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3 years ago

5.401×10 −1 as a standard form.

Mathematics

2 answers:

zvonat [6]3 years ago

4 0

A negative exponent means to move the decimal point to the left the number of the exponent.

10^-1 you would move the decimal 1 place to the left.

5.401 x 10^-1 = 0.5401

stellarik [79]3 years ago

3 0

$\huge\boxed{0.5401}$

We have the equation...

$\boxed{5401*10^-^1=(?)}$

Multiply 5401 by 10⁻¹.

$\boxed{5401*10^-^1=0.5401}$

Therefore, your final answer is $\boxed{0.5401}$

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What are the zeros of the function f(x) = (3x^2 + 9x + 6) / (3x - 6) ?

cricket20 [7]

3x^2 + 9x + 6 = 0
3x^2 + 3x + 6x + 6 = 0
3x(x + 1) + 6(x + 1) = 0
(3x + 6)(x + 1) = 0
3x + 6 = 0 and x + 1 = 0
3x = -6 and x = -1
x = -2 and x = -1

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3 years ago

If there are 9 stars for every 2 hearts, how many stars for 4 hearts?

Archy [21]

Answer:

18 stars

Step-by-step explanation:

We can use a ratio to solve

9 stars x

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2 hearts 4 hearts

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9*4 = 2x

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3 0

3 years ago

Read 2 more answers

What is 7.06 divide 0.353

solong [7]

7.06 ➗ 0.353= 20

Hope this helps! :3

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3 years ago

A 14-pound bag of All-Natural dog food is $39.75. If the prices of different weighted bags of this brand of dog food are proport

tekilochka [14]

Answer:

$19.88

Step-by-step explanation:

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8 0

3 years ago

Let X1 and X2 be independent random variables with mean μand variance σ².

My name is Ann [436]

Answer:

a) $E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu$

So then we conclude that $\hat \theta_1$ is an unbiased estimator of $\mu$

$E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu$

So then we conclude that $\hat \theta_2$ is an unbiased estimator of $\mu$

b) $Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}$

$Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}$

Step-by-step explanation:

For this case we know that we have two random variables:

$X_1 , X_2$ both with mean $\mu = \mu$ and variance $\sigma^2$

And we define the following estimators:

$\hat \theta_1 = \frac{X_1 + X_2}{2}$

$\hat \theta_2 = \frac{X_1 + 3X_2}{4}$

Part a

In order to see if both estimators are unbiased we need to proof if the expected value of the estimators are equal to the real value of the parameter:

$E(\hat \theta_i) = \mu , i = 1,2$

So let's find the expected values for each estimator:

$E(\hat \theta_1) = E(\frac{X_1 +X_2}{2})$

Using properties of expected value we have this:

$E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu$

So then we conclude that $\hat \theta_1$ is an unbiased estimator of $\mu$

For the second estimator we have:

$E(\hat \theta_2) = E(\frac{X_1 + 3X_2}{4})$

Using properties of expected value we have this:

$E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu$

So then we conclude that $\hat \theta_2$ is an unbiased estimator of $\mu$

Part b

For the variance we need to remember this property: If a is a constant and X a random variable then:

$Var(aX) = a^2 Var(X)$

For the first estimator we have:

$Var(\hat \theta_1) = Var(\frac{X_1 +X_2}{2})$

$Var(\hat \theta_1) =\frac{1}{4} Var(X_1 +X_2)=\frac{1}{4} [Var(X_1) + Var(X_2) + 2 Cov (X_1 , X_2)]$

Since both random variables are independent we know that $Cov(X_1, X_2 ) = 0$ so then we have:

$Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}$

For the second estimator we have:

$Var(\hat \theta_2) = Var(\frac{X_1 +3X_2}{4})$

$Var(\hat \theta_2) =\frac{1}{16} Var(X_1 +3X_2)=\frac{1}{4} [Var(X_1) + Var(3X_2) + 2 Cov (X_1 , 3X_2)]$

Since both random variables are independent we know that $Cov(X_1, X_2 ) = 0$ so then we have:

$Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}$

7 0

3 years ago