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seropon [69]
3 years ago
11

Using Euclid's division algorithm prove that : 847,2160 are co-primes

Mathematics
1 answer:
AysviL [449]3 years ago
5 0
Definition of eo-primes or relatively primes: Two numbers are said to be co-prime or relatively prime If their HCF IS 1 Hence to prove 847 and 2160 as co-prime numbers we will find their HCF and which should be 1

New steps to find HCF will be as under

2160 = 847 x 2+ 466
847 = 466 ×1 +381
466 = 381 x 1 + 85
381 =85 x 4+ 41
85 =41 x 2+3
41 =3 x 13+ 2
3 =2 x 1+1
2 =1 x 2+0

Therefore, the HCF=1 Hence, the numbers are co-primes (relatively prime).
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horsena [70]

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  

5 0
4 years ago
"More than 60% of the students scored below the mean on the last test." Can this statement be true? Why or why not?
Lelu [443]
The mean is the average of a set of numbers.

So if the remaining 40% of students score 100 points, and the 60% score like 50 points, then the mean would still be 70 points because of the 40% of students that scored super high.

So yes, it can be true.
8 0
4 years ago
Read 2 more answers
Solve the equations using addition or subtraction.<br> 5.5=-2+d
tia_tia [17]

Answer: d=7.5

Step-by-step explanation: 5.5=−2+d                                                                     d−2+2=5.5+2                                                                                                             d=7.5                  

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Step-by-step explanation:

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