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Svet_ta [14]
2 years ago
12

Find the percent change from the first value to the second 50;30

Mathematics
1 answer:
Alex787 [66]2 years ago
5 0

Answer:

-40%

Step-by-step explanation:

NV = new value

OV = old value

percent change = (NV - OV)/(OV) × 100%

percent change = (30 - 50)/(50) × 100%

percent change = -20/50 × 100%

percent change = -40%

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The lifetime of a battery in a certain application is normally distributed with mean μ = 16 hours and standard deviation σ = 2 h
DaniilM [7]

Answer:

Probability that a battery will last more than 19 hours is 0.0668.

Step-by-step explanation:

We are given that the lifetime of a battery in a certain application is normally distributed with mean μ = 16 hours and standard deviation σ = 2 hours.

<em>Let X = lifetime of a battery in a certain application</em>

So, X ~ N(\mu=16,\sigma^{2} =2^{2})

The z-score probability distribution for normal distribution is given by;

               Z = \frac{  X -\mu}{\sigma}  ~ N(0,1)

where, \mu = mean lifetime = 16 hours

            \sigma = standard deviation = 2 hours

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, the probability that a battery will last more than 19 hours is given by = P(X > 19 hours)

  P(X > 19) = P( \frac{  X -\mu}{\sigma} > \frac{19-16}{2} ) = P(Z > 1.50) = 1 - P(Z \leq 1.50)

                                              = 1 - 0.9332 = 0.0668

<em>Now, in the z table the P(Z </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1.50 in the z table which has an area of 0.9332.</em>

Hence, the probability that a battery will last more than 19 hours is 0.0668.

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