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Gelneren [198K]
3 years ago
9

A line with slope of -1/10 passes thorugh the point (1,8). What is its equation in point slope form?

Mathematics
2 answers:
Gennadij [26K]3 years ago
7 0

Answer:

Hello:

note :  

The equation of a linear function in point-slope form is y – y1 = m(x – x1)

The point is A (x1 , y1)

in this exercice : m = - 1/10    and   x1 = 1      y1 = 8

the equation in point slope form is :  y – 8 = (-1/10)(x – 1)



Alisiya [41]3 years ago
5 0

ABOUT POINT SLOPE FORM:

  • y - Y1 = m(x - X1)
  • m represents the slope
  • Y1 & X1 is a point
  • The form allows you to identify the slope & the point on the line

y - Y1 = m(x - X1)

y - 8 = -1/10(x - 1) -- IN POINT SLOPE FORM


Hope this helps you!!! :)

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Five males with an​ X-linked genetic disorder have one child each. The random variable x is the number of children among the fiv
Dahasolnce [82]

Answer and Explanation:

Given : Five males with an​ X-linked genetic disorder have one child each. The random variable x is the number of children among the five who inherit the​ X-linked genetic disorder.

To find :

a. Does the table show a probability distribution?

b. Find the mean and standard deviation of the random variable x.

Solution :

a) To determine that table shows a probability distribution we add up all six probabilities if the sum is 1 then it is a valid distribution.

\sum P(X)=0.029+0.147+0.324+0.324+0.147+0.029

\sum P(X)=1

Yes it is a probability distribution.

b) First we create the table as per requirements,

x      P(x)         xP(x)           x²            x²P(x)

0    0.029         0              0                0

1     0.147        0.147           1              0.147

2    0.324       0.648         4              1.296

3    0.324       0.972         9              2.916

4    0.147        0.588        16              2.352

5    0.029       0.145         25            0.725

   ∑P(x)=1      ∑xP(x)=2.5               ∑x²P(x)=7.436

The mean of the random variable is

\mu=\sum xP(x)=2.5

The standard deviation of the random sample is

Variance=\sigma^2

\sigma^2=\sum x^2P(x)-\mu^2

\sigma^2=7.436-(2.5)^2

\sigma^2=7.436-6.25

\sigma^2=1.186

\sigma=\sqrt{1.186}

\sigma=1.08

Therefore, The mean is 2.5 and the standard deviation is 1.08.

5 0
2 years ago
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