13 units from (10,6)

Vlad1618 [11]

Problem 1

Answer: Yes a triangle can be formed with side lengths 5,5,3

--------------------

Explanation:

Take any two sides and add them up. If we get a sum larger than the third side, then a triangle is possible. This is the triangle inequality theorem.

5+5 = 10 is larger than 3

5+3 = 8 is larger than 5

This shows a triangle is possible. The triangle is isosceles because two sides are the same length.

==========================================================

Problem 2

Answer: Yes a triangle can be formed with side lengths 8,8,8

--------------------

Explanation:

This triangle is equilateral because all sides are the same length.

Take any two sides, add them up, and you'll find the sum is larger than the third side.

==========================================================

Problem 3

Answer: No a triangle cannot be formed with sides 7,8,15

--------------------

Explanation:

Yes it is true that

8+15 = 23 is larger than 7

7+15 = 22 is larger than 8

but

7+8 = 15 is not larger than 15

So this means a triangle is not possible.

Imagine you had 3 strips of paper that were 7 inches, 8 inches and 15 inches long. The 7 and 8 inch strips add to 15 inches, but those two smaller strips only form a straight line. We cannot pull them so a triangle forms. If we did, then the third side would be smaller than 15 inches. This is one way to get hands on to see why the triangle inequality theorem works.

==========================================================

Problem 4

Answer: Yes a triangle is possible with side lengths 5,6,10

--------------------

Explanation:

Same idea as before. We can take any two sides, add them, and get a sum larger than the third side. A triangle is possible.

5+6 = 11 is larger than 10
6+10 = 16 is larger than 5
5+10 = 15 is larger than 6

4 0

3 years ago

3. Which graph indicates that one of the runners started 10 meters ahead of the other?

KATRIN_1 [288]

Answer:

graph A

Step-by-step explanation:

4 0

3 years ago

A random sample of size 15 is selected from a normal population. The population standard deviation is unknown. Assume the null h

taurus [48]

Answer:

Interval for t statistic = (-1.345, 1.345)

Step-by-step explanation:

given that a random sample of size 15 is selected from a normal population. The population standard deviation is unknown.

Since population std deviation is unknown and also sample size <30, we have to use t test only for testing of hypothesis

Given it is a two tailed test at 10% significance level.

Degree of freedom = 15-1 =14

t critical value for df 14 two tailed is

1.345

Hence our test statistic should lie between

(-1.345, 1.345)

Then null hypothesis is accepted

Other wise it is rejected.

4 0

3 years ago

What is the value of x? A. 19.8 B. 27.2 C. 12.9 D. 9.4 E. 11.6

PolarNik [594]

Answer:

D

Step-by-step explanation:

sin36/x=sin90/16

x=9.4

4 0

3 years ago

8.36 Football and Brain Size: Randomization Test Exercise 8.35 describes a study examining hippocampus volume in the brain betwe

erma4kov [3.2K]

Answer:

Since the calculated F= 2.8657 does not lie in the critical region F (2,2)≥ 19 we conclude that there no difference between the means.

Since the calculated F= 205.0819 lies in the critical region F(1,2) ≥ 18.51 we conclude that there is a difference between the standard deviations of the three groups.

The F- statistic for the observed data is 0.101 as calculated by F- test in excel.

Step-by-step explanation:

To conduct an ANOVA randomization test we perform anova without replication in excel.

The data for the test is

Mean Standard Deviation Sample Size

Never

Played Football 7602.5 1074 25

FB with no History 6459.2 779.9 25

FB with Concussion 5734.6 593.4 25

Total 6598.8 1133.5 75

Anova: Two-Factor Without Replication

SUMMARY Count Sum Average Variance

Row 1 2 8676.5 4338.25 21310656.13

Row 2 2 7239.1 3619.55 16127224.25

Row 3 2 6328 3164 13215968.72

Column 1 3 19796.3 6598.766667 886871.7433

Column 2 3 2447.3 815.7666667 58708.90333

ANOVA Table

Source Sum of Degrees Mean

of Variation Square of freedom Squares

SS df MS F P-value

B/w Rows 1401945.703 2 700972.8517 2.8657 0.259

B/w Col 50164633.5 1 50164633.5 205.0819 0.00484

Error 489215.59 2 244607.795

Total 52055794.79 5

Set up the null and alternate hypotheses :

H0 ; there's no difference between the means and

there's no difference between the standard deviations

against the claim

Ha: there's a difference between the means and

there's a difference between the standard deviations

The significance level is chosen at 0.05

The test statistics to use are

F1= estimated variance from Between Means/ estimated variance from Error SS

F1= s1²/s3²

F1= estimated variance from Between Standard Deviations / estimated variance from Error SS

F2- s2²/s3²

with υ1= (2,2) and υ2(1,2) degrees of freedom.

The critical regions for F (2,2)≥ 19 for 0.05 significance level

and F(1,2) ≥ 18.51 for 0.05 significance level

Since the calculated F= 2.8657 does not lie in the critical region F (2,2)≥ 19 we conclude that there no difference between the means.

Since the calculated F= 205.0819 lies in the critical region F(1,2) ≥ 18.51 we conclude that there is a difference between the standard deviations of the three groups.

The F- statistic for the observed data is 0.101 as calculated by F- test in excel.

7 0

2 years ago