<h3>
Answer: -1.5</h3>
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Explanation:
Draw a horizontal line from W to the y axis. You should arrive somewhere between -1 and -2. While this isn't exact, it looks like we should arrive right at the middle of -1 and -2; therefore we should get to -1.5
The y coordinate of W is -1.5
Keep in mind this is based on the assumption we reach the halfway point. Unfortunately, W is not on any horizontal grid lines to be able to determine exactly where W is along the y axis.
Answer:
JMK = 54
JKH = 126
HLK = 90
HJL = 27
LHK = 63
JLK = 27
Step-by-step explanation:
Just use 180 for total angles in each triangle. Vertical angles mean the opposite to 126 is 126 and the other two that make up the inner vertices are (360- 126*2)/2 = 54
So, 126, `16, 54 , 54 are the inner angles in the circle. Just go from there.
Answer:
One
Step-by-step explanation:
Add a "t" to "one" and you get "tone", a musical sound.
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).
Answer:
18) Area= 5*5/2=25/2=12.5 unit ^2
19) Area=AB^2V3/4=8a^2*V3/4=2V3a^2
Step-by-step explanation:
18. A(-3,0)
B(1,-3)
C(4,1)
AB=V(-3-1)^2+(0+3)^2=V16+9=V25=5
AC=V(-3-4)^2+(0-1)^2=V49+1=V50=5V2
BC=V(1-4)^2+(-3-1)^2=V9+16=V25=5
so AB=BC=5
and AC^2=AB^2+BC^2
so trg ABC is an isosceles right angle triangle (<B=90)
Area= 5*5/2=25/2=12.5 unit ^2
19. A(a,a)
B(-a,-a)
C(-V3a, V3a)
AB=V(a+a)^2+(a+a)^2=V4a^2+4a^2=V8a^2
AC=V(a+V3a)^2+(a-V3)^2=Va^2+2a^2V3+3a^2+a^2-2a^2V3+3a^2=V8a^2
BC=V(-a+V3a)^2+(-a-V3a)^2=V8a^2
so AB=AC=BC
Area=AB^2V3/4=8a^2*V3/4=2V3a^2
