Answer: the person would have to visit a total of twelve times for it to add up to $48
Step-by-step explanation:
48-15=33
then divide the 33 by 3 and you get 11 plus the 15 dollar membership and visit will get u 12 visits for $48.
The three consecutive integers are 5, 6, and 7.
What are consecutive integers?
Whole numbers that follow one another without a gap are known as consecutive integers. A few examples of consecutive integers are 15, 16, and 17, 1,2 and 3, and so on.
Finding the Consecutive Integers
Let the three consecutive integers be a, a + 1, and a + 2.
These three consecutive integers are to be such that two times the first integer when added to three times the second integer minus the third integer, it is equal to 21.
⇒ 2a + 3(a+1) - (a+2) = 21
2a + 3a + 3 - a -2 = 21
5a - a + 3 - 2 = 21
4a + 1 = 21
4a = 20
a = 20/4
a = 5
∴ a+1 = 5+1
a+1 = 6
And, a+2 = 5 + 2
a +2 =7
Hence, the three consecutive integers come out to be 5, 6, and 7
Learn more about consecutive integers here:
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The factors of
375: 1; 3; 5; 15; 25, 75, 125, 375
66: 1; 2; 3; 6, 11, 22, 33, 66
33: 1, 3, 11, 33
Step-by-step explanation:
6.) 8p = 8x9 = 72
8.) 4(d+7) = 4(-2+7) = 4(5) = 20
can't read all of 10 to give an answer
Answer:
the probability is 2/9
Step-by-step explanation:
Assuming the coins are randomly selected, the probability of pulling a dime first is the number of dimes (4) divided by the total number of coins (10).
p(dime first) = 4/10 = 2/5
Then, having drawn a dime, there are 9 coins left, of which 5 are nickels. The probability of randomly choosing a nickel is 5/9.
The joint probability of these two events occurring sequentially is the product of their probabilities:
p(dime then nickel) = (2/5)×(5/9) = 2/9
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<em>Alternate solution</em>
You can go at this another way. You can list all the pairs of coins that can be drawn. There are 90 of them: 10 first coins and, for each of those, 9 coins that can be chosen second. Of these 90 possibilities, there are 4 dimes that can be chosen first, and 5 nickels that can be chosen second, for a total of 20 possible dime-nickel choices out of the 90 total possible outcomes.
p(dime/nickel) = 20/90 = 2/9
