Answer:
x < 3 and x > -1
Step-by-step explanation:
Step 1: Subtract 2 from both sides.
Step 2: Solve absolute value.
- We know x - 1 < 2 and x - 1 > -2.
Condition 1:
Condition 2:
Therefore, the answer is x < 3 and x > -1.
Answer:
Option A and C is correct.
Step-by-step explanation:
Discount is defined as a reduced price on something being sold or at a price lower than that item is normally sold for.
For a 20% discount,
Given:
Initial prices = $ d
Discounted price = % discount × original/initial cost
= 20/100 × d
= 0.2 × d
Selling price = original cost - discounted price
= d - 0.2d
= 0.8 × d
= 0.8d
Answer:
75
Step-by-step explanation:
Answer:
x = 0.25
Step-by-step explanation:
When logs are added together, they are actually multiplied and then the logs taken of the product.
That sentence is actually correct, but you are going to have to read it a couple of times. You might understand it if I actually just solve the problem.
ln(2x) + ln(2) = 0 Combine the two subjects to make 1 ln.
ln (2)(2x) = 0 Now take the antilog
ln(4x) = 0
antilog ln(4x) = e^0 e^0 = 1
4x = 1 See your last problem.
x = 1/4
Now the question is "What's the answer?" It might be 1/4 but I doubt it. A better choice would be x = 1/4 or x = 0.25
I'd try the last one first.
Answer:
The Answer is 76.
Step-by-step explanation:
Given the normal distribution " 10% of employees (rated) exemplary, 20% distinguished, 40% competent, 20% marginal, and 10% unacceptable'', we can see that exemplary employees are top 10% rated employees.
We have the formula for normal distribution:
z=(X-M)÷σ
where z is the <em>minimum z-score </em>for top 10% employee, X is the <em>minimum </em>score for top 10% employee, M is the <em>mean</em> of the score distribution, σ is the <em>standard deviation</em> of the score distribution.
The z-score we are looking for is the value "a" that separates the highest 10% from the lowest 90% i.e. P(z≤a)=0.90
If we look at z-table, corresponding value for a is 1.28155
We can now put the values in the formula:
1.28155=
So X=(1.28155×20)+50=75.631
Therefore minimum score for exemplary employee is 76.