The result can be shown in both exact and decimal forms.Exact Form:<span><span>−<span>14</span></span><span>-<span>14</span></span></span>Decimal Form:<span>−<span>0.25
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<span>The 1 in the hundred thousands place is the value of the middle of 1 and 4.</span>
Answer:
The worth of the car after 6 years is £2,134.82
Step-by-step explanation:
The amount at which Dan buys the car, PV = £2200
The rate at which the car depreciates, r = -0.5%
The car's worth, 'FV', in 6 years is given as follows;

Where;
r = The depreciation rate (negative) = -0.5%
FV = The future value of the asset
PV = The present value pf the asset = £2200
n = The number of years (depreciating) = 6
By plugging in the values, we get;

The amount the car will be worth which is its future value, FV after 6 years is FV ≈ £2,134.82 (after rounding to the nearest penny (hundredth))
Answer:
-9 x 1 x 1 x 1 x 2 = (-18)
-9 + 1 + 1 + 1 + 2 = (-4)
Answer:
H0: μ ≤ 1.30
H1: μ > 1.30
|Test statistic | > 1.833 ; Reject H0
Test statistic = 3.11
Yes
Pvalue = 0.006
Step-by-step explanation:
H0: μ ≤ 1.30
H1: μ > 1.30
Samples, X ; 1.36,1.35,1.33, 1.66, 1.58, 1.32, 1.38, 1.42, 1.90, 1.54
Xbar = 14.84 / 10 = 1.484
Standard deviation, s = 0.187 (calculator)
Decison rule :
|Test statistic | > TCritical ; reject H0
df = n - 1 = 10 - 1 = 9
Tcritical(0.05; 9) = 1.833
|Test statistic | > 1.833 ; Reject H0
Test statistic :
(xbar - μ) ÷ (s/√(n))
(1.484 - 1.30) ÷ (0.187/√(10))
0.184 / 0.0591345
Test statistic = 3.11
Since ;
|Test statistic | > TCritical ; We reject H0 and conclude that water consumption has increased
Pvalue estimate using the Pvalue calculator :
Pvalue = 0.006