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4 years ago

Slope of (0,1) and (3,0)

Mathematics

2 answers:

Marysya12 [62]4 years ago

5 0

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \sf{ \frac{ - 1}{3}}}}}}$

Step-by-step explanation:

Let the points be A and B

Let A ( 0 , 1 ) be ( x₁ , y₁ ) and B ( 3 , 0 ) be ( x₂ , y₂ )

Finding the slope of the points

$\boxed{ \sf{slope = \frac{y2 - y1}{x2 - x1} }}$

$\longrightarrow{ \sf{slope = \frac{0 - 1}{ 3 - 0}}}$

$\longrightarrow{ \sf{slope = \frac{ - 1}{3}}}$

Hope I helped!

Best regards! :D

pychu [463]4 years ago

5 0

Rise of run every time you ave slope in points an you want to solve do y^2-y^1 Over x^2-x^1

Step-by-step explanation:

this is how 0-1 over 3-0 gives you

-1/3

and that is your answer

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In ΔQRS, q = 890 cm, s = 810 cm and ∠S=120°. Find all possible values of ∠Q, to the nearest degree.

Lelechka [254]

Answer: no possible triangles

Step-by-step explanation:

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3 years ago

In words how do you write 1000

valkas [14]

Easy... :)

1,000= One thousand.

Thus, your answer. :D

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4 years ago

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Decrease 42 by 15%. it's for maths and it's decreasing percenages

sleet_krkn [62]

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4 years ago

A college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of

Alla [95]

Answer:

80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

Step-by-step explanation:

We are given that a college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of hours they watch per week. The results are provided below;

Hours of TV per week (X): 6, 14, 13, 6, 16, 10, 19, 4, 5, 5, 18, 8, 7, 14, 8, 8, 9, 12, 6, 5.

Firstly, the Pivotal quantity for 80% confidence interval for the true average is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean number of hours of TV watched per week = $\frac{\sum X}{n}$ = 9.65

s = sample standard deviation = $\sqrt{\frac{\sum (X -\bar X)^{2} }{n-1} }$ = 4.61

n = sample of people = 20

$\mu$ = true average number of hours of TV watched per week

Here for constructing 80% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.

So, 80% confidence interval for the true average, $\mu$ is ;

P(-1.33 < $t_1_9$ < 1.33) = 0.80 {As the critical value of t at 19 degrees of

freedom are -1.33 & 1.33 with P = 10%}

P(-1.33 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.33) = 0.80

P( $-1.33 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.33 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.33 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.33 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.33 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.33 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $9.65-1.33 \times {\frac{4.61}{\sqrt{20} } }$ , $9.65+1.33 \times {\frac{4.61}{\sqrt{20} } }$ ]

= [8.28 hours, 11.02 hours]

Therefore, 80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

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3 years ago

Can so one help ?? picture shown

Lorico [155]

To multiply rational expressions you just multiply tops and multiply the bottoms so
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4 years ago

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