The simplified ratio that correctly compares 27 feet to 36 feet is 3:4
Step-by-step explanation:
Step 1: Simplify the ratio by finding common factor
Step 2: Here, in this case the lowest common factor is 3. Divide both numerator and denominator by 3.
⇒ 27/36 = 9/12 (on dividing by 3)
⇒ 9/12 = 3/4 (again on dividing by 3)
There are 100 squares in the grid.
63 are shaded.
Number of squares shaded is the numerator (number on top of the decimal).
Total number of squares is the denominator (number at the bottom of the decimal).

Whenever the denominator of the fraction is 100, you can just take the numerator and put a decimal behind the number.
Therefore, the decimal represented would be 0.63.
20 divided by 6 =3 1/3
so the answer is 3 1/3
Answer:
a) 0.0853
b) 0.0000
Step-by-step explanation:
Parameters given stated that;
H₀ : <em>p = </em>0.6
H₁ : <em>p = </em>0.6, this explains the acceptance region as;
p° ≤
=0.63 and the region region as p°>0.63 (where p° is known as the sample proportion)
a).
the probability of type I error if exactly 60% is calculated as :
∝ = P (Reject H₀ | H₀ is true)
= P (p°>0.63 | p=0.6)
where p° is represented as <em>pI</em><em> </em>in the subsequent calculated steps below
= P ![[\frac{p°-p}{\sqrt{\frac{p(1-p)}{n}}} >\frac{0.63-p}{\sqrt{\frac{p(1-p)}{n}}} |p=0.6]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bp%C2%B0-p%7D%7B%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%20%3E%5Cfrac%7B0.63-p%7D%7B%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%20%7Cp%3D0.6%5D)
= P ![[\frac{p°-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} >\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} ]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bp%C2%B0-0.6%7D%7B%5Csqrt%7B%5Cfrac%7B0.6%281-0.6%29%7D%7B500%7D%7D%7D%20%3E%5Cfrac%7B0.63-0.6%7D%7B%5Csqrt%7B%5Cfrac%7B0.6%281-0.6%29%7D%7B500%7D%7D%7D%20%5D)
= P ![[Z>\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500} } } ]](https://tex.z-dn.net/?f=%5BZ%3E%5Cfrac%7B0.63-0.6%7D%7B%5Csqrt%7B%5Cfrac%7B0.6%281-0.6%29%7D%7B500%7D%20%7D%20%7D%20%5D)
= P [Z > 1.37]
= 1 - P [Z ≤ 1.37]
= 1 - Ф (1.37)
= 1 - 0.914657 ( from Cumulative Standard Normal Distribution Table)
≅ 0.0853
b)
The probability of Type II error β is stated as:
β = P (Accept H₀ | H₁ is true)
= P [p° ≤ 0.63 | p = 0.75]
where p° is represented as <em>pI</em><em> </em>in the subsequent calculated steps below
= P ![[\frac{p°-p} \sqrt{\frac{p(1-p)}{n} } }\leq \frac{0.63-p}{\sqrt{\frac{p(1-p)}{n} } } | p=0.75]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bp%C2%B0-p%7D%20%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%20%7D%20%7D%5Cleq%20%5Cfrac%7B0.63-p%7D%7B%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%20%7D%20%7D%20%7C%20p%3D0.75%5D)
= P ![[\frac{p°-0.6} \sqrt{\frac{0.75(1-0.75)}{500} } }\leq \frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bp%C2%B0-0.6%7D%20%5Csqrt%7B%5Cfrac%7B0.75%281-0.75%29%7D%7B500%7D%20%7D%20%7D%5Cleq%20%5Cfrac%7B0.63-0.75%7D%7B%5Csqrt%7B%5Cfrac%7B0.75%281-0.75%29%7D%7B500%7D%20%7D%20%7D%20%5D)
= P![[Z\leq\frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]](https://tex.z-dn.net/?f=%5BZ%5Cleq%5Cfrac%7B0.63-0.75%7D%7B%5Csqrt%7B%5Cfrac%7B0.75%281-0.75%29%7D%7B500%7D%20%7D%20%7D%20%5D)
= P [Z ≤ -6.20]
= Ф (-6.20)
≅ 0.0000 (from Cumulative Standard Normal Distribution Table).
Distance Formula: d = √(x₂ - x₁)² + (y₂ - y₁)²
(-6 , 2) and (8 , 10)
x₁ y₁ x₂ y₂
d = √(8 - (-6))² + (10 - 2)²
d = √(8+6)² + (10 - 2)²
d = √(14)² + (8)²
d = √(196) + (64)
d = √ 260
d = 2√65
I think this is right, but I am not sure.