Are 2/5 and 6/10 are eqivalent

Mathematics

2 answers:

jeka57 [31]4 years ago

3 0

No, because 2/5 in decimal form is 0.4 and 6/10 in decimal form is 0.6, so it's obviously not equal.

Hope I helped! <3

schepotkina [342]4 years ago

3 0

Yes they are equivalent because if you put 6/10 as its simplest form it becomes 2/5

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Write a linear function f with the values f(-4) = 0 and f(0) =0 please help

Sergeeva-Olga [200]

$f(\stackrel{x_1}{-4})=\stackrel{y_1}{0}\qquad f(\stackrel{x_2}{0})=\stackrel{y_2}{0}~\hfill (\stackrel{x_1}{-4}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{0}-\stackrel{y1}{0}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{(-4)}}}\implies \cfrac{0}{0+4}\implies \cfrac{0}{4}\implies 0$

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6 0

3 years ago

The following information was obtained from independent random samples taken of two populations. Assume normally distributed pop

Volgvan

Answer:

1. The 95% confidence interval for the difference between means is (-5.34, 11.34).

2. The standard error of (x-bar1)-(x-bar2) is 4.

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{9.2195^2}{10}+\dfrac{9.4868^2}{12}}\\\\\\s_{M_d}=\sqrt{8.5+7.5}=\sqrt{16}=4$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the difference between means.

The sample 1, of size n1=10 has a mean of 45 and a standard deviation of √85=9.2195.

The sample 2, of size n2=12 has a mean of 42 and a standard deviation of √90=9.4868.

The difference between sample means is Md=3.

$M_d=M_1-M_2=45-42=3$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{9.2195^2}{10}+\dfrac{9.4868^2}{12}}\\\\\\s_{M_d}=\sqrt{8.5+7.5}=\sqrt{16}=4$

The critical t-value for a 95% confidence interval is t=2.086.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_{M_d}=2.086 \cdot 4=8.34$