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lyudmila [28]
3 years ago
5

An experimenter has conducted a single‐factor experiment with four levels of the factor, and each factor level has been replicat

ed six times. The computed value of the F F ‐statistic is F 0 = 3.26 F 0 = 3.26 . Find bounds on the P P ‐value.

Mathematics
1 answer:
djverab [1.8K]3 years ago
6 0

Answer:

Step-by-step explanation:

You can find your answer in attached document.

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5. A carton of ice cream holds 5 cups of ice cream. One serving of ice cream is 1 cups.
Anna35 [415]

Answer:

5 servings

Step-by-step explanation:

1 serving is 1 cup so 5 servings is 5 cups

4 0
3 years ago
Kevin has 5 fish in his fish tank. Jasmine has 4 times as many fish as Kevin does. How many fish does jasmine have?
cupoosta [38]
5×4=20
Jasmine has 20 fish


hope this helps
4 0
3 years ago
Read 2 more answers
Can you please help me ​
Tanzania [10]

Answer:

The order of the number of the answers from top to bottom in the boxes are: 5, 6, 3, 1, 4, 2

7 0
3 years ago
Question 5
sergey [27]

Explanation : If 66 is a factor of this unknown value, then 22 must be as well considering that 22 is a factor off 66. Let's say that this large value is 330. It is a multiple of 66, as 66 * 5 = 330. At the same time 22 * 15 = 330, so 330 is a multiple of 22 as well - or vice versa, 12 is a factor of 330.

We can also tell that 15, 22 fit into 330 through another approach. 22 * 3 = 66, and 66 * 5 = 330, so 5 * 3 = 15 - the same value. This proves that 22 will always be a factor of a value that is the factor of 66.

3 0
3 years ago
Read 2 more answers
This is the same bonus problem as Lesson 12. This time, you have to use the method of Lagrange Multipliers to solve it. A rectan
MariettaO [177]

Answer:

length=10 ft , width = 10 ft and height = 5 ft

Step-by-step explanation:

using Lagrange multipliers , we have the main function that is the Area A of the tank ,

A(x,y,z) = x*y + 2*x*z + 2*y*z

constrained to the Volume V(x,y,z) = x*y*z=a = 500 ft³

using Lagrange multipliers

Ax - λ*Vx = 0 →  (y + 2*z) - y*z*λ = 0 → λ= 1/z + 2/y

Ay - λ*Vy = 0 → (x + 2*z) - x*z*λ = 0 → λ= 1/z + 2/x

Az - λ*Vz = 0 → (2*x + 2*y) - x*y*λ = 0  → λ= 2/x + 2/y

V =a →  x*y*z=a

adding the first and second equations

2*λ= 1/z + 2/y + 1/z + 2/x = 2/z + λ

λ = 2/z → z= 2/λ

therefore

λ= 1/z + 2/y = λ/2 + 2/y

λ/2 = 2/y → y= 4/λ

and similarly x=4/λ

then

x*y*z=a

2/λ*4/λ*4/λ= a

32/λ³ = a

λ = ∛32/a

therefore

z= 2/λ = 2*∛a/32 =   2*∛(500/32) = 5

y= 4/λ = 2*5 = 10

x=10

therefore the dimensions that minimize the area are x=10 , y=10 and z=5 (length=10 ft , width = 10 ft and height = 5 ft)

5 0
3 years ago
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