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1 year ago

A television is on sale for 25% off the regular

Mathematics

1 answer:

zlopas [31]1 year ago

8 0

Answer:

Step-by-step explanation:

First subtract $339.95 by 25% which will equal $254.96525. Then add 15% by $254.96525 which will equal $293.206875. And always remember to round up.

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Eli and Angela are saving money to buy their grandparents an anniversary gift. Eli has saved 8 more than 1/3 of Angela's savings

Trava [24]

Answer:

$10

Step-by-step explanation:

Let a = the amount Angela has saved.

Let e = the amount Eli has saved.

"Eli has saved $8 more than 1/3 of Angela's savings."

e = a/3 + 8

"If they each save $10 more." then

Eli will have: e + 10

Angela will have a + 10

"Eli will have saved $4 more than Angela's savings."

e + 10 = a + 10 + 4

This equation simplifies to: e = a + 4

We have a system of two equations in two variables.

e = a/3 + 8

e = a + 4

Since both equations are solved for e, we just equate the right sides.

a/3 + 8 = a + 4

Subtract a/3 from both sides. Subtract 4 from both sides.

a/3 - a/3 + 8 - 4 = a - a/3 + 4 - 4

4 = (2/3)a

Multiply both sides by 3/2

(3/2)4 = (3/2)(2/3)a

6 = a

a = 6

Angela has $6.

e = a + 4 = 6 + 4 = 10

Eli has $10.

4 0

3 years ago

Please help me this is for my finals!!

Morgarella [4.7K]

Answer:

Step-by-step explanation:

7 0

3 years ago

Why does the fraction 20/25 not belong with 1/20, 2/3 and 5/4?

devlian [24]

I’m not 100% sure but I think it’s because 20/25 can still be simplified further, while the other 3 can’t.

7 0

3 years ago

The area of a circle with radius r is given by A = π r2. Find the area of a circle with radius 7 centimeters. Use 3.14 for π..

scZoUnD [109]

Step-by-step explanation:

Hey there!

Given;

Radius of a circle (r) = 7cm

Now;

We know that;

Area of circle (a) = π*(r^2)

= 3.14*(7^2)

= 3.14*49

= 153.86 cm^2

Therefore, the area of a circle is 153.86 cm^2.

Hope it helps...

4 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