What is greater three sets of 5-2 or two sets of 2+3? Draw diagrams

What are the vertex and x-intercepts of the graph of the function given below?

Travka [436]

In order to determine the vertex of this, you can complete the square. To do that, first set the equation equal to 0, then move the -35 over to the other side by adding. That gives us $x^{2} -2x=35$ . Now we can complete the square. Do this by taking half of the linear term, squaring it, and adding it in to both sides. Our linear term is 2x. Half of 2 is 1, and 1 squared is 1. So we add 1 to both sides, creating something that looks like this: $x^{2} -2x+1=35+1$ . We will do the math on the right and get 36, and the left will be expressed as the perfect square binomial we created by doing this whole process. $(x-1)^2=36$ . Now move the 36 over by subtraction and set it back to equal y and your vertex is apparent. It is (1, -36). You find the x-intercepts when y = 0. That means you need to set your original equation equal to zero and factor it. The easiest, surest way to do this is to use the quadratic formula. Doing that gives us x values of 7 and -5. And you're done!

7 0

2 years ago

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Stacey and Michele both babysit in the evening.staceys hours rate is $10 for one child plus $2per additional child Michael hourl

lara31 [8.8K]

They have two children...
8 + 3c = 10 + 2c
c = 2

7 0

3 years ago

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Help me please !! Tell me the algebraic expression to 1 please !!!

Anuta_ua [19.1K]

You need to distribute the 4 to both the n and the positive 5 so you’re answer would be 4n + 20.

5 0

3 years ago

C = 12d for d solve the equation for the indicated variable

kkurt [141]

Answer:

c/12=d

Step-by-step explanation:

what is the indicated variable?

6 0

2 years ago

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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

2 years ago