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IceJOKER [234]
3 years ago
15

Choose the inverse of y = x2 - 10x.

Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
3 0

=Y2-10Y

We move all terms to the left:

-(Y2-10Y)=0

We add all the numbers together, and all the variables

-(+Y^2-10Y)=0

We get rid of parentheses

-Y^2+10Y=0

We add all the numbers together, and all the variables

-1Y^2+10Y=0

a = -1; b = 10; c = 0;

Δ = b2-4ac

Δ = 102-4·(-1)·0

Δ = 100

The delta value is higher than zero, so the equation has two solutions

We use following formulas to calculate our solutions:

Y1=−b−Δ√2aY2=−b+Δ√2a

Δ‾‾√=100‾‾‾‾√=10

Y1=−b−Δ√2a=−(10)−102∗−1=−20−2=+10

Y2=−b+Δ√2a=−(10)+102∗−1=0−2=0

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A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. ​(a) What must be true
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Answer:

(a) The distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b) The value of P(\bar X is 0.7642.

(c) The value of P(\bar X\geq 69.1) is 0.3670.

Step-by-step explanation:

A random sample of size <em>n</em> = 10 is selected from a population.

Let the population be made up of the random variable <em>X</em>.

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(a)

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. <em>n</em> = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

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And the standard deviation of the distribution of the sample mean is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74

Thus, the distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b)

Compute the value of P(\bar X as follows:

P(\bar X

                    =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the value of P(\bar X is 0.7642.

(c)

Compute the value of P(\bar X\geq 69.1) as follows:

Apply continuity correction as follows:

P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)

                    =P(\bar X>69.6)

                    =P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})

                    =P(Z>0.34)\\=1-P(Z

Thus, the value of P(\bar X\geq 69.1) is 0.3670.

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