Can someone aswer this Please

Im confused, help please?

professor190 [17]

I think it’s g(0)=0, g(1)=1, and g(-1)=1

6 0

3 years ago

Read 2 more answers

Identity Structure find the rule for each function table.

Andreyy89

Answer:

The rule for given table is y = 4x - 1.

Step-by-step explanation:

From the given table it is noticed that the function passing through the points (1,3), (2,7) and (3,11).

If a line passing through the two points, then the equation of line is

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

The equation of line is

$y-3=\frac{7-3}{2-1}(x-1)$

$y-3=\frac{4}{1}(x-1)$

$y-3=4(x-1)$

$y-3=4x-4$

Add 3 both sides.

$y=4x-4+3$

$y=4x-1$

Therefore the rule for given table is y = 4x - 1.

6 0

3 years ago

Read 2 more answers

Using what you know about similar figure, solve for X

ale4655 [162]

Since the scale factor is 3:1 you can multiply the length of the side of the triangle to the right by 3 to make the ratio 3:3

the equation would be:
2x+5 = 45

subtract 5 from both sides
2x = 40

divide both sides by 2

x = 20

hope this helps

3 0

3 years ago

The distance between flaws on a long cable is exponentially distributed with mean 12 m.

Elden [556K]

Answer:

(a) The probability that the distance between two flaws is greater than 15 m is 0.2865.

(b) The probability that the distance between two flaws is between 8 and 20 m is 0.3246.

(c) The median is 8.322.

(d) The standard deviation is 12.

(e) The 65th percentile of the distances is 12.61 m.

Step-by-step explanation:

The random variable X can be defined as the distance between flaws on a long cable.

The random variable X is exponentially distributed with mean, μ = 12 m.

The parameter of the exponential distribution is:

$\lambda=\frac{1}{\mu}=\frac{1}{12}=0.0833$

The probability density function of X is:

$f_{X}(x)=0.0833e^{-0.0833x};\ x\geq 0$

(a)

Compute the probability that the distance between two flaws is greater than 15 m as follows:

$P(X\geq15)=\int\limits^{\infty}_{15}{0.0833e^{-0.0833x}}\, dx\\=0.0833\times \int\limits^{\infty}_{15}{e^{-0.0833x}}\, dx\\=0.0833\times |\frac{e^{-0.0833x}}{-0.0833}|^{\infty}_{15}\\=e^{0.0833\times 15}\\=0.2865$

Thus, the probability that the distance between two flaws is greater than 15 m is 0.2865.

(b)

Compute the probability that the distance between two flaws is between 8 and 20 m as follows:

$P(8\leq X\leq20)=\int\limits^{20}_{8}{0.0833e^{-0.0833x}}\, dx\\=0.0833\times \int\limits^{20}_{8}{e^{-0.0833x}}\, dx\\=0.0833\times |\frac{e^{-0.0833x}}{-0.0833}|^{20}_{8}\\=e^{0.0833\times 8}-e^{0.0833\times 20}\\=0.51355-0.1890\\=0.32455\\\approx0.3246$

Thus, the probability that the distance between two flaws is between 8 and 20 m is 0.3246.

(c)

The median of an Exponential distribution is given by:

$Median=\frac{\ln (2)}{\lambda}$

Compute the median as follows:

$Median=\frac{\ln (2)}{\lambda}$

$=\farc{0.69315}{0.08333}\\=8.322$

Thus, the median is 8.322.

(d)

The standard deviation of an Exponential distribution is given by:

$\sigma=\sqrt{\frac{1}{\lambda^{2}}}$

Compute the standard deviation as follows:

$\sigma=\sqrt{\frac{1}{\lambda^{2}}}$

$=\sqrt{\frac{1}{0.0833^{2}}}\\=12.0048\\\approx 12$

Thus, the standard deviation is 12.

(e)

Let x be 65th percentile of the distances.

Then, P (X < x) = 0.65.

Compute the value of x as follows:

$\int\limits^{x}_{0}{0.0833e^{-0.0833x}}\, dx=0.65\\0.0833\times \int\limits^{x}_{0}{e^{-0.0833x}}\, dx=0.65\\0.0833\times |\frac{e^{-0.0833x}}{-0.0833}|^{x}_{0}=0.65\\-e^{-0.0833x}+1=0.65\\-e^{-0.0833x}=-0.35\\-0.0833x=-1.05\\x=12.61$

Thus, the 65th percentile of the distances is 12.61 m.

4 0

3 years ago

Please help!!!!! I’ve been stuck for like hours

BARSIC [14]

$10^2 + b^2=15^2\\\\100+b^2=225\\\\b^2=125\\\\b =\sqrt{125}\\\\b \approx \boxed{11.2}$

6 0

2 years ago