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3 years ago

A family has two cars. The first car has a fuel efficiency of

1 answer:

8 0

The first car consumed 21 gallons while the second car consumed 49 gallons. here is the how it's done.
for the first car gallons consumed is 15/50x70= 21 gallons
the second car consumed 35/50x70=49 gallons

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I need you show me you answer for number 47 it's a or b or c

nikitadnepr [17]

The answer for question 47 is A

3 0

3 years ago

An investment website can tell what devices are used to access their site. The site managers wonder whether they should enhance

lbvjy [14]

Answer: a) Unimodal and symmetric

b) 0.26

c) 0.038

Step-by-step explanation:

Given: Sample size of investors (n)= 131

True proportion of smartphone users(p) =26%

a) Since sampling distribution for the sample proportion is approximately normal when n is larger.

Normal distribution is Unimodal and symmetric.

So correct option : Unimodal and symmetric

b) mean of this sampling distribution = p = 0.26

c) standard deviation of the samplingdistribution = $\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.26\times (1-0.26)}{131}}=\sqrt{0.00146870229008}$

$=0.0383236518364\approx0.038$

6 0

3 years ago

Add: (7-5i) (-10+2i)

Mice21 [21]

Answer:

-10i^(2) + 64i - 70

5 0

3 years ago

If a is a positive integer and b is a negative integer, which expression must be positive?

nydimaria [60]

So, let's solve all of them and see:

a - (-b) = a + b ⇒ can be either positive or negative, depending on the numbers

a + (-b) = a - b ⇒ can either be positive or negative, depending on the numbers

a * (-b) = a * (-b) ⇒ negative answer no matter what

a / (-b) = a / (-b) ⇒ negative answer no matter what

So the answer is:

<em>"None of the above"</em>.

Hope it helped,

BioTeacher101

(If you have any questions, feel free to ask them in the comments)

4 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:

$E(\hat \theta_i) = \mu , i = 1,2$