Answer:
a. 54.05 Mpbs.
b. 2.745... standard deviations.
c. The z-score is 2.745....
d. The carrier's highest data speed is significantly high.
Step-by-step explanation:
a. The difference between the highest measured data speed and the mean is 72.6 - 18.55 = 54.05 Mbps.
b. The amount of standard deviations of 54.05 Mbps is equal to this value divided by the standard deviations, so we yield
standard deviations.
c. The z-score is equal to the difference between the mean and a data point in standard deviations, so the z-score is 2.745....
d. 2.745... is not between -2 and 2, so the carrier's highest data speed is not insignificant - so it's significantly high.
Answer would be 39 and thats no cap
I think it’s 1/2 divided by 2
If
N = a (mod 10)
N = b (mod 13)
gcd(10,13) = 1
then
N = 10 bx + 13 ay (mod 130)
Where
10x + 13y = 1
<span>-> </span>(10x + 13)
(mod 2) = 1 (mod 2)
<span>-> </span>y (mod 2) = 1
y = -3, x = 4
-> N = 40b – 39a
(mod 130)
It is given that ra + sb
should be non-negative:
N = 40b – 39a (mod 130)
N = 40b + (130 – 39)a (mod 130)
N = 40b + 91a (mod 130)
Therefore, N modulo 130, in terms of a and b is: <span>N = 40b + 91a
(mod 130).</span>