Answer:
£1837.5
Step-by-step explanation:
Given data
Cost of car P= £2100.
Rate r= 2.2%
Time t= 6 years
Now we want to find the worth after 6 years, let us apply the compound interest expression but this time for depreciation
A= P(1-r)^t
Substitute
A= 2100(1-0.022)^6
A= 2100*(0.978)^6
A= 2100*0.875
A= £1837.5
Hence the amount of the car after 6 years is £1837.5
Answer:
slope: 5/2
Step-by-step explanation:
Answer:
<h2>A, B, D, E</h2>
Step-by-step explanation:
2(2x + 1) = (2)(2x) + (2)(1) = 4x + 2 → A
<em>used distributive property</em>
2(2x + 1) = 2(1 + 2x) → B
<em>used commutative property</em>
2(2x + 1) = (2x + 1) + (2x + 1) = 2x + 1 + 2x + 1 → D
<em>used 2a = a + a</em>
2(2x + 1) = 4x + 2 = x + x + x + x + 1 + 1 → E
<em>used 4x = x + x + x + x and 2 = 1 + 1</em>
Answer:
a) 0.6636
b) 0.0951
c) 0,9474
d) 0.0047
e) 0.9957
f) 0.1308
Step-by-step explanation:
We look in tables z values and then we see carefully aereas inside normal curve
a) P[- 1.46 < z < 0.63 ] point 1.46 from table 0.0721 this s th area from value -1.46 to the left . And the value z = 0.63 corresond to the area 0.7357 which includes the area between 1.46 to the left tail, then we have to subtarct and get 0.6636 .
P[- 1.46 < z < 0.63 ] = 0.6636 66.36 %
b) P [ 0 < z < 1.31 ] we just need the area for point 1.31 that is 0.0951
P [ 0 < z < 1.31 ] = 0.0951 9.51 %
c) P [z > - 1.62 ] = 1 - 0.0526
P [z > - 1.62 ] = 0,9474 94.74 %
d) P[z < - 2.6 ] = 0.0047 0.47 %
e) P [ z < 2.63 ] = 1 - 0.0043
P [ z < 2.63 ] = 0.9957 99.57 %
f) P [ -2.58 < z < -1.1 ] = 0.1357 - 0.0049 =
P [ -2.58 < z < -1.1 ] = 0.1308 13.08 %
Answer:
y = - 35
Step-by-step explanation:
Expressing as a fraction , that is
= 5 ( multiply both sides by - 7 to clear the fraction )
y = - 35
