An accelerated life test on a large number of type-d alkaline batteries revealed that the mean life for a particular use before they failed is 19.0 hours. the distribution of the lives approximated a normal distribution. the standard deviation of the distribution was 1.2 hours. about 95% of the batteries failed between what two values?

Step-by-step explanation:

Let X be the accelerated life test on a large number of type-D alkaline batteries revealed that the mean life for a particular use before they failed

X is NOrmal with mean =19, sigma = 1.2

For 95% critical value = ±1.96

Margin of error = ±1.96*1.2

=±2.352

Confidence interval 95% = Mean ±2.352

=(16.648, 21.352)

20

step-by-step explanation:

x > 1.3

step by step:

8.5−0.9x> 9.8−1.9x

step 1: simplify both sides of the inequality.

−0.9x+8.5> −1.9x+9.8

step 2: add 1.9x to both sides.

−0.9x+8.5+1.9x> −1.9x+9.8+1.9x

1x+8.5> 9.8

step 3: subtract 8.5 from both sides.

1x+8.5−8.5> 9.8−8.5

1x> 1.3

step 4: divide both sides by 1.

1x/1 > 1.3/1