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Reika [66]
4 years ago
12

the length of a pool is 10 meters more than twice the width. The area of the pool is 1500 square meters. What is the length of t

he pool?
Mathematics
2 answers:
Valentin [98]4 years ago
5 0
Length is 1010, width is 494.
Arte-miy333 [17]4 years ago
4 0
Let's make the length L and the width w. So, here's what we know:

L = 2w + 10

Lw = 1500

Substitute L in as 2w + 10:

w(2w + 10) = 1500.

Simplify.

2w squared +10w = 1500

Make it a quadratic equation, w = 25 or -30. Since it can't be negative, w = 25.

1500/25 = 60.

Length = 60
Width = 25


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Answer:

angle JKL is 106

Step-by-step explanation:

an isosceles triangle means two sides are the same and the bottom two angles are the same length. So you would equal the two equations and solve it to get x. Then replace one of the equation's x with the number you got, which would be 5, and solve that. Since both bottom angle are the same, the other angle would be the same degree, 37*. A triangle is 180* total, so subtract that by two 37's and you'll get the answer.

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3 years ago
What number should replace the letter x in the "Seed color" row?<br> 152<br> 207<br> 224<br> 2,001
Reptile [31]

Answer:c

Step-by-step explanation:

4 0
3 years ago
Are these lines parallel or not: L1: (1,2), (3,1), and L2: (0,-1), (2,0)
Phoenix [80]

First find the slope of the first line

L1

m = ( y2-y1)/(x2-x1)

m = ( 1-2)/(3-1)

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Now find the slope of the second line

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m= ( y2-y1)/(x2-x1)

= ( 0 - -1)/(2 -0)

= (0+1)/(2-0)

= 1/2

Parallel lines have the same slope

These lines do not have the same slope, they are different by a negative sign.

These lines are not parallel.

5 0
1 year ago
The length of a rectangle is 2 feet longer than its width the perimeter is 16 feet. find the dimensions of the rectangle.
sergiy2304 [10]

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4 0
3 years ago
Let X1, X2, ... , Xn be a random sample from N(μ, σ2), where the mean θ = μ is such that −[infinity] &lt; θ &lt; [infinity] and
Sliva [168]

Answer:

l'(\theta) = \frac{1}{\sigma^2} \sum_{i=1}^n (X_i -\theta)

And then the maximum occurs when l'(\theta) = 0, and that is only satisfied if and only if:

\hat \theta = \bar X

Step-by-step explanation:

For this case we have a random sample X_1 ,X_2,...,X_n where X_i \sim N(\mu=\theta, \sigma) where \sigma is fixed. And we want to show that the maximum likehood estimator for \theta = \bar X.

The first step is obtain the probability distribution function for the random variable X. For this case each X_i , i=1,...n have the following density function:

f(x_i | \theta,\sigma^2) = \frac{1}{\sqrt{2\pi}\sigma} exp^{-\frac{(x-\theta)^2}{2\sigma^2}} , -\infty \leq x \leq \infty

The likehood function is given by:

L(\theta) = \prod_{i=1}^n f(x_i)

Assuming independence between the random sample, and replacing the density function we have this:

L(\theta) = (\frac{1}{\sqrt{2\pi \sigma^2}})^n exp (-\frac{1}{2\sigma^2} \sum_{i=1}^n (X_i-\theta)^2)

Taking the natural log on btoh sides we got:

l(\theta) = -\frac{n}{2} ln(\sqrt{2\pi\sigma^2}) - \frac{1}{2\sigma^2} \sum_{i=1}^n (X_i -\theta)^2

Now if we take the derivate respect \theta we will see this:

l'(\theta) = \frac{1}{\sigma^2} \sum_{i=1}^n (X_i -\theta)

And then the maximum occurs when l'(\theta) = 0, and that is only satisfied if and only if:

\hat \theta = \bar X

6 0
3 years ago
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