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2 years ago

(2,0) (4,5) write an equation that represents the line

Mathematics

1 answer:

Yuliya22 [10]2 years ago

4 0

Answer:

the equation of the line in slope-intercept form would be y=5/2x-5

Step-by-step explanation:

hope this helped!! :D

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The number of flaws per square yard in a type of carpet material varies with mean 1.7 flaws per square yard and standard deviati

maw [93]

Let x be a random variable representing the mean number of flaws per square yard inspected
P(x > 1.8) = P(z > (1.8 - 1.7)/0.8) = P(z > 0.125) = P(z < 0.125) = 0.5497

7 0

3 years ago

ASAP someone please use the Pythagorean Theorem to solve for the missing side of these two right triangles.

solong [7]

Answer:

9) x = 5

10) x = 4

Step-by-step explanation:

a^2 + b^2 = c^2

9) 6^2 + x^2 = 8^2

- 36 + x^2 = 64

- 36 - 36

_______________

x^2 = 28

Square root them

Squareroot x^2 is x

Square root 28 is 5.29 round is 5

so x = 5

*Correct me if im wrong*

10) 3^2 + x^2 = 5^2

9 + x^2 = 25

- 9 - 9

____________________

x^2= 16

Square root them

square root of x ^2 is x and square root of 16 is 4

so x = 4

6 0

3 years ago

So if each equation for x x / a - 5 equals B

Akimi4 [234]

Answer:

Sorry ineed point ehhh thnks po sa points:)

6 0

2 years ago

Which of the following statements is true about

lions [1.4K]

Answer:

The y-axis stays the same

Step-by-step explanation:

6 0

2 years ago

There are 4 people at a party. Consider the random variable X=’number of people having the same birthday ’ (match only month, N=

yulyashka [42]

Answer:

S = {0,2,3,4}

P(X=0) = 0.573 , P(X=2) = 0.401 , P(x=3) = 0.025, P(X=4) = 0.001

Mean = 0.879

Standard Deviation = 1.033

Step-by-step explanation:

Let the number of people having same birth month be = x

The number of ways of distributing the birthdays of the 4 men = (12*12*12*12)

The number of ways of distributing their birthdays = 12⁴

The sample space, S = { 0,2,3,4} (since 1 person cannot share birthday with himself)

P(X = 0) = $\frac{12P4}{12^{4} }$

P(X=0) = 0.573

P(X=2) = P(2 months are common) P(1 month is common, 1 month is not common)

P(X=2) = $\frac{3C2 * 12P2}{12^{4} } + \frac{4C2 * 12P3}{12^{4} }$

P(X=2) = 0.401

P(X=3) = $\frac{4C3 * 12P2}{12^{4} }$

P(x=3) = 0.025

P(X=4) = $\frac{12}{12^{4} }$

P(X=4) = 0.001

Mean, $\mu = \sum xP(x)$

$\mu = (0*0.573) + (2*0.401) + (3*0.025) + (4*0.001)\\\mu = 0.879$

Standard deviation, $SD = \sqrt{\sum x^{2} P(x) - \mu^{2}} \\SD =\sqrt{ [ (0^{2} * 0.573) + (2^{2} * 0.401) + (3^{2} * 0.025) + (4^{2} * 0.001)] - 0.879^{2}}$

SD = 1.033

4 0

3 years ago