Answer: 5 and 14.
Step-by-step explanation:
We know that the Raiders and Wildcats both scored the same number of points in the first quarter so let a,a+d,a+2d,a+3d be the quarterly scores for the Wildcats. The sum of the Raiders scores is a(1+r+r^{2}+r^{3}) and the sum of the Wildcats scores is 4a+6d. Now we can narrow our search for the values of a,d, and r. Because points are always measured in positive integers, we can conclude that a and d are positive integers. We can also conclude that $r$ is a positive integer by writing down the equation:
a(1+r+r^{2}+r^{3})=4a+6d+1
Now we can start trying out some values of r. We try r=2, which gives
15a=4a+6d+1
11a=6d+1
We need the smallest multiple of 11 (to satisfy the <100 condition) that is 1 (mod 6). We see that this is 55, and therefore a=5 and d=9.
So the Raiders' first two scores were 5 and 10 and the Wildcats' first two scores were 5 and 14.
Answer:
(221.39, 300.61) and (255.2223, 266.7777)
Step-by-step explanation:
Given that X, the lengths of pregnancies in a small rural village are normally distributed with a mean of 261 days and a standard deviation of 17 days
Middle 98% would lie on either side of the mean with probability ±2.33 in the std normal distribution on either side of 0
Corresponding we have x scores as
Between
and ![261+2.33*17](https://tex.z-dn.net/?f=261%2B2.33%2A17)
i.e. in the interval = (221.39, 300.61)
If sample size = 47, then std error of sample would be
![\frac{17}{\sqrt{47} }](https://tex.z-dn.net/?f=%5Cfrac%7B17%7D%7B%5Csqrt%7B47%7D%20%7D)
So 98% of pregnancies would lie between
and ![262+2.33*\frac{17}{\sqrt{47} }](https://tex.z-dn.net/?f=262%2B2.33%2A%5Cfrac%7B17%7D%7B%5Csqrt%7B47%7D%20%7D)
= (255.2223, 266.7777)
Answer:
6^2
Step-by-step explanation:
We know a^b / a^c = a^(b-c)
6^5 / 6^3 = 6^(5-3) = 6^2
-5/11 is what I got with a fraction calculator.