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KatRina [158]
3 years ago
8

Which table of ordered pairs represents a propositional relationship ?

Mathematics
2 answers:
brilliants [131]3 years ago
8 0

Answer:

C. is the best answer if i am wrong then i am sorry

Step-by-step explanation:

Angelina_Jolie [31]3 years ago
3 0
U divided it and that it
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15 points please IF NOT YOUR COMMENT WOULD BE REPORTED ALSO YOUR ACCOUNT pls thank you
olga55 [171]

the answer to this question is C

6 0
3 years ago
Write word problems that represent each way you can use a remainder in a division problem include Solutions
Galina-37 [17]
Skittles come in packages of 10. Arthurate 25 Skittles. How many whole boxes didhe eat and how many Skittles does he haveleft? 

Answer: He ate 2 and 1/2 




7 0
3 years ago
Read 2 more answers
The director of admissions at the University of Maryland, University College is concerned about the high cost of textbooks for t
nadezda [96]

Answer:

a. There is evidence that the population mean is above $300.

b. There is no evidence that the population mean is above $300.

c. There is no evidence that the population mean is above $300.

d. The director could ask for cheaper similar books.

Step-by-step explanation:

Let X be the random variable that represents the cost of textbooks. We have observed n = 25 values, \bar{x} = 315.4 and s = 43.20. We suppose that X is normally distributed.

We have the following null and alternative hypothesis

H_{0}: \mu = 300 vs H_{1}: \mu > 300 (upper-tail alternative)

We will use the test statistic

T = \frac{\bar{X}-300}{S/\sqrt{25}} and the observed value is

t_{0} = \frac{315.4 - 300}{43.20/\sqrt{25}} = 1.7824.

If H_{0} is true, then T has a t distribution with n-1 = 24 degrees of freedom.

a. The rejection region is given by RR = {t | t > t_{0.9}} where t_{0.9} = 1.3178 is the 90th quantile of the t distribution with 24 df, so, RR = {t | t > 1.3178}. Because the observed value satisty 1.7824 > 1.3178, there is evidence that the population mean is above $300.

b. If s = 75, then the observed value is t_{0} = \frac{315.4 - 300}{75/\sqrt{25}} = 1.0267. The rejection region for a 0.05 level of significance is RR = {t | t > t_{0.95}} where t_{0.95} = 1.7108 is the 95th quantile of the t distribution with 24 df, so, RR = {t | t > 1.7108}. Because the observed value does not fall inside the rejection region, there is no evidence that the population mean is above $300.

c. If \bar{x} = 305.11 and s = 43.20, the observed value is t_{0} = \frac{305.11 - 300}{43.20/\sqrt{25}} =  0.5914. For RR = {t | t > 1.3178} we have that the observed value does not fall inside RR, therefore, there is no evidence that the population mean is above $300.

d. Because the director of admissions is concerned about the high cost of textbooks, and there is evidence that the population mean of costs is above $300, the director could ask for cheaper similar books.

8 0
3 years ago
o check the effect of cold temperature on the elasticity of two brands of rubber bands, one box of Brand A and one box of Brand
const2013 [10]

Answer:

No this is a not good experimental design

Step-by-step explanation:

In an experiment, we seek to establish cause an effect relationship. The effect of one variable on another is examined while keeping other variables constant. A control often establishes the validity of the experiment.

Now the ten rubber bands were selected at random from each box. The experimental group was put in a freezer while the control group was maintained at room temperature.

Comparison of the mean stretch before breakage of the rubber bands in both groups establishes the effect of cold temperature on elasticity of rubber bands.

However, this is not a good experimental design because the sample rubber bands should have been picked from different boxes of brand A and B and not from the same box.

Secondly, samples from the two brands should have been put in the freezer and kept at room temperature. That is, ten rubber bands from A are put on the freezer and another 10 are left at room temperature. 10 rubber bands from B are put in the freezer and another 10 are left at room temperature.

The mean elasticity of the both groups can now be meaningfully compared from the data obtained.

3 0
3 years ago
MATH
Cerrena [4.2K]

Answer:

<u>Corresponding Angles Theorem</u>

When a straight line intersects 2 parallel lines, the angles in the same relative position are congruent (equal).

<u>Alternate Exterior Angles Theorem</u>

When a straight line intersects 2 parallel lines, the alternate exterior angles are congruent (equal).

<u>Vertical Angle Theorem</u>

When two straight lines intersect, the vertical angles are congruent (equal).

<h3><u>Part A</u></h3>

Q1.   As  s ║ c  we can apply the Corresponding Angles Theorem:

⇒ 11x - 5 = 116

⇒ 11x - 5 + 5 = 116 + 5

⇒ 11x = 121

⇒ 11x ÷ 11 = 121 ÷ 11

⇒ x = 11

Q2.   As  s ║ c  we can apply the Alternative Exterior Angles Theorem:

⇒ 12x - 4 = 148

⇒ 12x - 4 + 4 = 148 + 4

⇒ 12x = 152

⇒ 12x ÷ 12 = 152 ÷ 12

⇒ x = 38/3 = 12.7 (nearest tenth)

<h3><u>Part B</u></h3>

Q1.   As  j ⊥ r  then the sum of the angles is 90°

⇒ 4x + 6x + 10 = 90

⇒ 10x + 10 - 10 = 90 - 10

⇒ 10x = 80

⇒ 10x ÷ 10 = 80 ÷ 10

⇒ x = 8

Q2.  As  j ⊥ r   we can apply the Vertical Angles Theorem:

⇒ 5x - 10 = x + 70

⇒ 5x - 10  + 10 = x + 70 + 10

⇒ 5x = x + 80

⇒ 5x - x = x + 80 - x

⇒ 4x = 80

⇒ 4x ÷ 4 = 80 ÷ 4

⇒ x = 20

6 0
2 years ago
Read 2 more answers
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