find three consecutive integers. such that 4 times of first integer is 18 more than the sum of the second and third integer
1 answer:
Three consecutive numbers are x, x+1 and x+2.
Four times the first integer is 4x
The sum of the second and third is (x+1)+(x+2)=2x+3.
So, we have
Subtract 2x from both sides:
Divide both sides by 2:
So, you can't have three consecutive integers such that four times the first is 18 more than the sum of the other two: the three numbers would be 10.5, 11.5, 12.5.
In fact, you have
and
You might be interested in
Answer:
Step-by-step explanation:
Based on the question we are given the percentages of each of the types of candies in the bag except for brown. Since the sum of all the percentages equals 75% and brown is the remaining percent then we can calculate that brown is (100-75 = 25%) 25% of the bag. Now we can show the probabilities of getting a certain type of candy by placing the percentages over the total percentage (100%).
- Brown:
- Yellow or Blue:
....add the numerators
- Not Green:
.... since the sum of all the rest is 80%
- Stiped:
.... there are 0 striped candies.
Assuming the <u><em>ratios/percentages</em></u> of the candies stay the same having an infinite amount of candy will not affect the probabilities. That being said in order to calculate consecutive probability of getting 3 of a certain type in a row we have to multiply the probabilities together. This is calculated by multiplying the numerators with numerators and denominators with denominators.
- 3 Browns:
- the 1st and 3rd are red while the middle is any. We multiply 15% * (total of all minus red which is 85%) * 15% like so.
- None are Yellow: multiply the percent of all minus yellow three times.
- At least 1 green: multiply the percent of green by 100% twice, since the other two can by any