle=2h

            Hence, Volume of medium candle V_2 is:

       V_2=\pi\times (2r)^2\times (2h)\\\\V_2=8\pi r^2h

  • Similarly the largest candle is one which is obtained by taking a scale factor of 3.

      Hence, radius of largest candle=3r

      Height of largest candle=3h

        Hence, Volume of largest candle V_3 is:

       V_3=\pi\times (3r)^2\times (3h)\\\\V_3=27\pi r^2h

The amount of wax required to create one set of candle is equal to total volume of all the three candles.

\text{Amount\ of\ wax}=V_1+V_2+V_3\\\\\\\text{Amount\ of\ wax}=\pi r^2h+8 \pi r^2h+27\pi r^2 h\\\\\\\text{Amount\ of\ wax}=36\pi r^2h

Now on putting the value of r and h in the expression we get:

\text{Amount\ of\ wax}=36\times \pi\times (0.5)^2\times 3\\\\\\\text{Amount\ of\ wax}=27\pi\ \text{cubic\ inches}

                  Hence, the answer is:

                       27π cubic inches.

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rusak2 [61]
Given: 1 cm = 2 3/4 ft ; kitchen length = 4 1/2 cm
2 3/4 = 11/4; 4 1/2= 9/2

11/4 x 9/2 = 99/8 =12 3/8

OR You could use decimals

2 3/4 = 2.75 ; 4 1/2 = 4.5
2.75 x 4.5 = 12.375
12.375 = 12 375/1000 or 12 3/8
5 0
2 years ago
Company F sells fabrics known as fat quarters, which are rectangles of fabric created by cutting a yard of fabric into four piec
jeka94

Answer:

a) Y 0 1 2

P(Y) 0.58 0.23 0.11

b) mean= 0.45, S.D= 0.6718

c) mean= 1.285, S.D= 8.74

Step-by-step explanation:

a) The following table shows the probability distribution of X:

X 0 1 2 3 4 or more

P(X) 0.58 0.23 0.11 0.05 0.03

Defect >2 = cannot be sold

Y = the number of defects on a fat quarter that can be sold by Company F.

Y = defect that can be sold

Y = Defect less or equal to 2 = 0,1,2

Probability distribution of the random variable Y:

Y 0 1 2

P(Y) 0.58 0.23 0.11

b) mean of Y (μ)

μ = Σ x*P(Y)

= (0*0.58) +(1*0.23)+(2*0.11)

= 0+0.23+0.22 = 0.45

Standard deviation of Y = σ

σ = Σ√(x-mean)^2*P(Y)

= Σ√[(x- μ )^2*P(Y)]

= √[(0-0.45)^2*0.58+ (1-0.45)^2*0.23 + (2-0.45)^2*0.11]

= √[0.11745 + 0.069575 +0.264275

= √(0.4513

σ = 0.6718

Company G:

σ for defect that be sold = 0.66

μ for defect that be sold = 0.40

Difference between μ of F and μ of G

= 0.45-0.40 = 0.05

Difference between σ of F and σ of G

= 0.67-0.66 = 0.01

Selling price of fat quarter without defect = $5

Discount per defect = $1.5

Selling price per defect = 5-1.5 = $3.5

Discount per 2 defect = $1.5*2 = $3

Selling price per defect = 5-3 = $2

Since defect to be sold cannot be greater than 2, let Y = 5,3,2

Probability distribution of the selling price Y:

Y 5 3 2

P(Y) 0.58 0.23 0.11

μ = (5*0.58) +(3.5*0.23)+(2*0.11)

μ = 2.9+0.805+0.22 =1.285

σ = Σ√[(x- μ )^2*P(Y)]

σ = √[(5-1.285)^2*0.58+ (3-1.285)^2*0.23 + (2-1.285)^2*0.11]

σ = 8.00+0.68+0.06 = 8.74

7 0
2 years ago
(b) we often read that iq scores for large populations are centered at 100. what percent of these 78 students have scores above
zheka24 [161]
Given that the<span> iq scores for large populations are centered at 100.

To get what percent of these 78 students have scores above 100 we conduct a normal distribution probability of the data.

P(x > 100) = P(z > (100 - 100)/sd) = P(z > 0) = 1 - P(z < 0) = 1 - 0.5 = 0.5 = 50%
</span>
5 0
2 years ago
To increase sales, a local donut shop began putting an extra donut in some of the boxes. Customers are unaware of which boxes ha
Bingel [31]

Answer:

Step-by-step explanation:

Hello!

Part A

First, determine your study variable:

X: Number of boxes with an extra donut in a sample of eight.

To see if the variable has a Binomial distribution you have to check if the binomial criteria are met:

1. The number of observation of the trial is fixed (In this case n = 8, the boxes each customer bought make the sample)

2. Each observation in the trial is independent, this means that none of the trials will affect the probability of the next trial (In this case, the amount of donuts in one box does not affect on the probability of the next box having an extra donut)

3. The probability of success in the same from one trial to another (In this case our "success" will be that the box has an extra donut, according to the owners claim that is 1/7; ρ=0,14)

So X≈ Bi (n;ρ)

Where n represents the sample (n=8) and ρ is the probability of success (ρ=0.14)

Part B

The mean of the binomial distribution is E(X)= nρ

E(X)= 8 * 0.14= 1.12

The mean of the distribution is also called the expected value. You'd expect that 1.12 boxes have an extra donut.

The variance of the binomial distribution is V(X)= nρ(1 - ρ)

V(X)= 8*0.14*(1 - 0.14)= 0.9632

Its square root is the standard deviation

√V(X)= 0.98

The standard deviation is a measure of dispersion, it indicates how much the values ​​of the distribution of the central value are separated. In this case, 0.98 indicates that the distribution of the number of boxes with an extra donut is far from the expected value.

Part C

Two of the eight customers buy a box with an extra donut, symbolically:

P(X=2) = P(X≤2) - P(X≤1)= 0.91 - 0.68 = 0.22

There is a 22% chance that two customers bought a box with an extra donut.

Compute:

P(X≥2)= 1 - P(X<2)= 1 - P(X≤1)= 1 - 0.68= 0.32

There is a 32% chance that two or more customers bought a box with an extra donnut.

I hope it helps!

3 0
1 year ago
Bernie helped design a magician costume for a school play. He used 5.75 feet of ribbon for the magician's wand. He used 11.75 fe
Nonamiya [84]

Answer:

The ribbon cost per foot $0.6 per foot

Step-by-step explanation:

Total number of ribbon used = (5.75 + 11.75) = 17.50 feet

17.50 feet of ribbon cost = $10.50

1 foot of ribbon cost = x

Cross Multiply

17.50 × x = 1 foot × $10.50

x = 1 foot × 10.50/17.50

x = $0.6

The ribbon cost per foot $0.6 per foot

8 0
1 year ago
Read 2 more answers
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