Answer:
expression a
Step-by-step explanation:
The given expression is 15+0.25(d−1).
let suppose,
15 = a
0.25(d−1) = b
we get a + b
It clearly indicates the given expression is sum of two entities, we can exclude option b and option d.
Now we are left with option a and c, for that we have to evaluate the term b
b = 0.25(d−1) <u>that is the additional amount after d days</u>
Therefore, expression a is correct.
Answer:
6(4x-1) is equivalent to 24x-6 and (2*12x)-(2*3)
24x+6 is equivalent to 6(4x+1) and (6*4x)+(6*1)
(4*6x)+(4*2) is equivalent to 24x+8 and 4(6x+2)
Step-by-step explanation:
An equivalent expression is an expression which either simplified or factored is equal to it.
6(4x-1) is equivalent to 24x-6 and (2*12x)-(2*3)
24x+6 is equivalent to 6(4x+1) and (6*4x)+(6*1)
(4*6x)+(4*2) is equivalent to 24x+8 and 4(6x+2)
Answer:
Therefore, we use the linear depreciation and we get is 17222.22 .
Step-by-step explanation:
From Exercise we have that is boat $250,000.
The straight line depreciation for a boat would be calculated as follows:
Cost boat is $250,000.
For $95,000 Deep Blue plans to sell it after 9 years.
We use the formula and we calculate :
(250000-95000)/9=155000/9=17222.22
Therefore, we use the linear depreciation and we get is 17222.22 .
Answer:
(a) 0.06154
(b) 0.2389
(c) 0.6052
(d) 2478
Step-by-step explanation:
probability density function of the time to failure of an electronic component in a copier (in hours) is
P(x) = 1/1076e^−x/1076
λ = 1/1076
A) A component lasts more than 3000 hours before failure:
P(x>3000) = 1 − e^−3000/1076
= 0.06154
B) A component fails in the interval from 1000 to 2000 hours:
P(1000>x>2000) =1 − e^−2000/1076 − 1 +e^−1000/1076 = e^−1000/1076 − e^−2000/1076 = 0.3948 − 0.1559
= 0.2389
C) A component fails before 1000 hours:
P(x<1000) = 0.6052
D) The number of hours at which 10% of all components have failed:
e^−x/1076 = 0.1
= −x/1076
= ln(0.1)
x =(2.3026)×(1076)
x = 2478
Answer:
60 maths books
Step-by-step explanation:
<u>Let the following be:</u>
- Maths books = m
- Science books = s
- History books = h
<u>And we have ratios and sum:</u>
- m/s = 3/2
- s/h = 1/3
- m+s+h = 220
<u>From the ratios we get:</u>
<u>Considering the above in the sum of books:</u>
- m+s+h = 220
- 3s/2 + s + 3s = 220
- 3s + 2s + 6s = 2*220
- 11s = 440
- s = 440/11
- s = 40
<u>Number of maths books:</u>
