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Simora [160]
4 years ago
7

-5} =4" align="absmiddle" class="latex-formula">
Mathematics
2 answers:
Brums [2.3K]4 years ago
5 0

\sqrt[3]{x-5}=4

Cube the equation.

x-5=64

Bring 5 to the other side.

x=69

⭐ Answered by Hyperrspace (Ace) ⭐

⭐ Brainliest would be appreciated, I'm trying to reach genius! ⭐

⭐ If you have questions, leave a comment, I'm happy to help! ⭐

zloy xaker [14]4 years ago
4 0

Answer:

X=69

Step-by-step explanation:

\left(\sqrt[3]{x-5}\right)^3=4^3

x-5=64

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kenny6666 [7]
So, it starts off with 10 half gallons. 10*1/2 is 5, so in other words she put in 5 gallons and paid 15.26. As an equation it can be represented by 5x=15.26 where x is the cost of the gallons which is what we're trying to find. We can solve by using the division property of equality and divide both sides by 5. This gets us x=3.052. So, it costs $3.052 exactly for a gallon of gasoline.
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3 years ago
What is the answer ?
mr Goodwill [35]

Answer:

$14476.13

Step-by-step explanation:

This is an example of a geometric sequence in which the common ratio (the number you multiply a term by to get the next one) is less than 1.

If the value of the car is decreasing by 12.5% each year, so each year, the value is 100% - 12.5% = 87.5% of the value in the previous year.

After 5 years, the value is 28750(0.875)^5.

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4 years ago
The triangular sail of a toy sailboat is supposed to be a right triangle. The manufacturer says the sides of the sail have lengt
Shtirlitz [24]
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3 years ago
Does anyone know how to do this?
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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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