Given : Monitors manufactured by TSI Electronics have life spans that have a normal distribution with a variance of $\sigma^2=2,250,000$ and a mean life span of $\mu=13,000$ hours.

Here , $\sigma=\sqrt{2250000}=1500$

Let x represents the life span of a monitor.

Then , the probability that the life span of the monitor will be more than 14,650 hours will be :-

$P(x>14650)=P(\dfrac{x-\mu}{\sigma}>\dfrac{14650-13000}{1500})\\\\=P(z>1.1)=1-P(z\leq1.1)\ \ [\because\ P(Z>z)=1-P(Z\leq z)]\\\\=1-0.8643339=0.1356661\approx0.1357$

Hence, the probability that the life span of the monitor will be more than 14,650 hours = 0.1357

6 0

3 years ago

25,000 bricks were used to build the new library. The total weight of the bricks was 30,000 pounds. What is the weight of each b

Lelechka [254]

I would go with 1.2 pounds

4 0

3 years ago

A dinner costs $89.75<br> Your share of the bill is 20%.<br> The best estimate of your share is:

dusya [7]

Answer:

17.95

Step-by-step explanation:

x over 20 89.75 over 100 then cross multiply

7 0

3 years ago