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3 years ago

Find g(x) where g(x) is the translation 3 units right of f(x)=x^2

Mathematics

1 answer:

Fed [463]3 years ago

6 0

Answer:

g(x) = x^2 -6x +9

Step-by-step explanation:

A function f(x) translated right h units and up k units will become ...

g(x) = f(x -h) +k

You want the function f(x) = x^2 to be translated right h=3 units, so it will become ...

g(x) = f(x -3) = (x -3)^2

g(x) = x^2 -6x +9

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A right rectangular prism is packed with cubes of side length fraction 1 over 5 inch. If the prism is packed with 15 cubes along

Ann [662]

Given:

A right rectangular prism is packed with cubes of side length $\dfrac{1}{5}$ inch.

Prism is packed with 15 cubes along the length, 6 cubes along the width, and 5 cubes along the height.

To find:

The volume of the prism

Solution:

Side length of each cube $=\dfrac{1}{5}$ inch.

15 cubes along the length, so length of prism is

$l=15\times \dfrac{1}{5}=3\text{ inches}$

6 cubes along the width, so width of prism is

$b=6\times \dfrac{1}{5}=\dfrac{6}{5}\text{ inches}$

5 cubes along the height, so height of prism is

$h=5\times \dfrac{1}{5}=1\text{ inch}$

Now,

Volume of rectangular prism is

$V=l\times b\times h$

$V=3\times \dfrac{6}{5}\times 1$

$V=\dfrac{18}{5}$

$V=\dfrac{3\times 5+3}{5}$

$V=3\dfrac{3}{5}\text{ cubic inches}$

Therefore, the correct option is B.

7 0

3 years ago

A fair coin is to be tossed 20 times. Find the probability that 10 of the tosses will fall heads and 10 will fall tails, (a) usi

lbvjy [14]

Using the distributions, it is found that there is a:

a) 0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

b) 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

c) 0.1742 = 17.42% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item a:

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

20 tosses, hence $n = 20$ .
Fair coin, hence $p = 0.5$ .

The probability is P(X = 10), thus:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{20,10}.(0.5)^{10}.(0.5)^{10} = 0.1762$

0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item b:

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.
The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

The probability of an exact value is 0, hence 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item c:

For the approximation, the mean and the standard deviation are:

$\mu = np = 20(0.5) = 10$

$\sigma = \sqrt{np(1 - p)} = \sqrt{20(0.5)(0.5)} = \sqrt{5}$

Using continuity correction, this probability is $P(10 - 0.5 \leq X \leq 10 + 0.5) = P(9.5 \leq X \leq 10.5)$ , which is the p-value of Z when X = 10.5 subtracted by the p-value of Z when X = 9.5.