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Softa [21]
3 years ago
6

A salon increased the sales by about 12.5% this week. If they sold 43 products last week, how many did they sell this week?

Mathematics
2 answers:
levacccp [35]3 years ago
6 0
43 + (43 \times 0.125) = 43\times 1.125\\\\= \boxed{\bf{ 48.375}}

The salon sold 48.375 products this week.

Your final answer is 48.375 products.
natulia [17]3 years ago
3 0
<span>A salon increased the sales by about 12.5% this week.
Question: If they sold 43 products last week, how many did they sell this week?
=> 43 products sold are there usual sale every week.
=> Now, this week, they were able to increase their sales up to 12.5%. Let’s solve to find out how many products were added to their sale this week.
=> 43 products * 12.5%
=> 43 * .125 = 5.4 products were added to their sale and got a total of:
=> 43 + 5.4 = 48.4 products

</span>



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professor190 [17]

Answer:

(- 5, 1 )

Step-by-step explanation:

- 6x - 14y = 16 → (1)

- 2x + 7y = 17 → (2)

Multiplying (2) by - 3 and adding to (1) will eliminate the x- term

6x - 21y = - 51 → (3)

Add (1) and (3) term by term to eliminate x

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2 years ago
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1 year 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})

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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}

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3 years ago
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Korvikt [17]

Answer:

102

Step-by-step explanation:

89 + a

Let a=13

89+13

102

4 0
3 years ago
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