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forsale [732]
3 years ago
6

F20 in-1 30 in. 13 in. 13 in.

Mathematics
1 answer:
scoray [572]3 years ago
7 0

Answer:

13

Step-by-step explanation:

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What is the missing constant term in the perfect square that starts x^2+10x?
max2010maxim [7]
<span>x^2+10x + 25 = (x + 5)^2

missing = 25

hope it helps</span>
4 0
3 years ago
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an initial population of 5 squirrels increases by 9% each year for 10 years.using x for years and y for the number of squirrels,
ivann1987 [24]

Answer:

Equation: y(1.09)^x

After 10 years there will be about 12 squirrels.

Step-by-step explanation:

y(1.09)^x

=5(1.09)^10

=(approx.) 12 squirrels

5 0
3 years ago
An airline charges an extra fee if a suitcase weighs more than 50 pounds. After packing, Li's suitcase weighs 47.75 pounds. Whic
Vinvika [58]

Answer:

47.75 + x Less-than-or-equal-to 50

= 47.75 + x ≤ 50

Step-by-step explanation:

Solving the above Question:

Not going over the 50 pound case mean, less than or equal to 50 pounds

Let the extra pound of weight be represented as x

Hence, the inequality equation that can be used to determine how much more weight can be added to the suitcase without going over the 50-pound weight limit =

47.75 + x ≤ 50

4 0
3 years ago
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MARKING POEPLE AS BRIANLIST ALL DAY if you answer the question and it is right thank u
7nadin3 [17]

Answer: AAS that's what it should be

Step-by-step explanation:

5 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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