For this case we have the following equation:
Step 1 of the solution is to make the distributive property.
We have then:
Step 1 is correct.
The step is to combine similar terms. Caleb wrote:
He wrongly combined the similar terms on the right side of the equation.
The correct solution is:
Answer: In step 2, the like terms were not combined on the right side of the equation.
C
given 1 /( 3 + √2 )
then rationalise the denominator by multiplying the numerator/ denominator by the conjugate of the denominator
the conjugate of 3 + √2 is 3 - √2
1 / (3 + √2 ) × (3 - √2 ) / (3 - √2 )
= (3 - √2 ) /( 9 - 2 ) = (3 - √2 ) / 7
Answer:
t= 12.9 years
Step-by-step explanation:
Value after t years = initial value ( 1 - r )^t
Where,
Value after t years= $5000
Initial value = $22,400
r= depreciation rate = 11%
t= length of time (years)
Value after t years = initial value ( 1 - r )^t
5000 = 22,400 ( 1 - 0.11)^t
5000 = 22,400(0.89)^t
Divide both sides by 22,400
(0.89)^t = 5000 / 22,400
(0.89)^t = 0.2232
Take the log of both sides
t log 0.89 = log 0.2232
t= log 0.2232 / log 0.89
= -0.6513 / -0.0506
= 12.87
t= 12.9 years
Answer:
Step-by-step explanation:
Suppose the time required for an auto shop to do a tune-up is normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - u)/s
Where
x = points scored by students
u = mean time
s = standard deviation
From the information given,
u = 102 minutes
s = 18 minutes
1) We want to find the probability that a tune-up will take more than 2hrs. It is expressed as
P(x > 120 minutes) = 1 - P(x ≤ 120)
For x = 120
z = (120 - 102)/18 = 1
Looking at the normal distribution table, the probability corresponding to the z score is 0.8413
P(x > 120) = 1 - 0.8413 = 0.1587
2) We want to find the probability that a tune-up will take lesser than 66 minutes. It is expressed as
P(x < 66 minutes)
For x = 66
z = (66 - 102)/18 = - 2
Looking at the normal distribution table, the probability corresponding to the z score is 0.02275
P(x < 66 minutes) = 0.02275
