In a 45-45-90 right triangle, what is the ratio of the length of one leg to the

Please help! I'm not sure on what I do.

Alisiya [41]

Answer:

slope= 6.25

Step-by-step explanation:

the slope represents the meters he travels per second

5 0

3 years ago

Read 2 more answers

Mrs. Hernandez has 12 girls and 13 boys in her 2nd grade class. If

spin [16.1K]

Answer:

It should be 12/25. But try 12/15

Step-by-step explanation:

Hope this helps.

5 0

3 years ago

A soccer game is 90 min long.36 min have passed what percentage of the game has passed?

telo118 [61]

The answer is 40%
Here is how I got it.
36/90* 100%= 2/5* 100%=2* 20%=40%

5 0

3 years ago

What is the equation of the line that passes through (0, 2) and (3, - 4)

andrey2020 [161]

I’m lookin for same thing bro

7 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