Ask question

Ask question

All categories

katovenus [111]

2 years ago

15

One cup of grated, fresh Parmesan cheese weighs 3 ounces. How many cups of grated cheese can you expect to get from a 4.5-poun

d block of fresh Parmesan?

1 answer:

Ksenya-84 [330]2 years ago

5 0

Answer:

9 cups

Step-by-step explanation:

If one pound is equal to 2 cups, then you would multiply 2 by 4, which gives you 8. If one pound equals 2 cups, then half a pound equals 1 cup. So since we already have 8 cups, you just add one more cup, which gives you 9 cups total

SUMMARY: You can expect to get 9 cups of grated parmesan cheese from a 4.5-pound block.

You might be interested in

Allan drew a polygon with 4 sides and 4 angles. All four sides are equal. None of the angles are right angles. What figure did A

Lady_Fox [76]

Answer: The figure drawn by Alan is a rhombus.

Step-by-step explanation:

Given: Allan drew a polygon with 4 sides and 4 angles with two properties:-

All four sides are equal.
None of the angles are right angles.

We know that only a rhombus is a flat shape with 4 equal straight sides and 4 angles .All sides have equal length. Its angles need not to be right angle.

Therefore, the figure drawn by Alan is a rhombus.

8 0

3 years ago

Read 2 more answers

Beeswax used in making candles is produced by honeybees produce 7 pounds of honey for each pound of wax they produce. How many p

Mkey [24]

Answer:

1,015

Step-by-step explanation:

7 pounds of honey equals 1 pound of wax

So if we have 145 pounds of was, we multiply 145 by 7

145x7= 1,015

5 0

2 years ago

Matemática como resolve matriz 3x5

koban [17]

Well, 1x5=5 and 2x5=10 so 3x5=15

8 0

3 years ago

Read 2 more answers

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

In the problem 12+58=70,12 and 58 are called factors true or false

Marat540 [252]

Answer:

false

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers