Answer:
The bulbs should be replaced each 1060.5 days.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
How often should the bulbs be replaced so that no more than 1% burn out between replacement periods?
This is the first percentile, that is, the value of X when Z has a pvalue of 0.01. So X when Z = -2.325.
The bulbs should be replaced each 1060.5 days.
Answer:
Sample Response: To write a two-variable equation, I would first need to know how much Maya’s allowance was. Then, I would need the cost of playing the arcade game and of riding the Ferris wheel. I could let the equation be cost of playing the arcade games plus cost of riding the Ferris wheel equals the total allowance. My variables would represent the number of times Maya played the arcade game and the number of times she rode the Ferris wheel. With this equation I could solve for how many times she rode the Ferris wheel given the number of times she played the arcade game.
Answer:
a)0.7
b) 10.03
c) 0.0801
Step-by-step explanation:
Rate of return Probability
9.5 0.1
9.8 0.2
10 0.3
10.2 0.3
10.6 0.1
a.
P(Rate of return is at least 10%)=P(R=10)+P(R=10.2)+P(R=10.6)
P(Rate of return is at least 10%)=0.3+0.3+0.1
P(Rate of return is at least 10%)=0.7
b)
Expected rate of return=E(x)=sum(x*p(x))
Rate of return(x) Probability(p(x)) x*p(x)
9.5 0.1 0.95
9.8 0.2 1.96
10 0.3 3
10.2 0.3 3.06
10.6 0.1 1.06
Expected rate of return=E(x)=sum(x*p(x))
Expected rate of return=0.95+1.96+3+3.06+1.06=10.03
c)
variance of the rate of return=V(x)=![sum(x^2p(x))-[sum(x*p(x))]^2](https://tex.z-dn.net/?f=sum%28x%5E2p%28x%29%29-%5Bsum%28x%2Ap%28x%29%29%5D%5E2)
Rate of return(x) Probability(p(x)) x*p(x) x²*p(x)
9.5 0.1 0.95 9.025
9.8 0.2 1.96 19.208
10 0.3 3 30
10.2 0.3 3.06 31.212
10.6 0.1 1.06 11.236
sum[x²*p(x)]=9.025+19.208+30+31.212+11.236=100.681
variance of the rate of return=V(x)=sum(x²*p(x))-[sum(x*p(x))]²
variance of the rate of return=V(x)=100.681-(10.03)²
variance of the rate of return=V(x)=100.681-100.6009
variance of the rate of return=V(x)=0.0801
Answer:
180/7
Step-by-step explanation:
Comment if you want it, but it's very long.
Answer:
1m
The greatest common factor is one. They also have an ‘m’ in common
Hope this helps
