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pickupchik [31]
3 years ago
12

4% of how many days is 56 days

Mathematics
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
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