Answer:
9
Step-by-step explanation:
All u have to do is think only about two denominators, 9 and 3
And then memorize the multiplication of any number out of these two, until u get a number which is divisible by the other number. Its like,
9×1= 9
i
is it divisible by 3?
9/3= 3
Oh yes, it is divisible by 3 with no remainings. So that is the least common denominator
Answer:
2.5% probability that a randomly selected book has fewer than 133 pages.
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 189 pages
Standard deviation = 28 pages
What is the probability that a randomly selected book has fewer than 133 pages?
133 = 189 - 2*28
So 133 is two standard deviations below the mean.
The Empirical Rule states that 95% of the measures are within 2 standard deviations of the mean. The other 5% is more than two standard deviations distant from the mean. The normal distribution is symmetric, which means that of those 5%, 2.5% are more than 2 standard deviations below the mean and 2.5% are more than 2 standard deviations above the mean.
This means that there is a 2.5% probability that a randomly selected book has fewer than 133 pages.
Answer: 199,990
Step-by-step explanation:
You can't round to the nearest tenth because you don't have a decimal added to the number.
I rounded to the nearest tens and got 199,990
So, is two integers, and they must be consecutive, meaning, the next one will have to be either 1 before the first or one after... anyhow.
let's say the first integer is "a".
then a consecutive integer to that one will be just 1 hop away, or say "a + 1", so there, those are the two integers.
now, we know that
![\bf \stackrel{\textit{square of the first is decreased by 17}}{a^2-17}~~=~~\stackrel{\textit{4 times the second}}{4(a+1)} \\\\\\ a^2-17=4a+4\implies a^2-4a-21=0\implies (a-7)(a+3)=0 \\\\\\ a= \begin{cases} \boxed{7}\\ -3 \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bsquare%20of%20the%20first%20is%20decreased%20by%2017%7D%7D%7Ba%5E2-17%7D~~%3D~~%5Cstackrel%7B%5Ctextit%7B4%20times%20the%20second%7D%7D%7B4%28a%2B1%29%7D%0A%5C%5C%5C%5C%5C%5C%0Aa%5E2-17%3D4a%2B4%5Cimplies%20a%5E2-4a-21%3D0%5Cimplies%20%28a-7%29%28a%2B3%29%3D0%0A%5C%5C%5C%5C%5C%5C%0Aa%3D%0A%5Cbegin%7Bcases%7D%0A%5Cboxed%7B7%7D%5C%5C%0A-3%0A%5Cend%7Bcases%7D)
so, is a positive integer, so it can't be -3.
what's the second integer? well is a + 1.
Answer:
hmm how about no
Step-by-step explanation:
thx for the free points! Hope your life gets better