Circumference = 2 x radius x Pi
Circumference = 2 x 0.8 x pi
Circumference = 1.6pi or 5.024 yards
area = r^2 x pi
Area = 0.8^2 x pi
Area = 0.64pi or 2.0096 square yards
The distance (d) between two points (x1,y1) and (x2,y2) is given by the formula
d = √ ((X2-X1)2+(Y2-Y1)2)
d = √ (-400--800)2+(300-200)2
d = √ ((400)2+(100)2)
d = √ (160000+10000)
d = √ 170000
The distance between the points is 412.310562561766
The midpoint of two points is given by the formula
Midpoint= ((X1+X2)/2,(Y1+Y2)/2)
Find the x value of the midpoint
Xm=(X1+X2)/2
Xm=(-800+-400)/2=-600
Find the Y value of the midpoint
Ym=(Y1+Y2)/2
Ym=(200+300)/2=250
The midpoint is: (-600,250)
Graphing the two points, midpoint and distance
P1 (-800,200)
P2 (-400,300)
Midpoint (-600,250)
The length of the black line is the distance between the points (412.310562561766)
Answer:
C. 0.68
Step-by-step explanation:
Given;
number of products sold in a day by toll-free sales line = 85 products
number of calls in a day = 125
The daily success rate of the sales line is given by the ratio of the total products in a day to total number of calls in a day.
The daily success rate of the sales line = total products sold / number of calls
The daily success rate of the sales line = 85 / 125
The daily success rate of the sales line = 0.68
Therefore, the daily success rate of the sales line is 0.68
Answer:
42.77
Step-by-step explanation:
The pattern is taking away 0.3. 43.07 minus 0.3 equals 42.77
Answer:
a) H0 : u = 28.5%
H1 : u < 28.5%
b) critical value = - 1.645
c) test statistic Z= - 1.41
d) Fail to reject H0
e) There is not enough evidence to support the professor's claim.
Step-by-step explanation:
Given:
P = 28.5% ≈ 0.285
X = 210
n = 800
Level of significance = 0.05
a) The null and alternative hypotheses are:
H0 : u = 28.5%
H1 : u < 28.5%
b) Given a 0.05 significance level.
This is a left tailed test.
The critical value =
The critical value = -1.645
c) Calculating the test statistic, we have:


Z = -1.41
d) Decision:
We fail to reject null hypothesis H0, since Z = -1.41 is not in the rejection region, <1.645
e) There is not enough evidence to support the professor's claim that the proportion of obese male teenagers decreased.