(r, theta)= (8, 3/2 pi)
r=(x^2 +y^2)^(1/2)
theta= 3/2 pi
x= r(costheta)
y=r(sintheta)
x=8(cos(3/2 pi))
y=8(sin(3/2 pi))
x=8(0)
y=8(-1)
x=0
y=-8
r=((-8)^2+(0)^2)^(1/2)
r=(64+0)^(1/2)
r=8
rectangular coordinates= (0,-8)
Answer:
Step-by-step explanation:
Researchers measured the data speeds for a particular smartphone carrier at 50 airports.
The highest speed measured was 76.6 Mbps.
n= 50
X[bar]= 17.95
S= 23.39
a. What is the difference between the carrier's highest data speed and the mean of all 50 data speeds?
If the highest speed is 76.6 and the sample mean is 17.95, the difference is 76.6-17.95= 58.65 Mbps
b. How many standard deviations is that [the difference found in part (a)]?
To know how many standard deviations is the max value apart from the sample mean, you have to divide the difference between those two values by the standard deviation
Dif/S= 58.65/23.39= 2.507 ≅ 2.51 Standard deviations
c. Convert the carrier's highest data speed to a z score.
The value is X= 76.6
Using the formula Z= (X - μ)/ δ= (76.6 - 17.95)/ 23.39= 2.51
d. If we consider data speeds that convert to z scores between minus−2 and 2 to be neither significantly low nor significantly high, is the carrier's highest data speed significant?
The Z value corresponding to the highest data speed is 2.51, considerin that is greater than 2 you can assume that it is significant.
I hope it helps!
Answer:
x = 11
Step-by-step explanation:
9x + 1 + 100 = 180
9x + 101 = 180
9x = 99
x = 11
Ray OE is the angle bisector of angle AOD
ray oc is the bisector of angle BOD
Answer:
99.7%
Step-by-step explanation:
we start by getting the z-scores for the values given
We can get the z-scores using the formula below;
z-scores = (x-mean)/SD
for 338.5;
z = (338.5-598.4)/86.7 = -2.99 approximately-3
for 858.5;
z = (858.5-598.4)/86.7 = 3
so we want to get the percentages that lie within 3 standard deviations from the mean
According to the empirical rule, the percentages within 3 SD of the mean is;
100-0.15-0.15 = 99.7%
