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Anon25 [30]
3 years ago
10

12/15 in simplest form

Mathematics
2 answers:
alex41 [277]3 years ago
3 0
\frac{12}{15} in simplest form:

First, find the greatest common factor (GCF) of 12, 15. 
Factors of 12: 1, 2, 3, 4, 6, 12
Factors of 15: 1, 3, 5, 15
The GCF is 3. 
Second, divide the numerator (12) by the GCF. / Your problem should look like: 12 ÷ 3 = 4
Third, divide the denominator (15) by the GCF. / Your problem should look like: 15 ÷ 3 = 5
Fourth, rewrite the fraction in the simplified form. / Your problem should look like: \frac{4}{5}

Answer: \frac{4}{5}

nexus9112 [7]3 years ago
3 0

Answer:

First, find the greatest common factor (GCF) of 12, 15. 

Factors of 12: 1, 2, 3, 4, 6, 12

Factors of 15: 1, 3, 5, 15

The GCF is 3. 

Second, divide the numerator (12) by the GCF. / Your problem should look like: 12 ÷ 3 = 4

Third, divide the denominator (15) by the GCF. / Your problem should look like: 15 ÷ 3 = 5

Fourth, rewrite the fraction in the simplified form. / Your problem should look like: 

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Use the sample information 11formula13.mml = 37, σ = 5, n = 15 to calculate the following confidence intervals for μ assuming th
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Answer & Step-by-step explanation:

The confidence interval formula is:

I (1-alpha) (μ)= mean+- [(Z(alpha/2))* σ/sqrt(n)]

alpha= is the proposition of the distribution tails that are outside the confidence interval. In this case, 10% because 100-90%

σ= standard deviation. In this case 5

mean= 37

n= number of observations. In this case, 15

(a)

Z(alpha/2)= is the critical value of the standardized normal distribution. The critical valu for z(5%) is 1.645

Then, the confidence interval (90%):

I 90%(μ)= 37+- [1.645*(5/sqrt(15))]

I 90%(μ)= 37+- [2.1236]

I 90%(μ)= [37-2.1236;37+2.1236]

I 90%(μ)= [34.8764;39.1236]

(b)

Z(alpha/2)= Z(2.5%)= 1.96

Then, the confidence interval (90%):

I 95%(μ)= 37+- [1.96*(5/sqrt(15)) ]

I 95%(μ)= 37+- [2.5303]

I 95%(μ)= [37-2.5303;37+2.5303]

I 95%(μ)= [34.4697;39.5203]

(c)

Z(alpha/2)= Z(0.5%)= 2.5758

Then, the confidence interval (90%):

I 99%(μ)= 37+- [2.5758*(5/sqrt(15))

I 99%(μ)= 37+- [3.3253]

I 99%(μ)= [37-3.3253;37+3.3253]

I 99%(μ)= [33.6747;39.3253]

(d)

C. The interval gets wider as the confidence level increases.

8 0
3 years ago
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