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earnstyle [38]
3 years ago
6

Multiply: 1/6 * 2/3

Mathematics
2 answers:
Romashka [77]3 years ago
5 0
Problem⬇
1/6 (2/3)
Work⬇
1(2) / 6(3) = 2/18
Now, we get to your result for this problem
Your answer is 1/9
-----------------------------------------------------------------
1/6 × 2/3 = 1/9

☺

AnnZ [28]3 years ago
3 0
2/18 reduced = 1/9 that's my answer
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Round each decimal 32.62 tenths place
Tresset [83]

Answer:

32.6

Step-by-step explanation:

Look at the decimal in the hundredths place in 32.62

If the number is 0-4, round down. If the number is 5-9, round up.

In 32.62, the number in the hundredths place is 2. So, we round down to 32.6

3 0
2 years ago
Whats 926 divided by 30
balandron [24]

30.87 ✅

\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}

\frac{926}{30}  \\  = 30 \frac{26}{30}  \\ ( \: or \: ) \\  = 30.8666 \\  = 30.87

\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{!}}}}}

4 0
2 years ago
NEED HELP PLEASE ASAP!!!!!!!!!!!!!!
DedPeter [7]

So you know that 30 + 30 + w + w = P. We also know the area is 240 square feet. Divide 240 by 30 to find the width of the area. 240 / 30 = 8. w = 8. So add 30 + 30 + 8 + 8 to find that P = 76. HOWEVER, remember that George is using part of his house as a wall instead of a fence. So subtract 8 from 76 to get 68 ft. George will need 68 ft. of fencing for his dog run.

4 0
3 years ago
Read 2 more answers
How do I count by twelfths
Korvikt [17]

12,  x 1 = 12

12 x 2 =24  

12 x 3 = 36

12 x 4 =48

12 x 5= 60  

12 x 6 =72  

 

12 x 7 = 84

 

12 x 8= 96

12 x 9= 108  

12 x 10 = 120  

 

12 x 11=132

12x 12= 144

hope this helps

8 0
3 years ago
Suppose a certain type of fertilizer has an expected yield per acre of mu 1 with variance sigma 2, whereas the expected yield fo
mart [117]

Answer:

See the proof below.

Step-by-step explanation:

For this case we just need to apply properties of expected value. We know that the estimator is given by:

S^2_p= \frac{(n_1 -1) S^2_1 +(n_2 -1) S^2_2}{n_1 +n_2 -2}

And we want to proof that E(S^2_p)= \sigma^2

So we can begin with this:

E(S^2_p)= E(\frac{(n_1 -1) S^2_1 +(n_2 -1) S^2_2}{n_1 +n_2 -2})

And we can distribute the expected value into the temrs like this:

E(S^2_p)= \frac{(n_1 -1) E(S^2_1) +(n_2 -1) E(S^2_2)}{n_1 +n_2 -2}

And we know that the expected value for the estimator of the variance s is \sigma, or in other way E(s) = \sigma so if we apply this property here we have:

E(S^2_p)= \frac{(n_1 -1 )\sigma^2_1 +(n_2 -1) \sigma^2_2}{n_1 +n_2 -2}

And we know that \sigma^2_1 = \sigma^2_2 = \sigma^2 so using this we can take common factor like this:

E(S^2_p)= \frac{(n_1 -1) +(n_2 -1)}{n_1 +n_2 -2} \sigma^2 =\sigma^2

And then we see that the pooled variance is an unbiased estimator for the population variance when we have two population with the same variance.

8 0
3 years ago
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