*see attachement for the triangles given
Answer:
PQ = 6.8
Step-by-step explanation:
Corresponding sides of similar shapes or figures are proportional to each other. Since ∆LMN ~ ∆OPQ, the ratio of their corresponding sides would therefore be equal. Thus:
LN/OQ = MN/PQ
LN = 5
OQ = 17
MN = 2
PQ = ?
Plug in the values
5/17 = 2/PQ
Cross multiply
5*PQ = 2*17
5*PQ = 34
PQ = 34/5
PQ = 6.8
Answer:
Dimensions to minimize surface are is 28 ft x 28 ft x 14 ft
Step-by-step explanation:
The Volume of a box with a square base of say;x cm by x cm and height
h cm is;
V = x²h
Now, the amount of material used is directly proportional to the surface area, hence we will minimize the amount of material by minimizing the surface area.
The formula for the surface area of the box described is given by;
A = x² + 4xh
However, we need A as a function of
only x, so we'll use the formula;
V = x²h
V = x²h = 10,976 ft³
So,
h = 10976/x²
So,
A = x² + 4x(10976/x²)
A = x² + 43904/x
So, to minimize the area, it will be at dA/dx = 0.
So,
dA/dx = 2x - 43904/x² = 0
Factorizing out, we have;
2x³ = 43904
x³ = 43904/2
x³ = 21952
x = ∛21952
x = 28 ft
since, h = 10976/x²
h = 10976/28² = 14 ft
Thus,dimension to minimize surface are is 28 ft x 28 ft x 14 ft
Answer:
and 
Step-by-step explanation:
Given


From the first statement, we have that:

Convert mm to cm


Divide both sides by 0.1

From the second statement, we have that:

Substitute 0.2 + t + 0.5h for Cost

Collect Like Terms:


So, the inequalities are:
and 
Answer:

And the standard error is given by:

And replacing we got:
![SE_[p]= \sqrt{\frac{0.06*(1-0.06)}{373}}= 0.0123](https://tex.z-dn.net/?f=%20SE_%5Bp%5D%3D%20%5Csqrt%7B%5Cfrac%7B0.06%2A%281-0.06%29%7D%7B373%7D%7D%3D%200.0123)
And we want to find this probability:

We can calculate the z score for this case and we got:

And using the normal distribution table or excel we got:

Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
Solution to the problem
For this case we can find the mean and standard error for the sample proportion with these formulas:

And the standard error is given by:

And replacing we got:
![SE_[p]= \sqrt{\frac{0.06*(1-0.06)}{373}}= 0.0123](https://tex.z-dn.net/?f=%20SE_%5Bp%5D%3D%20%5Csqrt%7B%5Cfrac%7B0.06%2A%281-0.06%29%7D%7B373%7D%7D%3D%200.0123)
And we want to find this probability:

We can calculate the z score for this case and we got:

And using the normal distribution table or excel we got:

Is there any other information?