Ask question

Ask question

All categories

3 years ago

13

Arthur earned $136 in three weeks . He goes back to school in one more week . He needs at least $189 to buy the new coat that he

wants for school . How much must Arthur earn in the next week?

2 answers:

nadya68 [22]3 years ago

7 0

He must earn 53 dollars

JulsSmile [24]3 years ago

5 0

He needs 53

53+136=189$

You might be interested in

the volume, v of a sphere is given by its radius, r, in the formula v=4/3(pi)(r^3) what is the formula for the radius given the

MrMuchimi

R=(3V4<span>Home: Kyle's ConverterKyle's CalculatorsKyle's Conversion Blog</span>Volume of a Sphere CalculatorReturn to List of Free Calculators<span><span>Sphere VolumeFor Finding Volume of a SphereResult:
523.599</span><span>radius (r)units</span><span>decimals<span> -3 -2 -1 0 1 2 3 4 5 6 7 8 9 </span></span><span>A sphere with a radius of 5 units has a volume of 523.599 cubed units.This calculator and more easy to use calculators waiting at www.KylesCalculators.com</span></span> Calculating the Volume of a Sphere:

Volume (denoted 'V') of a sphere with a known radius (denoted 'r') can be calculated using the formula below:

V = 4/3(PI*r3)

In plain english the volume of a sphere can be calculated by taking four-thirds of the product of radius (r) cubed and PI.

You can approximated PI using: 3.14159. If the number you are given for the radius does not have a lot of digits you may use a shorter approximation. If the radius you are given has a lot of digits then you may need to use a longer approximation.

Here is a step-by-step case that illustrates how to find the volume of a sphere with a radius of 5 meters. We'll u

π)⅓

4 0

3 years ago

I need help with #28-31 please

Aliun [14]

(28) 2x+10+x-4=21
(29) 3x-16+4x-8=60
(30) 2x-8+3x-10=17

6 0

3 years ago

A plant grows 3 centimeters in 10 days. How much does it grow per day? How can I answer my question as a fraction in simplest fo

KiRa [710]

It grows .3 cm per day

7 0

3 years ago

A group of friends went to lunch, and their bill was $132.50. An 18% gratuity was added to the bill. What was the total amount o

Serggg [28]

Answer:

$156.35

Step-by-step explanation:

we know that

A group of friends went to lunch, and their bill was $132.50

In this problem

$132.50 represent the 100%

100%+18%=118%

so

using proportion

Find out the total amount that represent 118%

Let

x ----> the total amount of the bill

$\frac{\$132.50}{100\%}=\frac{x}{118\%}\\\\x=132.50(118)/100\\\\x=\$156.35$

7 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