The total cost of transactions for Belinda is $899.
<u>Solution:</u>
Given that, Belinda used an online broker who charged a flat fee of $8.99 per transaction.
Belinda bought 100 shares of of BMN Inc, stock at \$12.85 a share,
We have to find what was the total cost of the transaction?
Now, we know that, transaction cost per transaction = $8.99
And, number of transactions = 100 shares.
Then, total cost of transaction = cost per transaction x number of transactions = 100 x 8.99 = $899
Hence, the total cost of transactions is $899
Answer:
D) 3x
Step-by-step explanation:
The difference of adjacent terms is ...
(-7x) -(-10x) = (-7+10)x = 3x . . . . the common difference
When you compare 649 and 645, you see that the first two digits are the same, and you compare until the last digit: in a way you need to do 3 comparisons, one for each digit (and after first and second you conclude that you still need to compare).
when you compare 645 and 738 you see immediately that the second number is bigger: the numbers differ already in the number of hundreds than they have, so you don't need to compare their tenths and one's, just the hundreds.
Answer:
60 is the measure of C in degrees
<span>(a) This is a binomial
experiment since there are only two possible results for each data point: a flight is either on time (p = 80% = 0.8) or late (q = 1 - p = 1 - 0.8 = 0.2).
(b) Using the formula:</span><span>
P(r out of n) = (nCr)(p^r)(q^(n-r)), where n = 10 flights, r = the number of flights that arrive on time:
P(7/10) = (10C7)(0.8)^7 (0.2)^(10 - 7) = 0.2013
Therefore, there is a 0.2013 chance that exactly 7 of 10 flights will arrive on time.
(c) Fewer
than 7 flights are on time means that we must add up the probabilities for P(0/10) up to P(6/10).
Following the same formula (this can be done using a summation on a calculator, or using Excel, to make things faster):
P(0/10) + P(1/10) + ... + P(6/10) = 0.1209
This means that there is a 0.1209 chance that less than 7 flights will be on time.
(d) The probability that at least 7 flights are on time is the exact opposite of part (c), where less than 7 flights are on time. So instead of calculating each formula from scratch, we can simply subtract the answer in part (c) from 1.
1 - 0.1209 = 0.8791.
So there is a 0.8791 chance that at least 7 flights arrive on time.
(e) For this, we must add up P(5/10) + P(6/10) + P(7/10), which gives us
0.0264 + 0.0881 + 0.2013 = 0.3158, so the probability that between 5 to 7 flights arrive on time is 0.3158.
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