B=48/7 you simplify 7/b(b) to 7b/6
Then multiply both sides by 6 so 8 x 6 =7b then simplify 8 x 6 =48
And divide both sides by 7 48\7 then switch sides b=48/7
d = distance between the two cities
v₁ = average speed while going from chicago to kansas city = 440 knots
t₁ = time taken to travel distance going from chicago to kansas city
time taken to travel distance going from chicago to kansas city is given as
t₁ = d/v₁
t₁ = d/440 eq-1
v₂ = average speed while going from kansas city to chicago = 110 knots
t₂ = time taken to travel distance going from kansas city to chicago
time taken to travel distance going from kansas city to chicago is given as
t₂ = d/v₂
t₂ = d/110 eq-2
Given that :
t₂ = t₁ + 3
using eq-1 and eq-2
(d/110) = (d/440) + 3
d = 440
Answer:
A: f(x)=10cos(2π/5 x)+10
Step-by-step explanation:
The coefficient of x in the cosine argument of the function will be 2π/period. Since the period is 5, the coefficient is 2π/5. This observation eliminates choices B and C.
The description "not a reflection of the parent function over the x-axis" means the multiplier of the cosine function is not negative, eliminating choice D.
The remaining choice A matches the description.
The number of tests that it would take for the probability of committing at least one type I error to be at least 0.7 is 118 .
In the question ,
it is given that ,
the probability of committing at least , type I error is = 0.7
we have to find the number of tests ,
let the number of test be n ,
the above mentioned situation can be written as
1 - P(no type I error is committed) ≥ P(at least type I error is committed)
which is written as ,
1 - (1 - 0.01)ⁿ ≥ 0.7
-(0.99)ⁿ ≥ 0.7 - 1
(0.99)ⁿ ≤ 0.3
On further simplification ,
we get ,
n ≈ 118 .
Therefore , the number of tests are 118 .
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Answer:-2.75
Step-by-step explanation: y+4=-4(x-5) y+4=-4x+5/4 y=-4x-2.75