Answer:
104
Step-by-step explanation:
There are two ways to do it, though we need to know how it is affected.
20+6=26
26+6=32
32+6=38
38+6=44
Thus, you are adding 6 for each term.
1. You do it manually...(yes.)
I'll number it one by one
1.20
2.26
3.32
4.38
5.44
6.50
7.56
8.62
9.68
10.74
11.80
12.86
13.92
14.98
15.104
The 15th term is 104.
2. This method is easier. As shown above, there is a pattern. We can apply it using this formula:
20+6(n-1)
You can get this formula from the facts that:
-you start off with an additional added 8
-you add 6 every time
-if we do it 6n then it would be incorrect, with an extra 6 for each
-the formula is correct; you can test it for terms 2,3,4,5 like this:
20+6=26
20+2x6=32
20+3x6=38
20+4x6=44
To find the 15th term, you can:
20+14x6=20+84=104.
Answer:
-32
Step-by-step explanation:
-4 + f(8) + 3 x g(-7) = ?
To find f(8), look at each of the f(x) function and find x = 8. The y value is the answer you will plug in for f(8). The same goes for g(-7), but you will be looking at the g(x) function at x = -7.
f(8) = -4
g(-7) = -8
Make sure you solve in the correct order using PEMDAS.
-4 + (-4) + 3 x (-8) = -32
Answer:
54.80 MPa to 55.92 MPa
Step-by-step explanation:
Sample mean fracture strength (x) = 55.36 MPa
Sample standard deviation (s) = 3.99 MPa
Sample size (n) = 196.
The upper and lower bounds for a 95% confidence interval are given by:
![U=x+1.960*\frac{s}{\sqrt{n}} \\L=x-1.960*\frac{s}{\sqrt{n}}](https://tex.z-dn.net/?f=U%3Dx%2B1.960%2A%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%20%5C%5CL%3Dx-1.960%2A%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D)
The upper and lower bounds of the confidence interval are;
![U=55.36+1.960*\frac{3.99}{\sqrt{196}}\\U= 55.92\\L=55.36-1.960*\frac{3.99}{\sqrt{196}}\\L= 54.80\\](https://tex.z-dn.net/?f=U%3D55.36%2B1.960%2A%5Cfrac%7B3.99%7D%7B%5Csqrt%7B196%7D%7D%5C%5CU%3D%2055.92%5C%5CL%3D55.36-1.960%2A%5Cfrac%7B3.99%7D%7B%5Csqrt%7B196%7D%7D%5C%5CL%3D%2054.80%5C%5C)
The 95% confidence interval for true average fracture strength is 54.80 MPa to 55.92 MPa
Answer:
No
Step-by-step explanation:
For a point to be the midpoint of a line segment, it must bisect it into two equal segments and be on the line segment (hence, colinear with the endpoints). All four B points are equidistant from points A and C, but aren’t colinear with A and C. Therefore, they aren’t all midpoints of line segment AC.
I hope this helps! :)
A direct variation suggest that the value of x in the equation would greatly affect the value of y such that when x is increasing, y also increases and the other way around. The equation for a direct variation is that,
y = kx
Substituting the given values in the ordered pair,
5 = k(4) ; k = 5/4