Answer:
C = 53D + 99
Step-by-step explanation:
The cost (C) is equal to $53 per day (53 multiplied by the number of days, D) plus the fixed deposit (+ 99)
Expressed as a linear equation below
C = 53D + 99
Answer:
The range of weights of the middle 99.7% of M&M’s is between 0.8187 and 1.0203.
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
Approximately 68% of the measures are within 1 standard deviation of the mean.
Approximately 95% of the measures are within 2 standard deviations of the mean.
Approximately 99.7% of the measures are within 3 standard deviations of the mean.
Sample mean:
Sample standard deviation:
What is the range of weights of the middle 99.7% of M&M’s?
By the Empirical Rule, within 3 standard deviations of the mean, so:
0.9195 - 3*0.0336 = 0.8187.
0.9195 + 3*0.0336 = 1.0203.
The range of weights of the middle 99.7% of M&M’s is between 0.8187 and 1.0203.
Answer:
X= -2
Step-by-step explanation:
Useful Log Rules:
- log(A*B) = log(A)+log(B) .......... log rule 1
- log(A/B) = log(A) - log(B) .......... log rule 2
- log(A^B) = B*log(A) ................... log rule 3
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Part (a)
All logs shown below are base 3.
log(500) = log(5*100)
log(500) = log(5*10^2)
log(500) = log(5)+log(10^2) .... use log rule 1
log(500) = log(5) + 2*log(10) .... use log rule 3
log(500) = 1.4650 + 2*2.096 ....... substitution
log(500) = 5.657
<h3>Answer: 5.657</h3>
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Part (b)
All logs shown below are base 3.
log(2) = log(10/5)
log(2) = log(10) - log(5) .... use log rule 2
log(2) = 2.096 - 1.4650 ....... substitution
log(2) = 0.631
<h3>Answer: 0.631</h3>
