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3 years ago

If f(x) = -x³ + x² + x, find f(-3).

Mathematics

2 answers:

yan [13]3 years ago

5 0

Answer:

Step-by-step explanation:

Substitute f(-3) in the problem and you'll get f(-3) = -x³ + x² + x.

Substitute the given value into the function and evaluate.

joja [24]3 years ago

5 0

Answer:

33.

Step-by-step explanation:

f(x) = -x³ + x² + x.

f(-3) = -(-3)³ + (-3)² + (-3)

= -(-27) + (9) - 3

= 27 + 9 - 3

= 36 - 3

= 33.

Hope this helps!

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saveliy_v [14]

Answer:

$\large\boxed{m=10.8=10\dfrac{4}{5}}$

Step-by-step explanation:

$\dfrac{5}{8}m=6\dfrac{3}{4}\qquad\text{convert the mixed number to an improper fraction}\\\\\dfrac{5m}{8}=\dfrac{6\cdot4+3}{4}\\\\\dfrac{5m}{8}=\dfrac{27}{4}\qquad\text{cross multiply}\\\\(5m)(4)=(8)(27)\\\\20m=216\qquad\text{divide both sides by 20}\\\\\dfrac{20m}{20}=\dfrac{216}{20}\\\\m=10.8$

$\bold{Check:}\\\\\dfrac{5}{8}\cdot10.8=\dfrac{5\!\!\!\!\diagup^1}{8}\cdot\dfrac{108}{10\!\!\!\!\!\diagup_2}=\dfrac{108:4}{16:4}=\dfrac{27}{4}=\dfrac{24+3}{4}=\dfrac{24}{4}+\dfrac{3}{4}=6\dfrac{3}{4}\\\\\bold{CORRECT}$

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3 years ago

In the 1992 presidential election, Alaska’s 40 election districts averaged 1956.8 votes per district for President Clinton. The

torisob [31]

Answer:

(a) The distribution of X is N (1956.8, 90.49²).

(b) The value 1956.8 is the population mean.

(d) The probability that a randomly selected district had between 1,800 and 2,000 votes for President Clinton is 0.6426.

(e) The third quartile for votes for President Clinton is 2018.3.

Step-by-step explanation:

The random variable X is defined the number of votes for President Clinton for an election district.

The information provided is:

$n=40\\\bar x=1956.8\\\sigma=572.3$

(a)

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

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

$\mu_{x}=\mu = \bar x=1956.8$

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

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{572.3}{\sqrt{40}}=90.49$

Thus, the distribution of X is N (1956.8, 90.49²).

(b)

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

$\mu_{x}=1956.8$

The mean value, 1956.8 is the population mean of the sampling distribution of sample mean.

This is because the law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample ( $\bar x$ ) approaches the whole population mean ( $\mu_{x}$ ).