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3 years ago

ABCD is a parallelogram. Find the indicated measure:

Mathematics

1 answer:

uranmaximum [27]3 years ago

7 0

Answer:

12 12 12 12 12 12 12 12 12 1 2 12 12

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HELPPPPPPP PLEASE i need to know the answer like right now pleaseee just look at the picture and please help

Marrrta [24]

I’m pretty sure it’s B. But I could be wrong.

7 0

2 years ago

Read 2 more answers

For Jane Jones is a single person with 2 exemptions. In a week she earns $212.12. Using the chart above, Jane's federal income t

SpyIntel [72]

Answer:

The other answer is INCORRECT the CORRECT answer is : 34.00

Step-by-step explanation:

In the chart it shows how much for the number of exemptions. The other answer only included one but the question asks for TWO. Therefore making it 34.00

6 0

3 years ago

The Ohio Department of Agriculture tested 203 fuel samples across the state

Rus_ich [418]

Answer:

$\hat p = \frac{14}{105}= 0.133$

And that represent the proportion of failures.

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.133 - 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.0475$

$0.133 + 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.2185$

The 99% confidence interval would be given by (0.0475;0.2185)

Step-by-step explanation:

Previous concept

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58$

The proportion estimated would be:

$\hat p = \frac{14}{105}= 0.133$

And that represent the proportion of failures.

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.133 - 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.0475$

$0.133 + 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.2185$

The 99% confidence interval would be given by (0.0475;0.2185)

3 0

3 years ago

Read 2 more answers

For number 6, evaluate the definite integral.

maks197457 [2]

$\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{(8+2x)^2}}\cdot dx\impliedby \textit{now, let's do some substitution}\\\\
-------------------------------\\\\
u=8+2x\implies \cfrac{du}{dx}=2\implies \cfrac{du}{2}=dx\\\\
-------------------------------\\\\$

$\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{u^2}}\cdot \cfrac{du}{2}\implies \cfrac{1}{2}\int\limits_{0}^{28}\ u^{-\frac{2}{3}}\cdot du\impliedby 
\begin{array}{llll}
\textit{now let's change the bounds}\\
\textit{by using } u(x)
\end{array}\\\\
-------------------------------\\\\
u(0)=8+2(0)\implies u(0)=8
\\\\\\
u(28)=8+2(28)\implies u(28)=64$

$\bf \\\\
-------------------------------\\\\
\displaystyle \cfrac{1}{2}\int\limits_{8}^{64}\ u^{-\frac{2}{3}}\cdot du\implies \cfrac{1}{2}\cdot \cfrac{u^{\frac{1}{3}}}{\frac{1}{3}}\implies \left. \cfrac{3\sqrt[3]{u}}{2} \right]_8^{64}
\\\\\\
\left[ \cfrac{3\sqrt[3]{(2^2)^3}}{2} \right]-\left[ \cfrac{3\sqrt[3]{2^3}}{2} \right]\implies \cfrac{12}{2}-\cfrac{6}{2}\implies 6-3\implies 3$

3 0

2 years ago

You have 19 digs for your current volleyball season. There are 5 games left in the season. You want to break your previous recor

Charra [1.4K]

Answer:We must get more than 7 digs in the remaining 5 games to break record of 26.

Step-by-step explanation:

Given situation:- Let x represents the number of digs which we must get in the remaining 5 games to break your record ,and we have 19 digs for our current volleyball season.

To break previous record of 26 digs in a season, we have to get more than 26 i.e. 19 digs added to x must be greater than 26

⇒x+19 ≥ 26

⇒x≥26-19⇒x≥7

Therefore,we must get more than 7 digs in the remaining 5 games to break record of 26.

4 0

3 years ago