Given:
n = 150, sample size
Denote the sample proportion by q (normally written as
).
That is,
q = 60/150 = 0.4, sample proportion.
At the 96% confidence level, the z* multiplier is about 2.082, and the confidence interval for the population proportion is
![q \pm z^{*}[ \frac{q(1-q)}{ \sqrt{n} } ]](https://tex.z-dn.net/?f=q%20%5Cpm%20z%5E%7B%2A%7D%5B%20%5Cfrac%7Bq%281-q%29%7D%7B%20%5Csqrt%7Bn%7D%20%7D%20%5D)
That is,
0.4 +/- 2.082* √[(0.4*0.6)/150]
= 0.4 +/- 0.0833
= (0.3167, 0.4833)
= (31.7%, 48.3%)
Answer: The 96% confidence interval is about (32% to 48%)
(5,5730) (7,6022)
slope = (6022 - 5730) / (7 - 5) = 292/2 = 146
y = mx + b
slope(m) = 146
(5,5730)...x = 5 and y = 5730
sub and find b, the y int
5730 = 146(5) + b
5730 = 730 + b
5730 - 730 = b
5000 = b
so ur equation is : y = 146x + 5000 <==
In additive inverse, the sum of two numbers is always 0
Answer: a=21, b=42, c=117
Step-by-step explanation:
Answer:
1) 81.67 cm² (Option A)
2) 28.26 m² (Option B)
3) 153.86 mi.² (Option D)
4) 329.90 m² (Option A)
5) 63.585 ft.² (Option C)
Step-by-step explanation:
<u>GENERAL FORMULAES TO BE USED IN THIS QUESTION :-</u>
Diameter of a circle with radius 'r' = 2r
Area of a circle with radius 'r' = πr²
<u>SOLUTION :-</u>
1) Diameter of the dinner plate = 10.2 cm
Radius of the dinner plate = 10.2/2 = 5.1 cm
⇒ Area of the circle = π(5.1)² = 26.01π cm² ≈ 81.67 cm²
2) Length of his leash = 3m
⇒ Area the dog has to run around = π(3)² = 9π m² = 28.26 m²
3) Radius of the circular range of storm = 7 mi.
⇒ Area of the circular range of storm = π(7)² = 49π mi.² = 153.86 mi.²
4) Diameter of carousel = 20.5 m
Radius of carousel = 20.5/2 = 10.25 m
⇒ Area of the carousel = π(10.25)² = 105.0625π m² ≈ 329.90 m²
5) Radius of the range of sprinkler = 4.5 ft.
⇒ Area of the range of sprinkler = π(4.5)² = 20.25π ft.² = 63.585 ft.²
