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postnew [5]
3 years ago
7

A and B are two cities.

Mathematics
2 answers:
likoan [24]3 years ago
8 0

Answer:

The bearing of B from A is 28 degrees, approximately

The bearing of A from B is 180+28=208 degrees, approximately.

Step-by-step explanation:

The reason for approximation is from distortion of image.

See attached image to understand how the measurement has been obtained.

lesya692 [45]3 years ago
4 0

Answer:

The answer is b on edgunity

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Len [333]

Answer:

7 1/8

Step-by-step explanation:

Assuming that the first 13 was a mistake then the equation would be 13 1/2 + 2n = 27 3/4

subtrace 13 1/2 from 27 3/4

14 1/4

divide by 2

7 1/8

7 0
3 years ago
In what ways are inequalities like equations ?
solmaris [256]

Answer:

Step-by-step explanation:

Inequalities and equations are alike in that they both involve two sides of a sign (whether it is an equal sign or an inequality sign) being compared. For example, both y = x + 2 and y > x + 2 are just ways of comparing the values of x and y.

7 0
3 years ago
Find the measure of angle NXO. The measure of angle MXO is 110*​
Marat540 [252]

Answer:

60º

Step-by-step explanation:

NXO = MXO - MXN

NXO = 110 - 50

NXO = 60

4 0
3 years ago
For the function f(x) = -4^-x + 5, if x→∞, then y → ____.
alexgriva [62]

y approaches 5.

As x approaches infinity, -4^(-x) approaches 0.

0 + 5 = 5

8 0
3 years ago
In the past, 35% of the students at ABC University were in the Business College, 35% of the students were in the Liberal Arts Co
sukhopar [10]

Answer:

a. proportions have not changed significantly

Step-by-step explanation:

Given

Business College= 35 %

Arts College= 35 %

Education College = 30%

Calculated

Business College = 90/300= 9/30= 0.3 or 30%

Arts College= 120/300= 12/30= 2/5= 0.4 or 40%

Education College= 90/300= 9/30 = 0.3 or 30%

First we find the mean and variance of the three colleges using the formulas :

Mean = np

Standard Deviation= s= \sqrt{npq\\}

Business College

Mean = np =300*0.3= 90

Standard Deviation= s= \sqrt{npq\\}=\sqrt{0.3*0.7*300}= 7.94

Arts College

Mean = np =300*0.4= 120

Standard Deviation= s= \sqrt{npq\\}=\sqrt{0.4*0.6*300}=  8.49

Education College

Mean = np =300*0.3= 90

Standard Deviation= s= \sqrt{npq\\}=\sqrt{0.3*0.7*300}= 7.94

Now calculating the previous means with the same number of students

Business College

Mean = np =300*0.35= 105

Arts College

Mean = np =300*0.35= 105

Education College:

Mean = np =300*0.3= 90

Now formulate the null and alternative hypothesis

Business College

90≤ Mean≥105

Arts College

105 ≤ Mean≥ 120

Education College

U0 : mean= 90     U1: mean ≠ 90

From these we conclude that the  proportions have not changed significantly meaning that it falls outside the critical region.

8 0
4 years ago
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