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Leviafan [203]
2 years ago
5

Triangle JKL is equilateral. One side of the triangle, JL, is a diameter of circle M. Which is true about line segments JK and K

L?
Both segments are tangent to circle M but are not chords.
One segment is tangent to circle M and one segment is a chord in circle M.
Both segments are chords in circle M but are not tangents.
Neither segment is a chord nor tangent to circle M.
Mathematics
2 answers:
antoniya [11.8K]2 years ago
7 0
It could be concluded that since line JL is a diameter of a circle, therefore, JK and KL are the tangents of the specific circle that JL is part of. By definition, a tangent is "a line that touches a curve at one point without touching it." Both these lines touches the edges of the circle M therefore they are tangents.
Agata [3.3K]2 years ago
4 0

Answer:

Just took the test, it's D!


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Consider coordinate system a, with the x axis along the plane. which forces lie along the axes?
Butoxors [25]
It seem like there are information missing on the question posted. Let me answer this question with all I know. So here is what I believe the answer is, the horizontal force.

Hope my answer would be a great help for you.    If you have more questions feel free to ask here at Brainly.
7 0
2 years ago
The equation r(t)= (2t)i + (2t-16t^2)j is the position of a particle in space at time t. Find the angle between the velocity and
ankoles [38]

Answer:

The answer is 135 degrees.

Step-by-step explanation:

As we are given the position. If we take the <u>derivative</u>, we get the velocity vector. If we take the <u>derivative</u> again, we find the acceleration vector of the particle.

r(t)=(2t)i+(2t-16t^{2})j

V(t)=2i+(2-32t)j

a(t)=-32j

At time t=0;

v(t)=2i+2j

a(t)=-32j

As i attach in the picture the angle between the velocity and acceleration vector is (45+90)=135 degrees

4 0
2 years ago
Two random samples are taken from private and public universities
kati45 [8]

Answer:

Step-by-step explanation:

For private Institutions,

n = 20

Mean, x1 = (43120 + 28190 + 34490 + 20893 + 42984 + 34750 + 44897 + 32198 + 18432 + 33981 + 29498 + 31980 + 22764 + 54190 + 37756 + 30129 + 33980 + 47909 + 32200 + 38120)/20 = 34623.05

Standard deviation = √(summation(x - mean)²/n

Summation(x - mean)² = (43120 - 34623.05)^2+ (28190 - 34623.05)^2 + (34490 - 34623.05)^2 + (20893 - 34623.05)^2 + (42984 - 34623.05)^2 + (34750 - 34623.05)^2 + (44897 - 34623.05)^2 + (32198 - 34623.05)^2 + (18432 - 34623.05)^2 + (33981 - 34623.05)^2 + (29498 - 34623.05)^2 + (31980 - 34623.05)^2 + (22764 - 34623.05)^2 + (54190 - 34623.05)^2 + (37756 - 34623.05)^2 + (30129 - 34623.05)^2 + (33980 - 34623.05)^2 + (47909 - 34623.05)^2 + (32200 - 34623.05)^2 + (38120 - 34623.05)^2 = 1527829234.95

Standard deviation = √(1527829234.95/20

s1 = 8740.22

For public Institutions,

n = 20

Mean, x2 = (25469 + 19450 + 18347 + 28560 + 32592 + 21871 + 24120 + 27450 + 29100 + 21870 + 22650 + 29143 + 25379 + 23450 + 23871 + 28745 + 30120 + 21190 + 21540 + 26346)/20 = 25063.15

Summation(x - mean)² = (25469 - 25063.15)^2+ (19450 - 25063.15)^2 + (18347 - 25063.15)^2 + (28560 - 25063.15)^2 + (32592 - 25063.15)^2 + (21871 - 25063.15)^2 + (24120 - 25063.15)^2 + (27450 - 25063.15)^2 + (29100 - 25063.15)^2 + (21870 - 25063.15)^2 + (22650 - 25063.15)^2 + (29143 - 25063.15)^2 + (25379 - 25063.15)^2 + (23450 - 25063.15)^2 + (23871 - 25063.15)^2 + (28745 - 25063.15)^2 + (30120 - 25063.15)^2 + (21190 - 25063.15)^2 + (21540 - 25063.15)^2 + (26346 - 25063.15)^2 = 1527829234.95

Standard deviation = √(283738188.55/20

s2 = 3766.55

This is a test of 2 independent groups. Let μ1 be the mean out-of-state tuition for private institutions and μ2 be the mean out-of-state tuition for public institutions.

The random variable is μ1 - μ2 = difference in the mean out-of-state tuition for private institutions and the mean out-of-state tuition for public institutions.

We would set up the hypothesis. The correct option is

-B. H0: μ1 = μ2 ; H1: μ1 > μ2

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

t = (34623.05 - 25063.15)/√(8740.22²/20 + 3766.55²/20)

t = 9559.9/2128.12528473889

t = 4.49

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [8740.22²/20 + 3766.55²/20]²/[(1/20 - 1)(8740.22²/20)² + (1/20 - 1)(3766.55²/20)²] = 20511091253953.727/794331719568.7114

df = 26

We would determine the probability value from the t test calculator. It becomes

p value = 0.000065

Since alpha, 0.01 > than the p value, 0.000065, then we would reject the null hypothesis. Therefore, at 1% significance level, the mean out-of-state tuition for private institutions is statistically significantly higher than public institutions.

4 0
2 years ago
The graph shows the distance in miles of a runner over x hours. What is the average rate of speed over the interval [9, 11]?
JulsSmile [24]
For this case what you should see is that for the interval [9, 11] the behavior of the function is almost linear.
 Therefore, we can find the average rate of change as follows:
 m = (y2-y1) / (x2-x1)
 m = (11-6) / (11-9)
 m = (5) / (2)
 m = 5/2
 Answer:
 the average rate of speed over the interval [9, 11] is: 
 D. 5 / 2
8 0
2 years ago
Read 2 more answers
Olivia counted the number of ladybugs on each plant in her garden, then made the graph below.
Leni [432]

Answer:

A) Roses , C) Alfalfa

Step-by-step explanation:

Each ladybug symbol = 5 ladybugs

Roses: 35 ladybugs

Lettuce: 15 ladybugs

Alfalfa: 25 ladybugs

Grape vines: 10 ladybugs

10 ladybugs go from lettuce to alfalfa. You end up with:

Roses: 35 ladybugs

Lettuce: 5 ladybugs

Alfalfa: 35 ladybugs

Grape vines: 10 ladybugs

Roses and alfalfa end up with 35 ladybugs each.

Answer: A) Roses , C) Alfalfa

5 0
2 years ago
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