The following histogram displays the distribution of battery life (in hours) for a certain battery model used in cell phones: suppose that battery life is a normal random variable with μ = 8 and σ = 1.2. using the standard deviation rule, what is the probability that a randomly chosen battery will last between 6.8 and 9.2 hours?

68.27%

Step-by-step explanation:

An empirical rule for a normal distribution states that 68.27% of data is within 1 standard deviation () of the mean (), 95.45% of the data is within 2 standard deviations of the mean, and 99.73% of the data is within 3 standard deviations of the mean.

The problem asks for the probability that a battery will last between 6.8 and 9.2 hours considering that =8 and =1.2.

So, if the distribution is normal, the 68.27% of data is within 1, that is between and , replacing the values we have

8-1.2=6.8 and 8+1.2=9.2

So, 68.27% of batteries will last between 6.8 and 9.2 hours

ad = ae + ed = y + 7

similar triangles therefor

ac/ab = ad/ae

6/4 = (y + 7) / y

cross multiply

6y = 4(y + 7)

6y = 4y + 28

2y = 28

y = 14

step-by-step explanation:

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: