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Afina-wow [57]
2 years ago
10

There are 3 rows of boxes on a pallet and 24 boxes on each row. How many boxes are there altogether?

Mathematics
2 answers:
fenix001 [56]2 years ago
4 0
Simple multiplication so 3 x 24 = 72 so there are 72 boxes all together.
sergij07 [2.7K]2 years ago
4 0
If there are 3 rows of boxes, and 24 on each, we then have 24 boxes, three times. This is then 24 + 24 + 24, or 72.
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1. 10 in / 3 ft

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Step-by-step explanation:

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16) 2 is what percent of 20
KIM [24]

Answer:

2 is 10% of 20.

Step-by-step explanation:

2/20=10.

7 0
2 years ago
10 POINTS!!! PLEASE ANSWER SOON! THANK YOU!
Aloiza [94]

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5 0
3 years ago
A dairy company gets milk from two dairies and then blends the milk to get the desired amount of butterfat. Milk from dairy I co
zubka84 [21]

Answer:

a) i The company should buy 40 gallons from dairy I and 60 gallons from dairy

ii) What is the maximum amount of​ butterfat? The total amount of butterfat from Diary I and Diary II = 3.12% + 1.93%

=5.05%

b.The excess capacity of dairy I is 10 ​gallons, and for dairy II it is 30 gallons.

Step-by-step explanation:

a. How much milk from each supplier should the company buy to get at most 100 gallons of milk with the maximum amount of​ butterfat?

From the question, we are told that:

Milk from dairy I costs ​$2.40 per​ gallon, Milk from dairy II costs ​$0.80 per gallon.

Let's represent:

Number of gallons of Milk from dairy I = x

Number of gallons of Milk from dairy II = y

At most ​$144 is available for purchasing milk.

$2.40 × x + $0.80 × y = 144

2.40x + 0.80y = 144........ Equation 1

x + y = 100....... Equation 2

x = 100 - y

2.40(100 - y) + 0.80y = 144

240 - 2.4y + 0.80y = 144

-1.60y = 144 - 240

-1.6y = -96

y = -96/-1.6

y = 60

From Equation 2

x + y = 100....... Equation 2

x + 60 = 100

x = 100 - 60

x = 40

Therefore, since number of gallons of Milk from dairy I = x and number of gallons of Milk from dairy II = y

The company should buy 40 gallons from dairy I and 60 gallons from dairy

II. What is the maximum amount of​ butterfat?

From the question

Dairy I can supply at most 50 gallons averaging 3.9​% ​butterfat,

50 gallons = 3.9% butterfat

40 gallons =

Cross Multiply

= 40 × 3.9/50

= 3.12%

Dairy II can supply at most 90 gallons averaging 2.9​% butterfat.

90 gallons of milk = 2.9% butter fat

60 gallons of milk =

Cross Multiply

= 60 × 2.9%/90

=1.9333333333%

≈ 1.93%

The total amount of butterfat from Diary I and Diary II = 3.12% + 1.93%

=5.05%

b. The solution from part a leaves both dairy I and dairy II with excess capacity. Calculate the amount of additional milk each dairy could produce.

Excess capacity of Diary I =

50 gallons - 40 gallons = 10 gallons

Excess capacity of Diary II =

90 gallons - 60 gallons = 30 gallons

Therefore, the excess capacity of dairy I is 10 ​gallons, and for dairy II it is 30 gallons.

3 0
2 years ago
If X and Y are independent continuous positive random
Leni [432]

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F_Z(z)=P(Z\le z)=P(X\le Yz)=\displaystyle\int_{\mathrm{supp}(Y)}P(X\le yz\mid Y=y)P(Y=y)\,\mathrm dy

F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}P(X\le yz)P(Y=y)\,\mathrm dy

where the last equality follows from independence of X,Y. In terms of the distribution and density functions of X,Y, this is

F_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy

Then the density is obtained by differentiating with respect to z,

f_Z(z)=\displaystyle\frac{\mathrm d}{\mathrm dz}\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy=\int_{\mathrm{supp}(Y)}yf_X(yz)f_Y(y)\,\mathrm dy

b) Z=XY can be computed in the same way; it has CDF

F_Z(z)=P\left(X\le\dfrac zY\right)=\displaystyle\int_{\mathrm{supp}(Y)}P\left(X\le\frac zy\right)P(Y=y)\,\mathrm dy

F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}F_X\left(\frac zy\right)f_Y(y)\,\mathrm dy

Differentiating gives the associated PDF,

f_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}\frac1yf_X\left(\frac zy\right)f_Y(y)\,\mathrm dy

Assuming X\sim\mathrm{Exp}(\lambda_x) and Y\sim\mathrm{Exp}(\lambda_y), we have

f_{Z=\frac XY}(z)=\displaystyle\int_0^\infty y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy

\implies f_{Z=\frac XY}(z)=\begin{cases}\frac{\lambda_x\lambda_y}{(\lambda_xz+\lambda_y)^2}&\text{for }z\ge0\\0&\text{otherwise}\end{cases}

and

f_{Z=XY}(z)=\displaystyle\int_0^\infty\frac1y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy

\implies f_{Z=XY}(z)=\lambda_x\lambda_y\displaystyle\int_0^\infty\frac{e^{-\lambda_x\frac zy-\lambda_yy}}y\,\mathrm dy

I wouldn't worry about evaluating this integral any further unless you know about the Bessel functions.

6 0
3 years ago
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