Here we must calculate what percentage of the whole price is represented by the discount amount, which is the difference between the original price and the final one. Hence we have: DiscountedAmount = 440 - 154 = 286. And knowing that 440 constitutes 100% of the price, and 286 constitutes x% of it, we get that x = 286/440 <=> x = 65/100 or 65%. So the answer is D.

3 0

3 years ago

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The lifetime of a certain type of battery is normally distributed with mean value 15 hours and standard deviation 1 hour. There

Scorpion4ik [409]

Answer:

33.68 hours

Step-by-step explanation:

The lifetime of a certain type of battery is normally distributed with mean value 15 hours and standard deviation 1 hour then

The lifetime of a battery package consisted of 4 batteries is normally distributed with mean value 60(15×4) hours and standard deviation 16 (1 × $4^{2}$ ) hour

If the total lifetime of all batteries in a package exceeds lifetime value for only 5% of all packages then P(z<Z)=0.05 where

Z is the corresponding z-score for the lifetime value that exceeds 5% of the all packages' lifetime value.

From z-table we can conclude that Z=-1.645

Z is also calculated as:

Z=-1.645= $\frac{X-M}{s}$ where

X is the cut-off time for exceeding 5%

M is the mean lifetime of packages (60)

s is the standard deviation of package lifetime values (16)

Solving the equation for X we get X=33.68

5 0

3 years ago