What is the y-coordinate of the point that divides the directed line segment from J to K into a ratio of 5:1? m Y 1 77](V₂ − V₁]
1 answer:
The y-coordinate of the point that divides the directed line segment from J to K into a ratio of 5:1 is 0.
<h3>How to carry out division of line segments?</h3>
From the attached line segment graph, we see the coordinates as;
J (1, -10) and K (7, 2)
Let us assume that, point L divides the directed line segment from J to K into a ratio of 5:1.
From the properties of coordinate geometry we know that, the coordinates of point that divides the line joining (x₁, y₁), (x₂, y₂) in m:n, are;
[(mx₂ + nx₁)/(m + n)], [(my₂ + ny₁)/(m + n)]
Plugging in the relevant values gives;
[(5 * 7 + 1 * 1)/(5 + 1)], [(5 * 2 + 1 * -10)/(5 + 1)]
⇒ (6, 0)
Therefore, the y-coordinate of the point that divides the directed line segment from J to K into a ratio of 5:1 is 0.
Read more about division of line segments at; brainly.com/question/17374569
#SPJ1
You might be interested in
Answer:
We conclude that the mean duration of 15-sided union rugby games has decreased after the meeting.
Step-by-step explanation:
We are given that Before the meeting, the mean duration of the 15-sided rugby game time was 3 hours, 5 minutes, that is, 185 minutes.
A random sample of 36 of the 15-sided rugby games after the meeting showed a mean of 179 minutes with a standard deviation of 12 minutes.
Let = <em><u>mean duration of 15-sided union rugby games after the meeting.</u></em>
So, Null Hypothesis, :
185 minutes {means that the mean duration of 15-sided union rugby games has increased or remained same after the meeting}
Alternate Hypothesis, :
< 185 minutes {means that the mean duration of 15-sided union rugby games has decreased after the meeting}
The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;
T.S. = ~
where, = sample mean duration of 15-sided union rugby games = 179 min
s = sample standard deviation = 12 minutes
n = sample of 15-sided rugby games = 36
So, <u><em>the test statistics</em></u> = ~
= -3
The value of t test statistics is -3.
<u>Now, at 1% significance level the t table gives critical value of -2.437 at 35 degree of freedom for left-tailed test.</u>
Since our test statistic is less than the critical value of t as -3 < -2.437, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.
Therefore, we conclude that the mean duration of 15-sided union rugby games has decreased after the meeting.