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3 years ago

7

Fourteen of the kids in a class are girls. Eight of the kids wear blue shirts. Two of the kids are neither girls nor wear a blue

shirt. If five of the kids are girls who wear blue shirts, how many kids are there in the class?

2 answers:

bearhunter [10]3 years ago

7 0

Not enough information to solve, but my best assumption would be 19 kids.

masha68 [24]3 years ago

7 0

Answer:

There arent enough parts for this problem to be solved..sorry!

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Maria walks at a rate of 3 meters per second on a moving walkway traveling at a speed of 1 meter per second in the opposite dire

sasho [114]

When you set up equations like this, look at the units for help. They cancel out during division and multiplication like numbers to give desired quantities. Rates can be reversed if need be (3 meters/sec or 1/3 sec/meter).

time × (distance / time) = distance
See.. time/time = 1; leaving distance

Subtract the rate of the opposing walkway...
t × (3 m/s - 1 m/s) = d
2t = d

Answer is then A.

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4 years ago

F(x)=x^2 reflected over x=2

klasskru [66]

Answer:

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Step-by-step explanation:

To reflect a graph, f(x) over the x-axis, you take -f(x). So if f(x)=x^2, then -f(x) is -x^2. Then g(x)=-x^2 is the reflection of your function f(x) over the x-axis.

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2 years ago

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Consider the simple linear regression model Yi=β0+β1xi+ϵi, where ϵi's are independent N(0,σ2) random variables. Therefore, Yi is

Virty [35]

Answer:

See proof below.

Step-by-step explanation:

If we assume the following linear model:

$y = \beta_o + \beta_1 X +\epsilon$

And if we have n sets of paired observations $(x_i, y_i) , i =1,2,...,n$ the model can be written like this:

$y_i = \beta_o +\beta_1 x_i + \epsilon_i , i =1,2,...,n$

And using the least squares procedure gives to us the following least squares estimates $b_o$ for $\beta_o$ and $b_1$ for $\beta_1$ :

$b_o = \bar y - b_1 \bar x$

$b_1 = \frac{s_{xy}}{s_xx}$

Where:

$s_{xy} =\sum_{i=1}^n (x_i -\bar x) (y-\bar y)$

$s_{xx} =\sum_{i=1}^n (x_i -\bar x)^2$

Then $\beta_1$ is a random variable and the estimated value is $b_1$ . We can express this estimator like this:

$b_1 = \sum_{i=1}^n a_i y_i$

Where $a_i =\frac{(x_i -\bar x)}{s_{xx}}$ and if we see careful we notice that $\sum_{i=1}^n a_i =0$ and $\sum_{i=1}^n a_i x_i =1$

So then when we find the expected value we got:

$E(b_1) = \sum_{i=1}^n a_i E(y_i)$

$E(b_1) = \sum_{i=1}^n a_i (\beta_o +\beta_1 x_i)$

$E(b_1) = \sum_{i=1}^n a_i \beta_o + \beta_1 a_i x_i$

$E(b_1) = \beta_1 \sum_{i=1}^n a_i x_i = \beta_1$

And as we can see $b_1$ is an unbiased estimator for $\beta_1$

In order to find the variance for the estimator $b_1$ we have this:

$Var(b_1) = \sum_{i=1}^n a_i^2 Var(y_i) +\sum_i \sum_{j \neq i} a_i a_j Cov (y_i, y_j)$

And we can assume that $Cov(y_i,y_j) =0$ since the observations are assumed independent, then we have this:

$Var (b_1) =\sigma^2 \frac{\sum_{i=1}^n (x_i -\bar x)^2}{s^2_{xx}}$

And if we simplify we got:

$Var(b_1) = \frac{\sigma^2 s_{xx}}{s^2_{xx}} = \frac{\sigma^2}{s_{xx}}$

And with this we complete the proof required.

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4 years ago

What is the answer??

Amanda [17]

Answer:

angle 1= 104 angle 2= 76

Step-by-step explanation:

angle 1 equals x+66 and angle 2 equals 2x so you set up an equation where they are the sum of 180 degrees because the angles are supplementry angles.

x+66+2x=180 set the sum of the angles to 180

3x+66=180 combind like terms

3x=144 subract 66 from both sides

x=38 divide 3 from both sides

Now you put x back into the measurments

angle 1= (x+66) angle 2= 2x

(38+66) 2×38

angle 1= 104 angle 2= 76

Or after you know angle 1 you can subtract angle 1 from 180 to get angle 2.

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3 years ago

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Helga [31]

Answer:

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3 years ago