1/9,1/3,5/9 find the rule

Given the functions f(x) = 7x + 13 and g(x) = x^2 + 2, which of the following functions represents f[g(x)] correctly?

Paladinen [302]

Answer:

f(g(x)) = x^2 + 27.

Step-by-step explanation:

To find f(g(x)) we replace the x in f(x) by g(x) and simplify:

f(g(x)) = 7(x^2 + 2) + 13

= 7x^2 + 14 + 13.

= 7x^2 + 27.

8 0

3 years ago

Convert -36 degrees to radians

dimulka [17.4K]

-36 degrees = 0.62831853 radians

3 0

2 years ago

Read 2 more answers

Hi need help for this maths question

Sauron [17]

a) If f(y) is a probability density function, then both f(y) ≥ 0 for all y in its support, and the integral of f(y) over its entire support should be 1. eˣ > 0 for all real x, so the first condition is met. We have

$\displaystyle \int_{-\infty}^\infty f(y) \, dy = \frac14 \int_0^\infty e^{-\frac y4} \, dy = -\left(\lim_{y\to\infty}e^{-\frac y4} - e^0\right) = \boxed{1}$

so both conditions are met and f(y) is indeed a PDF.

b) The probability P(Y > 4) is given by the integral,

$\displaystyle \int_{-\infty}^4 f(y) \, dy = \frac14 \int_0^4 e^{-\frac y4} \, dy = -\left(e^{-1} - e^0\right) = \frac{e - 1}{e} \approx \boxed{0.632}$

c) The mean is given by the integral,

$\displaystyle \int_{-\infty}^\infty y f(y) \, dy = \frac14 \int_0^\infty y e^{-\frac y4} \, dy$

Integrate by parts, with

$u = y \implies du = dy$

$dv = e^{-\frac y4} \, dy \implies v = -4 e^{-\frac y4}$

Then

$\displaystyle \int_{-\infty}^\infty y f(y) \, dy = \frac14 \left(\left(\lim_{y\to\infty}\left(-4y e^{-\frac y4}\right) - \left(-4\cdot0\cdot e^0\right)\right) + 4 \int_0^\infty e^{-\frac y4} \, dy\right)$

$\displaystyle \cdots = \int_0^\infty e^{-\frac y4} \, dy$

$\displaystyle \cdots = -4 \left(\lim_{y\to\infty} e^{-\frac y4} - e^0\right) = \boxed{4}$

8 0

1 year ago

The amounts of electricity bills for all households in a city have a skewed probability distribution with a mean of $139 and a

ZanzabumX [31]

Answer:

P (within $6 of 4) = 0.9164

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions 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:

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

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

Mean of $139 and a standard deviation of $30.

This means that $\mu = 139, \sigma = 30$

Random sample of 75 households

This means that $n = 75, s = \frac{30}{\sqrt{75}} = 3.464$

75 > 30, which means that the sampling distribution is approximately normal.

Find the probability that the mean amount of electric bills for a random sample of 75 households selected from this city will be within $6 of the population mean.

This is the pvalue of Z when X = 139 + 6 = 145 subtracted by the pvalue of Z when X = 139 - 6 = 133.

X = 145

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$