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3 years ago

When j = 32, m = 2, and p = 1/4. If j varies directly with m and inversely with p, what is the constant of variation?

2 answers:

5 0

J = mk/p where k is the constant of variation.
32 = 2*k / (1/4)
Multiply both sides by 1/4.
8 = 2k
Divide both sides by 2.
4 = k

finlep [7]3 years ago

4 0

J = constant*m/p

32 = constant*2/(1/4)= 8 * constant,

constant=4!

Got it?

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Given:

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$\sigma_B=5$

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To find:

The variance. of combined set.

Solution:

Formula for variance is

$\sigma^2=\dfrac{\sum (x_i-\overline{x})^2}{n}$ ...(i)

Using (i), we get

$\sigma_A^2=\dfrac{\sum (x_i-\overline{x}_A)^2}{n_A}$

$(3)^2=\dfrac{\sum (x_i-\overline{x}_A)^2}{6}$

$9=\dfrac{\sum (x_i-\overline{x}_A)^2}{6}$

$54=\sum (x_i-\overline{x}_A)^2$

Similarly,

$\sigma_B^2=\dfrac{\sum (x_i-\overline{x}_B)^2}{n_B}$

$(5)^2=\dfrac{\sum (x_i-\overline{x}_B)^2}{4}$

$25=\dfrac{\sum (x_i-\overline{x}_B)^2}{4}$

$100=\sum (x_i-\overline{x}_B)^2$

Now, after combining both sets, we get

$\sigma^2=\dfrac{\sum (x_i-\overline{x}_A)^2+\sum (x_i-\overline{x}_B)^2}{n_A+n_B}$

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$\sigma^2=\dfrac{154}{10}$

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Therefore, the variance of combined set is 15.4.

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2 years ago

What did I do wrong I really need help!!!!!!!!!!!!!!!!!!!

Leviafan [203]

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3 years ago

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2 years ago

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