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4 years ago

Figure ABCD is graphed on a coordinate plane.

Mathematics

2 answers:

Lesechka [4]4 years ago

6 0

20.9 Is your answer

Phoenix [80]4 years ago

3 0

1. BC is 1+3=4 (units)

2. AD is 3+5= 8 (units)

3. Draw the altitude CH from point C, as shown in the figure.

4. Then triangle CHD is a right triangle, with hypothenuse CD, and sides HD and CH.

5. $CD^2=HD^2+CH^2$

$CD^2=2^2+4^2=4+16=20$

$CD= \sqrt{20}= \sqrt{4*5}= \sqrt{4} * \sqrt{5}=2 \sqrt{5}= 4.47$

6. AB=CD=4.47 because the trapezoid is isosceles.

7. P= BC+AD+AB+CD=4+8+4.47+4.47=20.9 (units)

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Pedro is doing math exercises from a software program he recently purchased. In the program, you have to get at least 70% of the

Nookie1986 [14]

The inequality would be
0.75q ≥ 0.70(q+1)

q is defined as the number of questions he answers after the first one. We are told he gets 75% of those correct; 75%=75/100=0.75. This gives us 0.75q.

Since he gains proficiency on the exercises, the total number he gets correct has to be at least 70%. This means the inequality would have the symbol greater than or equal to, as it cannot be less and have him gain proficiency.

He has already answered 1 question and answers q more; this gives us a total of q+1. Since he gains proficiency, the cutoff was 70%; 70%=70/100=0.70. This gives us the expression 0.70(q+1).

Our total inequality would then be 0.75q ≥ 0.70(q+1)

3 0

3 years ago

Read 2 more answers

A $3000 loan has an annual interest rate of 6.6% on the amount borrowed. How much time has elapsed if the interest is now $1386?

san4es73 [151]

Answer:

Step-by-step explanation:

time=interest

P*R

time=1386*100

6.6 * 3000

time=7

7 0

2 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

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

seropon [69]

Answer:

$h'(x)=\frac{3r^{2}}{2\sqrt{r^3+5}}$

Step-by-step explanation:

1) The Fundamental Theorem of Calculus in its first part, shows us a reciprocal relationship between Derivatives and Integration

$g(x)=\int_{a}^{x}f(t)dt \:\:a\leqslant x\leqslant b$

2) In this case, we'll need to find the derivative applying the chain rule. As it follows:

$h(x)=\int_{a}^{x^{2}}\sqrt{5+r^{3}}\therefore h'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left (\int_{a}^{x^{2}}\sqrt{5+r^{3}}\right )\\h'(x)=\sqrt{5+r^{3}}\\Chain\:Rule:\\F'(x)=f'(g(x))*g'(x)\\h'=\sqrt{5+r^{3}}\Rightarrow h'(x)=\frac{1}{2}*(r^{3}+5)^{-\frac{1}{2}}*(3r^{2}+0)\Rightarrow h'(x)=\frac{3r^{2}}{2\sqrt{r^3+5}}$

3) To test it, just integrate:

$\int \frac{3r^{2}}{2\sqrt{r^3+5}}dr=\sqrt{r^{3}+5}+C$

5 0

3 years ago

A 25-ounce solution is 20% alcohol. If 50 ounces of

creativ13 [48]

Answer:

20% =1/5

1/5 out of 25= 1/5*25/1=25/5=5

So 5oz -amount of ounces of alcohol in solution.

Solution is then 5oz alcohol + 20oz water =25oz

Add 50oz water

25(already existing) + 50 =75oz New solution

Still only 5oz of alcohol

5/75 =0.06666

Convert to percentage by multiplying result with 100.

Answer: 6.666%

4 0

3 years ago