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Eduardwww [97]
2 years ago
6

Which is larger? 7% or 0.7 and explain how you know.

Mathematics
2 answers:
Katyanochek1 [597]2 years ago
8 0
7% means you divide 7 by 100.

7 / 100 = .07.

Which is larger?
.07 is smaller than .7 so that means that .7 is larger than 7%.

Explain how you know.
I know because I found the decimal form of 7% which is .07 and I know that .7 is larger than .07.
disa [49]2 years ago
7 0
They are equal because you can measure the percentage as a decimal with a maximum of 1.
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Your math teacher gives you an extra credit problem to figure out her age. Half of her age three years
MatroZZZ [7]

Answer:

Teacher is 27 years old

Step-by-step explanation:

(1/2)(t - 3) = (1/3)(t + 9)

multiply by 6 to clear the fraction

3(t - 3) = 2(t + 9)

Distribute

3t - 9 = 2t + 18

Subtract 2t from both sides

t - 9 = 18

Add 9 to both sides

t = 27

Teacher is 27 years old

6 0
2 years ago
Are 6/10 and 6/30 multiples of 3/10
Fantom [35]
10 is not a multiple of 3 so no 6/10 is not multiples for 3/10 and 6 is not a mutiple of 10 so it can't be also the same with 6/30 although 6/30 is multiples for 3 but 6 is not a mutiple for 10.
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3 years ago
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B. Construct Arguments Tom says his friends ate the same amount of vegetable pizza as pepperoni pizza. How could that be true?​
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Step-by-step explanation:

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2 years ago
Read 2 more answers
Help on math please!
DedPeter [7]
(13•4) + (3•8) +3 = 79
The total cost is $79
6 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
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