Find total number of integers.
Find how many integers is divisible by 2.
Eliminate even numbers.
11, 13, 15,..., 57, 59
This array contains 51 - 26 = 25 numbers.
Eliminate numbers before the first number divisible by 3 and after the last number divisible by 3.
15, 17, 19,..., 55, 57
This array contains 25 - 3 = 22 numbers.
Now we should eliminate numbers divisible by 3: 15, 21, 27...
There are 8 such numbers.
Therefore, there are 25 - 8 = 17 numbers that <span>can be evenly divided by neither 2 nor 3</span>
Answer:
65.4
Step-by-step explanation:
60*.09 = 5.4
60+5.4 = 65.4
Step-by-step explanation:
A study was to be undertaken to determine if a particular training program would improve physical fitness. A sample of 31 university students was selected to be enrolled in the fitness program. The researchers wished to determine if there was evidence that their sample of students differed from the general population of untrained subjects. The sample mean is 47.4 and a standard deviation of 5.3. The 98% confidence interval is determined and is given as, (45.2, 49.6) .
If the level of confidence is changed to 95%, then the confidence interval will become shorter but the p-value will not change because it is calculated using the test statistic. So the correct answer is (a).
Answer:
Step-by-step explanation:
Answer:
What is the probability that a randomly selected family owns a cat? 34%
What is the conditional probability that a randomly selected family doesn't own a dog given that it owns a cat? 82.4%
Step-by-step explanation: We can use a Venn (attached) diagram to describe this situation:
Imagine a community of 100 families (we can assum a number, because in the end, it does not matter)
So, 30% of the families own a dog = .30*100 = 30
20% of the families that own a dog also own a cat = 0.2*30 = 6
34% of all the families own a cat = 0.34*100 = 34
Dogs and cats: 6
Only dogs: 30 - 6 = 24
Only cats: 34 - 6 = 28
Not cat and dogs: 24+6+28 = 58; 100 - 58 = 42
What is the probability that a randomly selected family owns a cat?
34/100 = 34%
What is the conditional probability that a randomly selected family doesn't own a dog given that it owns a cat?
A = doesn't own a dog
B = owns a cat
P(A|B) = P(A∩B)/P(B) = 28/34 = 82.4%