Ok so let me help you in here:
<span>After she pays $500 toward her medical expenses for the year, her insurance will pay 80% of the $12,500 cost. So
12,500 x .80 = 10,000. Then 12,500-10,000 = $2,500.
So you pay $2500 and the insurance company pays $10,000.
</span>Hope this is useful
Use the slope formula below:

To form an equation of a line - we need to find a slope and the y-intercept from y = mx+b. We are given two points which we can substitute in the formula.

We have finally got the slope. Next is to find the y-intercept. First we rewrite the equation of y = mx+b by substituting the slope.

The equation above is the slope-intercept form. Substitute m = -7/6 in the equation.

Since the graph passes through (0,0) which is an origin point. In y = mx+b if the graph passes through origin point, that means the b-value is 0. Therefore:

Answer
Hope this helps and let me know if you have any doubts! Good luck on your assignment!
Answer:
50 degrees
Step-by-step explanation:
40+50=130
130-180
50
Well, if you do this in reverse
23+17 = 40
40-23 = 17
40 would be your answer.
Answer:
The 95% confidence interval for the population mean span of boys from participating schools in Canada is between 154.3 cm and 160.3 cm.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used 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 = 100 - 1 = 99
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 99 degrees of freedom(y-axis) and a confidence level of
. So we have T = 1.9842
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 157.3 - 3 = 154.3 cm
The upper end of the interval is the sample mean added to M. So it is 157.3 + 3 = 160.3 cm.
The 95% confidence interval for the population mean span of boys from participating schools in Canada is between 154.3 cm and 160.3 cm.