2(4)+6
8+6
=14
you’re welcome :)
29 = 4 + .75b
subtract 4
25 = .75b
divide .75
b = 33.3333333
b = 33 braids
Answer negative 9 is the answer
Step-by-step explanation:
y=x+14 line 1
y=3x+2 line 2
These are both the equation of lines written in slope intercept form
y=mx+b where m is the slope and the point (0,b) is the y intercept.
The first line has a slope of m=1. The 2nd line has a slope of m=3
Since these lines have different slopes, they are not parallel, thus they will cross at some point. What you have to determine is where the lines cross, which will be a point (x,y) that is on both lines.
We already have y solved in terms of x from either equation so we can use substitution to solve the system.
Since y=x+14 from line 1, put x+14 in place of y in the equation of line 2.
x+14=3x+2
solve for x.
Subtract x from both sides...
14= 3x-x+2
14=2x+2
subtract 2 from both sides
14-2=2x
12=2x
divide both sides by 2
6=x
We now have the x value of the common point. Plug the value 6 in for x in one of the original equations and solve for y.
y=6+14
y=20
These two lines cross at the point (6,20) which is a point the two lines have in common.
Hope I helped (SharkieOwO)
Answer:
The 90% confidence interval for the mean test score is between 77.29 and 85.71.
Step-by-step explanation:
We have the standard deviation for the sample, so we use the t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 25 - 1 = 24
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of
. So we have T = 2.064
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 81.5 - 4.21 = 77.29
The upper end of the interval is the sample mean added to M. So it is 81.5 + 4.21 = 85.71.
The 90% confidence interval for the mean test score is between 77.29 and 85.71.