How many distinct seven-digit phone numbers can you create from the numbers 4, 4, 5, 6, 7, 7, 7? (e.g. 445-6777 is one example)?
1 answer:
You have 7 numbers, two numbers 4 , one 5, one 6 and three numbers 7.
From these 7 numbers you can create
phone numbers, but they are not all different.
If you have two numbers 4, then you have to divide obtained number by 2! (the number of all possible permutations of two numbers 4) and since you have three numbers 7 you should divide the obtained number by 3! (<span>the number of all possible <span>permutations of three numbers 7</span>). Then</span><span /><span>
</span><span>
</span><span>
.</span>
Answer: You can create 420 different phone numbers.<span />
From these 7 numbers you can create
If you have two numbers 4, then you have to divide obtained number by 2! (the number of all possible permutations of two numbers 4) and since you have three numbers 7 you should divide the obtained number by 3! (<span>the number of all possible <span>permutations of three numbers 7</span>). Then</span><span /><span>
</span><span>
</span><span>
Answer: You can create 420 different phone numbers.<span />
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Answer:
0.632
Step-by-step explanation:
Given that a homeowner is three times as likely to purchase additional jewelry coverage as additional electronics coverage
If probability of purchasing additional electronics coverage = p, then prob of purchasing jewelry coverage = 3p
The two events are independent hence joint probability is product of these two.
i.e. P(both) =
This is given as 0.2
the probability that a homeowner purchases exactly one coverage.
= Prob he purchases I + prob he purchases II-2(prob he purchases both)
Answer:
P = 0.0215 = 2.15%
Step-by-step explanation:
First we need to convert the values of 900 and 975 to standard scores using the equation:
Where z is the standard value, x is the original value, is the mean and
is the standard deviation. So we have that:
standard value of 900:
standard value of 975:
Now, we just need to look at the standard distribution table (z-table) for the values of z = 2 and z = 3:
z = 2 -> p_2 = 0.9772
z = 3 -> p_3 = 0.9987
We want the interval between 900 and 975 hours, so we need the interval between z = 2 and z = 3, so we just need to subtract their p-values:
P = p_3 - p_2 = 0.9987 - 0.9772 = 0.0215
So the probability is 0.0215 = 2.15%