Answer: 8359
Step-by-step explanation:
The formula for sample size needed for interval estimate of population proportion :-
Given : The significance level : 
Critical value : 
Previous estimate of proportion : 
Margin of error : 
Now, the required sample size will be :-
Hence, the required sample size = 8359
The dot product of the two vectors will be given as follows:
v*w=(-7i+4j)*(-6i+5j)
=(-7*(-6))i+(4*5)j
=-42i+20j
Hence the answer is:
v*w=(-42i+20j)
Answer:
13.567%
Step-by-step explanation:
We solve the above question, using z score formula
z = (x-μ)/σ, where
x is the raw score = 23.1 ounces
μ is the population mean = 22.0 ounces
σ is the population standard deviation = 1.0 ounce
More than = Greater than with the sign = >
Hence, for x > 23.1 ounces
z = 23.1 - 22.0/1.0
= 1.1
Probability value from Z-Table:
P(x<23.1) = 0.86433
P(x>23.1) = 1 - P(x<23.1)
P(x>23.1) = 1 - 0.86433
P(x>23.1) = 0.13567
Converting to percentage
= 0.13567 × 100
= 13.567%
Therefore, the percentage of regulation basketballs that weigh more than 23.1 ounces is 13.567%
Answer:
G=0.0728
Step-by-step explanation:
The expression given is:
This expression has the format of a compounded interest at a rate G for a time T. If B = 3,455, A = 1,969, and T = 8, the value of the rate G is determined by:
![3,455=1,969*(1+G)^8\\G = \sqrt[8]{\frac{3,455}{1,969}}-1\\G=0.0728](https://tex.z-dn.net/?f=3%2C455%3D1%2C969%2A%281%2BG%29%5E8%5C%5CG%20%3D%20%5Csqrt%5B8%5D%7B%5Cfrac%7B3%2C455%7D%7B1%2C969%7D%7D-1%5C%5CG%3D0.0728)
The rate G is 0.0728.
