Answer:
y = (1/3)x + 4
Step-by-step explanation:
Two points on this line are (0, 4) and (3, 5).
As we move from the first point to the second, x increases by 3 and y increases by 1. Thus, the slope, m, of the line is m = rise / run = 1/3.
Use the slope-intercept equation: y = mx + b.
If we use the data from the point (0, 4), we get:
4 = (1/3)(0) + b, so that b = 4. The desired equation is y = (1/3)x + 4.
Answer:
14 square units
Step-by-step explanation:
There are several ways we can find the area. Probably the easiest is to cut the kite in half vertically and find the area of each triangle. The area of the kite will be double that.
The height of the kite is 7 units, and the width is 4 units. So each triangle will have a base of 7 and height of 2.
A = 1/2 bh
A = 1/2 (7) (2)
A = 7
The area of the kite is double that, so:
2A = 14
Step-by-step explanation:
Let's say R is the initial radius of the sphere, and r is the radius at time t.
The volume of the sphere at time t is:
V = 4/3 π r³
Taking derivative with respect to radius:
dV/dr = 4π r²
This is a maximum when r is a maximum, which is when r = R.
(dV/dr)max = 4π R²
This is 4 times the sphere's initial great circle area, but not the great circle circumference. The problem statement contains an error.
We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:
The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:
![CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack](https://tex.z-dn.net/?f=CI%28%5Cmu%29%3D%5Clbrack%20x-Z_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ccdot%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%5B%5D%7Bn%7D%7D%2Cx%2BZ_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ccdot%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%5B%5D%7Bn%7D%7D%5Crbrack)
Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:
![CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack](https://tex.z-dn.net/?f=CI%28%5Cmu%29%3D%5Clbrack30.0-Z_%7B0.99%7D%5Ccdot%5Cfrac%7B2.40%7D%7B%5Csqrt%5B%5D%7B28%7D%7D%2C30.0%2BZ_%7B0.99%7D%5Ccdot%5Cfrac%7B2.40%7D%7B%5Csqrt%5B%5D%7B28%7D%7D%5Crbrack)
Where (from tables):
Finally, the interval at 98% confidence level is:
