The answer to this question would be: 1 5/6
To change a decimal number into fraction, you need to divide the number on the right of decimal point with 1. In this case, the number is 0.83.
This number is hard since .83 doesn't have many factors. To find the answer you can try to multiply the decimal with some number until it close to 1(no decimal left)
0.83* 2= 1.66
0.83* 3= 2.49 ---> close to half, if you find this number, you can try to double it
0.83* 4= 3.32
0.83* 5= 4.15
0.83* 6= 4.98---> close to 1, that means there is high probability that the number can be divided by 6
0.83 would be 4.98/6, but if we assume that the number is 0.8333...... then 0.83 would be 5/6. So, 1.83 would be 1 5/6
We have two set of solutions , ( 
, + 3 ) , ( - + 3 , - + 3).
<u>Step-by-step explanation:</u>
We have , the following equations:
and 
. Let's substitute value of y in bottom equation:
⇒
⇒ 
⇒ 
Roots of quadratic equation are given by: a = 1, b= 1 , c = -7
⇒
⇒ 
⇒ 
which is actually x = ± .
We know that ,
⇒ y = ± + 3.
We have two set of solutions , ( , + 3 ) , ( - + 3 , - + 3).
Answer:
The picture is black, is it just me?
Answer:
To provide a range of values that, with a certain measure of confidence, contains the population parameter of interest.
To provide a range of values that has a certain large probability of containing the population parameter of interest.
Step-by-step explanation:
For estimating a parameter value, it is important to know the statement or degree of confidence that the interval contains the parameter value along with knowing the point estimate and the amount of possible error in the point estimate (which is the interval likely to contain the parameter value).Thus , an interval estimate of a population parameter is the confidence interval with a statement of confidence that the interval contains the parameter value. The confidence interval is a range of values around the statistic that are believed to contain, with a certain probability (say 99%) the true value of that statistic or the population value.
