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Irina18 [472]
3 years ago
12

Use long division to find the quotient​ Q(x) and the remainder​ R(x) when​ P(x) is divided by​ d(x) and express​ P(x) in the for

m ​d(x) times •​Q(x)plus+​R(x).
​P(x=x^3 + 2x^2 - 11 x + 471
​ d(x)=x+9

​P(x)=​(x+9) (blank) + blank

Mathematics
1 answer:
lilavasa [31]3 years ago
4 0

Answer:

P(x)=(x+9)(x^2-7x+52)+3

Step-by-step explanation:

We are given that

P(x)=x^3+2x^2-11x+471

d(x)=x+9

We have to find the value of Q(x) and Remainder R(x).

Quotient, Q(x)=x^2-7x+52

Remainder,R(x)=3

We know that

Dividend=Divisor\times Quotient+Remainder

P(x)=(x+9)(x^2-7x+52)+3

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Answer:

50 from home to school and 55 from school to home.

Step-by-step explanation:

55 is 5 more then 50.

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3 years ago
An economist uses the price of a gallon of milk as a measure of inflation. She finds that the average price is $3.82 per gallon
salantis [7]

Answer:

(a) The standard error of the mean in this experiment is $0.052.

(b) The probability that the sample mean is between $3.78 and $3.86 is 0.5587.

(c) The probability that the difference between the sample mean and the population mean is less than $0.01 is 0.5754.

(d) The likelihood that the sample mean is greater than $3.92 is 0.9726.

Step-by-step explanation:

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

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

\mu_{\bar x}=\mu

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

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

n=40\\\mu=\$3.82\\\sigma=\$0.33

As <em>n</em> = 40 > 30, the distribution of sample mean is \bar X\sim N(3.82,\ 0.052^{2}).

(a)

The standard error is the standard deviation of the sampling distribution of sample mean.

Compute the standard deviation of the sampling distribution of sample mean as follows:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

    =\frac{0.33}{\sqrt{40}}\\\\=0.052178\\\\\approx 0.052

Thus, the standard error of the mean in this experiment is $0.052.

(b)

Compute the probability that the sample mean is between $3.78 and $3.86 as follows:

P(3.78

                               =P(-0.77

Thus, the probability that the sample mean is between $3.78 and $3.86 is 0.5587.

(c)

If the difference between the sample mean and the population mean is less than $0.01 then:

\bar X-\mu_{\bar x}

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

P(\bar X

                    =P(Z

Thus, the probability that the difference between the sample mean and the population mean is less than $0.01 is 0.5754.

(d)

Compute the probability that the sample mean is greater than $3.92 as follows:

P(\bar X>3.92)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{3.92-3.82}{0.052})

                    =P(Z

Thus, the likelihood that the sample mean is greater than $3.92 is 0.9726.

3 0
3 years ago
1
myrzilka [38]

Answer:

sorry I can't answer

Step-by-step explanation:

pls thanks me

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denis-greek [22]

Let X be the random variable denoting the number of girls that are born. Then X has a binomial distribution on n=125 babies and probability of getting a girl p=0.5. The probability we want is

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serg [7]
1014 hope thts the awnser it seemed to obvious
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