Answer:
The numerical value of the correlation between number of classes skipped and grade index is 0.60.
Step-by-step explanation:
In statistics R² is a measure that is used to determine the proportion of variation in the dependent variable that can be explained by the independent variable. The value of R² is compute by squaring the correlation coefficient (<em>r</em>) between the dependent and independent variable.
In this case the dependent variable is the grades and the independent variable is the number of classes skipped.
The R² between the grades and the number of classes skipped is 36% or 0.36.
Compute the value of <em>r</em> as follows:
Thus, the correlation between number of classes skipped and grade index is ±0.60.
Answer:
.
Step-by-step explanation:
Let the unknown fraction be ,
where, x and y both are prime numbers less than 20.
Now, it is given that adding 1 to both numerator and denominator will make the fraction .
Thus,
=
2(x + 1) = y + 1
2x + 2 = y + 1
2x + 1 = y.
Clearly if x will be any odd number , two times x will be odd and adding 1 to it will result in even number and y should be even number , which is not possible as only even prime is 2.
Thus , x should be the even prime which is 2.
And y will be 5.
Thus the required fraction is .
In order to find this answer we have to manipulate the equation algebraically so that we end up with a statement that x = something. We can manipulate the problem however we need to as long as we do the same thing to both sides of the equation (this keeps the equation true).
-8x/5 + 1/6 = -5x/3
first lets get the x's on one side of the equals sign by adding 8x/5 to each side.
1/6 = -5x/3 + 8x/5
Ok, now let's add the x's together (we need common denominators)
1/6 = -25x/15 + 24x/15
1/6 = -x/15
Now lets get x by itself by multiplying each side by 15
1/6 * 15 = -x
15/6 = -x
3/2 = -x
Multiply each side by -1 to make the x positive.
-3/2 = x
This expression simplified is 9n^4+8n^3+4
Answer:
Pr(X >42) = Pr( Z > -2.344)
= Pr( Z< 2.344) = 0.9905
Step-by-step explanation:
The scenario presented can be modeled by a binomial model;
The probability of success is, p = 0.65
There are n = 80 independent trials
Let X denote the number of drivers that wear a seat belt, then we are to find the probability that X is greater than 42;
Pr(X > 42)
In this case we can use the normal approximation to the binomial model;
mu = n*p = 80(0.65) = 52
sigma^2 = n*p*(1-p) = 18.2
Pr(X >42) = Pr( Z > -2.344)
= Pr( Z< 2.344) = 0.9905