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rusak2 [61]
3 years ago
8

Evaluate the expression \dfrac{x^5}{x^2} x 2 x 5 ​ start fraction, x, start superscript, 5, end superscript, divided by, x, squa

red, end fraction for x=2x=2x, equals, 2.
Mathematics
1 answer:
Elza [17]3 years ago
7 0

Answer:

  8

Step-by-step explanation:

Fill in the variable value and do the arithmetic.

  \dfrac{2^5}{2^2}=\dfrac{32}{4}=8

___

Of course, the fraction can be simplified first:

  \dfrac{x^5}{x^2}=x^{5-2}=x^3\\\\2^3=8

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Which is the solution to the inequality y+15 <3?
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3 years ago
What is the probability that a value falls in the interval μ ± 2σ, that computes the probability P(-2 &lt; Z &lt; 2). Round to f
bija089 [108]

Answer:

0.9544    

Step-by-step explanation:

P(-2 < Z < 2) means that Z has mean 0 and standard deviation 2.

P(−2 < Z < 2) = F(2) - F(-2)

Using the Z - table,

F(2) = 0.9772

and F(-2) = 0.0228

Thus,

P(−2 < Z < 2) = 0.9772 - 0.0228 = 0.9544

This means that data within two standard deviation is 95%.

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Read 2 more answers
When fishing off the shores of Florida, a spotted seatrout must be between 10 and 30 inches long before it can be kept; otherwis
Mademuasel [1]

Answer:

There is a 97.585% probability that a fisherman catches a spotted seatrout within the legal limits.

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}

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

The lengths of the spotted seatrout that are caught, are normally distributed with a mean of 22 inches, and a standard deviation of 4 inches, so \mu = 22, \sigma = 4.

What is the probability that a fisherman catches a spotted seatrout within the legal limits?

They must be between 10 and 30 inches.

So, this is the pvalue of the Z score of X = 30 subtracted by the pvalue of the Z score of X = 10

X = 30

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

Z = \frac{30 - 22}{4}

Z = 2

Z = 2 has a pvalue of 0.9772

X = 10

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

Z = \frac{10 - 22}{4}

Z = -3

Z = -3 has a pvalue of 0.00135.

This means that there is a 0.9772-0.00135 = 0.97585 = 97.585% probability that a fisherman catches a spotted seatrout within the legal limits.

3 0
3 years ago
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