Let Ted be x.
Ed is 7 years older = x + 7
Ed = (3/4)Ted
(x + 7) = (3/4)x
x + 7 = 3x/4
x - 3x/4 = -7
x/4 = -7
x = -28, Ted = -28 years.
(x + 7) = -28 + 7 = -21, Ed = -21 years
Goodness. We had negative numbers for the ages, well does that make sense? No it doesn't.
Our answer is correct. But the sense in the question is lacking. The question has been wrongly set.
<span>We might assume negative ages to mean before they came into the world, before birth! </span>
Answer:
the probability that a sample of the 35 exams will have a mean score of 518 or more is <em> 0.934 </em>or<em> 93.4%</em>.
Step-by-step explanation:
This is s z-test because we have been given a sample that is large (greater than 30) and also a standard deviation. The z-test compares sample results and normal distributions. Therefore, the z-statistic is:
(520 - 518) / (180/√35)
= 0.0657
Therefore, the probability is:
P(X ≥ 0.0657) = 1 - P(X < 0.0657)
where
- X is the value to be standardised
Thus,
P(X ≥ 0.0657) = 1 - (520 - 518) / (180/√35)
= 1 - 0.0657
= 0.934
Therefore, the probability that a sample of the 35 exams will have a mean score of 518 or more is <em>0.934 or 93.4%</em>.
Answer:
a. 12
b. 7.200 and 2.683
Step-by-step explanation:
The computation is shown below:
Given that
P = 0.40 and n = 30
a)
The expected value of received e-mails is
= n × p
= 30 × 0.4
= 12
b)
The variance of emails received is
= n × p × (1 - p)
= 30 × 0.4 × 0.6
= 7.200
Now
The standard deviation of emails received is
= sqrt(variance)
= 2.683
We simply applied the above formula
