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Lady bird [3.3K]
3 years ago
8

Solve the system. x+3y=1 -2x+7=-23

Mathematics
1 answer:
Viefleur [7K]3 years ago
5 0
X+ 3y = 1  -----(first equation)
-2x+7 = -23 ---(second equation)
-2x = -23-7
x = 30/2
x = 15

Now, Substitute the value of x in first equation,
15 + 3y = 1
3y = 1 - 15
y = -14/3

In short, Final Answers are: x = 15 & y = -14/3

Hope this helps!
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Simplify complex fraction 3/4 + 1/3/1/6 + 3/4
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A flashlight battery manufacturer makes a model of battery whose mean shelf life is three years and four months, with a standard
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Answer:

17,065 of those batteries can be expected to last between three years and one month and three years and seven months

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem:

I will calculate the time in months. Each year has twelve months.

Mean shelf life is three years and four months, with a standard deviation of three months. So

\mu = 3*12 + 4 = 40

\sigma = 3

Proportion lasting between three years and one month and three years and seven months:

This is the pvalue of Z when X = 3*12 + 7 = 43 subtracted by the pvalue of Z when X = 3*12 + 1 = 37

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Z = \frac{X - \mu}{\sigma}

Z = \frac{43 - 40}{3}

Z = 1

Z = 1 has a pvalue of 0.8413.

X = 37

Z = \frac{X - \mu}{\sigma}

Z = \frac{37 - 40}{3}

Z = -1

Z = -1 has a pvalue of 0.1587.

0.8413 - 0.1587 = 0.6826

Out of 25,000 batteries:

68.26% of the batteries are expected to last between three years and one month and three years and seven months.

0.6826*25000 = 17,065

17,065 of those batteries can be expected to last between three years and one month and three years and seven months

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