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pogonyaev
3 years ago
11

Two separate bacteria populations grow each month and are represented by the

Mathematics
1 answer:
ASHA 777 [7]3 years ago
5 0

Answer:

Option 4.) month 4

Step-by-step explanation:

we have

f(x)=3^{x} ----> equation A

g(x)=7x+6 ----> equation B

Solve the system by graphing

Remember that the solution is the intersection point both graphs

The solution is the point (3,27)

That means -----> For x=3 months, the populations in both functions are the same (the bacteria populations is 27)

therefore

For x> 3 months the f(x) population is greater then the g(x) population

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a=00110111 = 00110_2 \times 2^{111_2} = 6 \times 2^{-1} = 3


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4 0
4 years ago
A set of test scores is normally distributed with a mean of 130 and a standard deviation of 30 What score is necessary to reach
nlexa [21]

Answer:

A score of 150.25 is necessary to reach the 75th percentile.

Step-by-step explanation:

Problems of normal distributions 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.

A set of test scores is normally distributed with a mean of 130 and a standard deviation of 30.

This means that \mu = 130, \sigma = 30

What score is necessary to reach the 75th percentile?

This is X when Z has a pvalue of 0.75, so X when Z = 0.675.

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

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X - 130 = 0.675*30

X = 150.25

A score of 150.25 is necessary to reach the 75th percentile.

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