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2 years ago

Is 6:1 equivalent to 12:3

Mathematics

2 answers:

Anika [276]2 years ago

7 0

Answer: Nope

Step-by-step explanation: No, because you can see they multiply 1 x 3 to get to 3, but if you multiply 6 x 3, it equals 18, not 12. (6 x 2 = 12) but they have to have the same denominator and numerator to be equivalent. Therefore, no, they are not equivalent.

WITCHER [35]2 years ago

3 0

Answer: No, absolutely not,

Step-by-step explanation: 6x2 = 12, so you have to multiply by 2 on each side. 1x2=2, so the equivilent ratio is 12:2!

You might be interested in

Find a number lower than 19 that has more factors than 19,21,23, and 25

Len [333]

12 has more factors the 19,21,,23, and 25.

7 0

3 years ago

The percentage of body fat of a random sample of 36 men aged 20 to 29 found a sample mean of 14.42. Find a 95% confidence interv

Rina8888 [55]

Answer:

$14.42-1.96\frac{6.95}{\sqrt{36}}=12.150$

$14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690$

So on this case the 95% confidence interval would be given by (12.150;16.690)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=14.42$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=6.95$ represent the population standard deviation

n=36 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

Now we have everything in order to replace into formula (1):

$14.42-1.96\frac{6.95}{\sqrt{36}}=12.150$

$14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690$

So on this case the 95% confidence interval would be given by (12.150;16.690)

5 0

3 years ago

Someone can help??

oee [108]

Answer:

Step-by-step explanation:

let cos^{-1}x=t

cos t=x

when x=1,cos t=1=cos 0

t \rightarrow 0

$\lim_{x \to 1} \frac{1-\sqrt{x}}{(cos ^{-1}x)^2 } \\= \lim_{t \to 0} \frac{1-\sqrt{cos~t}}{t^2} \times \frac{1+\sqrt{cos~t}}{1+\sqrt{cos ~t}} \\= \lim_{t \to 0} \frac{1-cos~t}{t^2(1+\sqrt{cos~t})}} \\= \lim_{t \to 0 }\frac{2 sin^2~\frac{t}{2}}{t^2(1+\sqrt{cos~t})}} \\= 2\lim_{t \to 0 }(\frac{sin~t/2}{\frac{t}{2} })^2 \times \frac{1}{4} \times \lim_{t \to 0 }\frac{1}{1+\sqrt{cos~t}} \\=\frac{2}4} \times 1^2 \times \frac{1}{1+\sqrt{cos~0}} \\=\frac{1}{2} \times \frac{1}{1+1} \\=\frac{1}{4}$