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lapo4ka [179]
3 years ago
7

11/6358 long agurithum

Mathematics
1 answer:
miss Akunina [59]3 years ago
8 0

Answer:

Least common multiple:

lcm (578; 11) = 6,358 = 2 × 11 × 172;

Numbers have no common prime factors: 6,358 = 578 × 11.

Step-by-step explanation:

Approach 1. Integer numbers prime factorization:

578 = 2 × 172;

11 is a prime number, it cannot be broken down to other prime factors;

Multiply all the prime factors, by the largest exponents.

Least common multiple:

lcm (578; 11) = 2 × 11 × 172;

lcm (578; 11) = 2 × 11 × 172 = 6,358

Numbers have no common prime factors: 6,358 = 578 × 11.

Integer numbers prime factorization

Approach 2. Euclid's algorithm:

Calculate the greatest (highest) common factor (divisor), gcf, hcf, gcd:

Step 1. Divide the larger number by the smaller one:

578 ÷ 11 = 52 + 6; Step 2. Divide the smaller number by the above operation's remainder:

11 ÷ 6 = 1 + 5; Step 3. Divide the remainder from the step 1 by the remainder from the step 2:

6 ÷ 5 = 1 + 1; Step 4. Divide the remainder from the step 2 by the remainder from the step 3:

5 ÷ 1 = 5 + 0; At this step, the remainder is zero, so we stop:

1 is the number we were looking for, the last remainder that is not zero.

This is the greatest common factor (divisor).

Least common multiple, formula:

lcm (a; b) = (a × b) / gcf, hcf, gcd (a; b);

lcm (578; 11) =

(578 × 11) / gcf, hcf, gcd (578; 11) =

6,358 / 1 =

6,358;

lcm (578; 11) = 6,358 = 2 × 11 × 172;

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Answer:

t=\frac{2.8-3}{\frac{0.5}{\sqrt{64}}}=-3.2    

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Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly lower than 3 grams at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=2.8 represent the sample mean

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\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

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State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is lower than 3 grams, the system of hypothesis would be:  

Null hypothesis:\mu \geq 3  

Alternative hypothesis:\mu < 3  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-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:  

t=\frac{2.8-3}{\frac{0.5}{\sqrt{64}}}=-3.2    

P-value

The first step is calculate the degrees of freedom, on this case:  

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Since is a one sided test the p value would be:  

p_v =P(t_{(63)}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly lower than 3 grams at 5% of signficance.  

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