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pickupchik [31]

3 years ago

12

4% of how many days is 56 days

1 answer:

Varvara68 [4.7K]3 years ago

7 0

4% of X days is 56. 0.04x=56, x=56/0.04, x=1400

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If x+1/x = 5 find the value of x^3 +1/x^3. pls solve

Korolek [52]

Answer:

1) solve x+1/x = 5

$\frac{x + 1}{x} = 5$

$x + 1 = 5x$

$x - 5x = - 1$

$- 4x = - 1$

$x = \frac{ -1}{ - 4}$

$x = \frac{1}{4}$

2) solve x³+1/x³

$x {}^{3} + \frac{1}{x {}^{3} }$

substitute x = 1/4 into the expression

$( \frac{1}{4} ) {}^{3} + \frac{1}{( \frac{1}{4}) {}^{3} }$

$\frac{1}{64} + \frac{1}{ \frac{1}{64} }$

$\frac{1}{64} + \frac{64 \times 1}{1}$

$\frac{1}{64} + 64$

$\frac{4097}{64} \: or \: 64.02$

5 0

3 years ago

Help Plzz<br>Which section of the function is increasing? <br>ОА <br>ОВ <br>ОС <br>OD

Rama09 [41]

Answer: A

Step-by-step explanation:

because i nsaid sooooo

3 0

2 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

3(5z-7)+2(9z-11)=4(8z-7)-111

NISA [10]

3(5z - 7) + 2(9z - 11) = 4(8z - 7) - 111
15z - 21 + 18z - 22 = 32z - 28 - 111
33z - 43 = 32z - 139
33z - 32z = -139 + 43
z = - 96

5 0

3 years ago

Read 2 more answers

if the cost of 3 chocolates and 2 cookies is $22 and that of 2 chocolates and 3 cookies is $18, what is the cost of cookies.

Vanyuwa [196]

Answer:$2

Step-by-step explanation:Express as two equations . Lex x Be chocolate and y be cookies.

3x+2y=22

2x+3y= 18

Common factor of 6 so times first ran by 2, second ran by 3 . Eliminate.

4 0

2 years ago

Read 2 more answers