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3 years ago

What is the slope of the line that passes through the points (-10,8)and (-15,-7)

Mathematics

2 answers:

balandron [24]3 years ago

5 0

Answer:

m = 3

Step-by-step explanation:

Use the slope formula to find the slope m .

Tasya [4]3 years ago

4 0

M=3 is the slope to find the m

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1. Write the equation for a line in point slope form that had a slope of 4 and passes through (1,1)

Dahasolnce [82]

Solution:

Note that:

Slope = m = 4
Given point: (x₁,y₁) = (1,1) [x₁ = 1; y₁ = 1]
Formula: y - y₁ = m(x - x₁)

Use the formula to convert the line in point slope form.

y - y₁ = m(x - x₁)
y - 1 = 4(x - 1)

The equation of the line (In point slope form) is y - 1 = 4(x - 1).

8 0

2 years ago

Read 2 more answers

O triplo de um mumero somado a 9 e igual a 30 .qual e esse numero

-Dominant- [34]

Do you think than i ve understood it right

3x+9=30

3x=21

x = 21/3 = 7

x = 7

hope helped

7 0

3 years ago

What is 12 to the 10th power? Added by 375?

dmitriy555 [2]

Answer:

the answer is 61917364599

Step-by-step explanation:

I hope this helps! :)

3 0

2 years ago

Read 2 more answers

Solve each equation MUST SHOW WORK!!!! 10 points.

Sergio039 [100]

Answer:

1. x=34

2. c=8

3. x=9

4. m=32

That the answer for all 4.

Step-by-step explanation: Hope this help :D

4 0

3 years ago

Read 2 more answers

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or

Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let X denote the event that the student has an “on” day, and let Y denote the

denote the event that he passes the examination. Then,

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

The events ( $Y|X$ ) follows a Binomial distribution with probability of success 0.80 and the events ( $Y|X^{c}$ ) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

$P(X)=2\cdot P(X^{c})$

Then,

$P(X)+P(X^{c})=1$

⇒

$2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}$

Then,

$P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}$

Compute the probability that the students passes if request an examination with 3 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}$

$=0.715$

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}$

$=0.734$

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

8 0

2 years ago