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Firdavs [7]
3 years ago
11

Which set of numbers can represent the side lengths, in inches, of an acute triangle? 4, 5, 7 5, 7, 8 6, 7, 10 7, 9, 12

Mathematics
2 answers:
yarga [219]3 years ago
4 0

Answer:

4 5 7

Step-by-step explanation:

i know it

Natalija [7]3 years ago
3 0

Answer:

Set B i.e., 5 , 7 , 8 represent the side of acute angled triangle.

Step-by-step explanation:

Given: Set of numbers.

To find: Set that represent sides of an acute angled traingle.

We use the following result:

When given 3 triangle sides then to determine if the triangle is acute angled , right angled or obtuse angled.

First find Square all 3 sides, then Sum the squares of the 2 shortest sides and then Compare the sum to the square of the last side.

if sum > Square of last side ⇒ it is Acute Triangle

if sum = Square of last side ⇒ it is Right Triangle

if sum < Square of last side ⇒ it is Obtuse Triangle  

a). 4  , 5 , 7

4² = 16 ,  5² = 25 , 7² = 49

16 + 25 = 41

∵ 41 < 49

⇒ It is an Obtuse Traingle.

b). 5 , 7 , 8

5² = 25 , 7² = 49 , 8² = 64

25 + 49 = 74

∵ 74 > 64

⇒ It is an acute Triangle.

c). 6 , 7 , 10

6² = 36 , 7² = 49 , 10² = 100

36 + 49 = 85

∵ 85 < 100

⇒ It is an Obtuse Triangle.

d). 7 , 9 , 12

7² = 49 , 9² = 81 , 12² = 144

49 + 81 = 130

∵ 130 < 144

⇒ It is an Obtuse Triangle.

Therefore, Set B i.e., 5 , 7 , 8 represent the side of acute angled triangle.

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Answer:

Case n =5

z=\frac{4.8-5}{\frac{0.5}{\sqrt{5}}}=-0.894  

Case n =15

z=\frac{4.8-5}{\frac{0.5}{\sqrt{15}}}=-1.549  

Case n = 40

z=\frac{4.8-5}{\frac{0.5}{\sqrt{40}}}=-2.530  

P value

Case n =5

p_v =P(z  

Case n =15

p_v =P(z  

Case n =40

p_v =P(z  

Step-by-step explanation:

Data given and notation  

\bar X=4.8 represent the sample mean

\sigma=0.5 represent the population standard deviation

n sample size  

\mu_o =5 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is lower than 5, the system of hypothesis would be:  

Null hypothesis:\mu \geq 5  

Alternative hypothesis:\mu < 5  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

Case n =5

z=\frac{4.8-5}{\frac{0.5}{\sqrt{5}}}=-0.894  

Case n =15

z=\frac{4.8-5}{\frac{0.5}{\sqrt{15}}}=-1.549  

Case n = 40

z=\frac{4.8-5}{\frac{0.5}{\sqrt{40}}}=-2.530  

P-value  

Since is a left tailed test the p value would be:  

Case n =5

p_v =P(z  

Case n =15

p_v =P(z  

Case n =40

p_v =P(z  

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