The claim is that the standard deviation of the speeds of drivers in Bradford is 8 mph. An experienced driver feels that this is very low. If a random speed check is conducted for 50 drivers, the standard deviation is 10.5 mph. Can we reject the null at

Null hypothesis is rejected. Standard deviation is > 8 miles per hour

Step-by-step explanation:

σ [Population standard deviation] = 8 , s [sample standard deviation] = s

n [no of items] = 8

H0 [Null] : σ = 8 ; H1 [Alternate - Right Tail] : σ > 8

χ2 = (n - 1) . s^2 / σ^2

= 49 x (10.5)^2 / 82 = 5402.25 / 64

χ2 = 84.410

df [degree of freedom] = n -1 = 50 - 1 = 49

P value (χ^2 49 > 84.410) = 0.00125

p = 0.0013

p < α ie 0.05

So, H0 is rejected

Hence we state that standard deviation is > 8 miles per hpur

answer: 0,6

step-by-step explanation:

this would make it the y intercept.

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752