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3 years ago

F(x)=x2−3x+9f(x)=x2−3x+9

Mathematics

1 answer:

shepuryov [24]3 years ago

5 0

You have copied double of each function, so
(f-g)(x) = x^2-3x+9-3x^3-x^2+4x+9

= a. −3x3−x2+x+18

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Use the confidence level and sample data to find a confidence interval for estimating the population μ. Round your answer to the

Levart [38]

Answer: Choice A.) 88.2 < μ < 93.0

=============================================================

Explanation:

We have this given info:

n = 92 = sample size
xbar = 90.6 = sample mean
sigma = 8.9 = population standard deviation
C = 99% = confidence level

Because n > 30 and because we know sigma, this allows us to use the Z distribution (aka standard normal distribution).

At 99% confidence, the z critical value is roughly z = 2.576; use a reference sheet, table, or calculator to determine this.

The lower bound of the confidence interval (L) is roughly

L = xbar - z*sigma/sqrt(n)

L = 90.6 - 2.576*8.9/sqrt(92)

L = 88.209757568781

L = 88.2

The upper bound (U) of this confidence interval is roughly

U = xbar + z*sigma/sqrt(n)

U = 90.6 + 2.576*8.9/sqrt(92)

U = 92.990242431219

U = 93.0

Therefore, the confidence interval in the format (L, U) is approximately (88.2, 93.0)

When converted to L < μ < U format, then we get approximately 88.2 < μ < 93.0 which shows that the final answer is choice A.

We're 99% confident that the population mean mu is somewhere between 88.2 and 93.0

7 0

3 years ago

someone, please help me asap, that would be great, offering 15 points, and please do not attach a link for the answer, thanks.

algol13

Pretty sure the answer is 288

7 0

3 years ago

. A recent report in a women magazine stated that the average age for women to marry in the United States is now 25 years of age

Setler [38]

Answer: 0.0136

Step-by-step explanation:

Given : Mean : $\mu=25$

Standard deviation : $\sigma=3.2$

Sample size : $n=50$

The formula to calculate the z-score :-

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

For x = 24

$z=\dfrac{24-25}{\dfrac{3.2}{\sqrt{50}}}=-2.21$

The p-value = $P(z\leq-2.21)= 0.0135526\approx0.0136$

Hence, the probability that the sample mean age for 50 randomly selected women to marry is at most 24 years = 0.0136

3 0

3 years ago

According to the article "Characterizing the Severity and Risk of Drought in the Poudre River, Colorado" (J. of Water Res. Plann

mihalych1998 [28]

Answer:

(a) P (Y = 3) = 0.0844, P (Y ≤ 3) = 0.8780

(b) The probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

Step-by-step explanation:

The random variable Y is defined as the number of consecutive time intervals in which the water supply remains below a critical value y₀.

The random variable Y follows a Geometric distribution with parameter p = 0.409.

The probability mass function of a Geometric distribution is:

$P(Y=y)=(1-p)^{y}p;\ y=0,12...$

(a)

Compute the probability that a drought lasts exactly 3 intervals as follows:

$P(Y=3)=(1-0.409)^{3}\times 0.409=0.0844279\approx0.0844$

Thus, the probability that a drought lasts exactly 3 intervals is 0.0844.

Compute the probability that a drought lasts at most 3 intervals as follows:

P (Y ≤ 3) = P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3)

$=(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409+(1-0.409)^{2}\times 0.409\\+(1-0.409)^{3}\times 0.409\\=0.409+0.2417+0.1429+0.0844\\=0.8780$

Thus, the probability that a drought lasts at most 3 intervals is 0.8780.

(b)

Compute the mean of the random variable Y as follows:

$\mu=\frac{1-p}{p}=\frac{1-0.409}{0.409}=1.445$

Compute the standard deviation of the random variable Y as follows:

$\sigma=\sqrt{\frac{1-p}{p^{2}}}=\sqrt{\frac{1-0.409}{(0.409)^{2}}}=1.88$

The probability that the length of a drought exceeds its mean value by at least one standard deviation is:

P (Y ≥ μ + σ) = P (Y ≥ 1.445 + 1.88)

= P (Y ≥ 3.325)

= P (Y ≥ 3)

= 1 - P (Y < 3)

= 1 - P (X = 0) - P (X = 1) - P (X = 2)

$=1-[(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409\\+(1-0.409)^{2}\times 0.409]\\=1-[0.409+0.2417+0.1429]\\=0.2064$

Thus, the probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

6 0

3 years ago

-3r+9=-4r+5 solve please help me

timofeeve [1]

Answer:

r=-4

make r the subject and solve

4 0

3 years ago