Using the t-distribution, it is found that the confidence interval is (14.2, 14.8).
<h3>What is a t-distribution confidence interval?</h3>
The confidence interval is:
In which:
- is the sample mean.
- s is the standard deviation for the sample.
The critical value, using a t-distribution calculator, for a <em>two-tailed 95% confidence interval</em>, with 18 - 1 = <em>17 df</em>, is t = 2.1098.
The other parameters are given by:
.
Hence, the bounds of the interval are given by:
The confidence interval is (14.2, 14.8).
More can be learned about the t-distribution at brainly.com/question/16162795
Answer:
29.) -14
45.) B
46.) D
The one after that.) C
Step-by-step explanation:
29.
3x-4=4x+10
3x-4x=10+4
-x=14
x=-14
45.
B because it says that the value of c or the number of cars cannot be greater than 5000, so it has to be 0 to 5000
46.
D because it doesn't set a limit on haw many days you can stay, rather it just talks about the the equation for payment.
The One After That.
C because the only time that the graph is decreasing is the time when there is a negative slope. That negative slope is only present in the interval given for answer C.
Answer:
one solution
Step-by-step explanation:
5x=2
Divide each side by 5
5x/5 = 2/5
x = 2/5
There is one solution
Answer:
The standard form.
Step-by-step explanation:
Between the standard form and the expanded form the equation that helps find the y-intercept most effectively is the Standard form. The standard form of an equation takes the form of;
Ax + by = C, where A, B and C are constants and x and y are the intercepts.
y- intercept is a point in which x = 0 and then y is determined.
Another equation that helps find the y-intercept most effectively is the POINT -SLOPE form which can be represented mathematically by;
y = mx + c, where c =y- intercept, m = slope = y2 - y1/x2 - x1 and x and y are the axis.
The standard form is efficient because it can describe both zero and constant slope lines while the expanded form can only describe the equation with either zero or constant slope lines.
The equation is actually
. Free fall is always -16t^2 as the position function. We are looking for how long it takes the object to hit the ground. In other words, the height of an object is 0 when it is laying on the ground, so how long (t) did it take to get there? We will then set that position equal to 0 and solve for t.
. If we subtract 1437 from both sides and divide by -16, we have
. Taking the square root of both sides gives us, rounded to the nearest tenth, t = 9.5 or t=-9.5. The 2 things in math that will never EVER be negative are time and distance/length, so -9.5 is out. That means that it took just about 9.5 seconds for the object to fall to the ground from a height of 1437 feet when pulled on by the force of gravity.