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3 years ago

please be serious

Mathematics

1 answer:

shusha [124]3 years ago

5 0

Answer:

y intercept is (0,5)

x intercept is (5,0)

Step-by-step explanation:

re arange the equation to y = -x + 5.

To find the y intercept, you plug in 0 for x and find out what the value of y is. To find the x intercept, you plug in 0 for y and find the value of x.

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Answer: 15
Step-step-explanation:

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2 years ago

Factor 28v^2-8v+36 Please help

pychu [463]

Answer:

<h2>36 divided by a Factor of 36 will equal another Factor of 36. How to find the Factors of 36 Since the Factors of 36 are all the numbers that you can evenly divide into 36, we simply divide 36 by all numbers up to 36 to see which ones result in an even quotient.</h2>

Step-by-step explanation:

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3 years ago

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Simplify the expression

SashulF [63]

Answer:

1/(c - d)

Step-by-step explanation:

Multiply numerator and denominator by cd, then factor the denominator and cancel the common factor. (The result is restricted to c+d ≠ 0.)

$\displaystyle\frac{\frac{1}{c}+\frac{1}{d}}{\frac{c}{d}-\frac{d}{c}}=\frac{d+c}{c^2-d^2}\\\\=\frac{d+c}{(c-d)(d+c)}=\frac{1}{c-d}$

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3 years ago

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Audrey, an astronomer is searching for extra-solar planets using the technique of relativistic lensing. Though there are believe

Stolb23 [73]

Answer:

Step-by-step explanation:

The model $N (t)$ , the number of planets found up to time $t$ , as a Poisson process. So, the $N (t)$ has distribution of Poison distribution with parameter $(\lambda t)$

The mean for a month is, $\lambda = \frac{1}{3}$ per month

$E[N(t)]= \lambda t\\\\=\frac{1}{3}(24)\\\\=8$

(Here. t = 24)

For Poisson process mean and variance are same,

$Var[N (t)]= Var[N(24)]\\= E [N (24)]\\=8$

(Poisson distribution mean and variance equal)

The standard deviation of the number of planets is,

$\sigma( 24 )] =\sqrt{Var[ N(24)]}=\sqrt{8}= 2.828$

For the Poisson process the intervals between events(finding a new planet) have independent exponential distribution with parameter $\lambda$ . The sum of $K$ of these independent exponential has distribution Gamma $(K, \lambda)$ .