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4 years ago

The segments shown below could form a triangle.OA. TrueOB. False

Mathematics

1 answer:

sveta [45]4 years ago

4 0

Answer:

Step-by-step explanation:

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Which one of these points lies on the line described by the equation below y - 5 = 6 ( x - 7 )

kenny6666 [7]

Answer:

the answer would be (7,5)

6 0

3 years ago

Joe can cut and split a cord of firewood in 3 fewer hours than Dwight can. When they work together, it takes them 2 hours. How

aalyn [17]

Answer:

Dwight will take 6 hours to finish the job alone and Joe will take 3 hours to finish the job alone.

Step-by-step explanation:

Let us assume the time taken by Dwight to split a cord of firewood = K hrs

So, the per hour rate of Dwight = $(\frac{1}{K})$

As, Joe uses 3 LESS hours then Dwight.

So, the time taken by Joe to split a cord of firewood = (K- 3) hrs

So, the per hour rate of Joe = $(\frac{1}{K-3})$

Now, when both of them wok together, it takes them 2 hours.

So, the per hour rate of BOTH of them = $(\frac{1}{2})$

⇒ Per hour rate of ( Dwight + Joe) = $(\frac{1}{2} )$

$\implies (\frac{1}{K}) + (\frac{1}{K-3}) = (\frac{1}{2})$

Now, solving for the value of K , we get:

$(\frac{1}{K}) + (\frac{1}{K-3}) = (\frac{1}{2})\\\implies \frac{(K-3) + K}{K (K-3)} = (\frac{1}{2})\\\implies 2(2K -3) = K^2 - 3K\\\implies k^2 - 3K -4K +6 = 0\\\implies K^2 - 7K + 6= 0\\\implies K^2 - 6K - K + 6= 0\\\implies K(K-6) -1(K - 6)= 0\\\implies (K-6)(K-1) = 0$

Implies either K = 6 Or K = 1

But if K = 1, (K-3) = 1- 3 = -2 hours would be A CONTRADICTION.

⇒ K = 6 hours

Hence, Dwight will take 6 hours to finish the job alone and Joe will take (k-3) = (6-3) = 3 hours to finish the job alone.

8 0

3 years ago

Of the population of all fruit flies we wish to give a 90% confidence interval for the fraction which possess a gene which gives

maks197457 [2]

Answer:

The margin of error for the 90% confidence interval is of 0.038.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

To this end we have obtained a random sample of 400 fruit flies. We find that 280 of the flies in the sample possess the gene.

This means that $n = 400, \pi = \frac{280}{400} = 0.7$

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

Give the margin of error for the 90% confidence interval.

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$M = 1.645\sqrt{\frac{0.7*0.3}{400}}$

$M = 0.038$

The margin of error for the 90% confidence interval is of 0.038.

8 0

3 years ago

Determine if the graph is symmetric about the x-axis, the y-axis, or the origin.

Alexandra [31]

Answer: y-axis.

Explanation:

The equation r = 3 - 4 sin θ may be graphed using cartesian coordinates (i.e. x and y) or polar coordinates.

The logical way is to graph in polar coordinates. I did it using a graphing software and obtained the attached graph.

Also, I attached the graph in cartesian coordinates.

Here you have a table that you can build by your self and use the pairs to draw the graph (either cartesian or polar coordinates).

 θ (rad) r = 3 - 4 sin θ 

0 3 
π/8 2.61731657 
 π/4 2.29289322 
3π/8 2.07612047
 π/2 2
 5π/8 2.07612047 
3π/4 2.29289322 
7π/8 2.61731657 
π 3 
9π/8 3.38268343 
5π/4 3.70710678 
11π/8 3.92387953 
3π/2 4 
13π/8 3.92387953 
7π/4 3.70710678 
15π/8 3.38268343 
2π 3 

In both of the attached graphs you can see that the left side is equal to the right side, but the upper side is not equal to the bottom side.

Therefore, the graph is symmetric about the y-axis.

7 0

3 years ago

Read 2 more answers

The 2008 Workplace Productivity Survey, commissioned by LexisNexis and prepared by WorldOne Research, included the question, "Ho

vitfil [10]

Answer:

Therefore, the sampling distribution of $\bar{x}$ is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.

The 95% interval estimate of the population mean $\mu$ is

LCL = 7.431 hours to UCL = 10.569 hours

Step-by-step explanation:

Let X be the number of hours a legal professional works on a typical workday. Imagine that X is normally distributed with a known standard deviation of 12.6.

The population standard deviation is

$\sigma = 12.6 \: hours$

A sample of 250 legal professionals was surveyed, and the sample's mean response was 9 hours.

The sample size is

$n = 250$

The sample mean is

$\bar{x} = 9 \: hours$

Since the sample size is quite large then according to the central limit theorem, the sample mean is approximately normally distributed.

The population mean would be the same as the sample mean that is

$\mu = \bar{x} = 9 \: hours$

The sample standard deviation would be

$s = {\frac{\sigma}{\sqrt{n} }$

Where is the population standard deviation and n is the sample size.

$s = {\frac{12.6}{\sqrt{250} }$

$s = 0.7969 \: hours$

Therefore, the sampling distribution of $\bar{x}$ is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.

The population mean confidence interval is given by

$\text {confidence interval} = \mu \pm MoE\\\\$

Where the margin of error is given by

$$ MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $ \\\\$

Where n is the sampling size, s is the sample standard deviation and is the t-score corresponding to a 95% confidence level.

The t-score corresponding to a 95% confidence level is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 250 - 1 = 249

From the t-table at α = 0.025 and DoF = 249

t-score = 1.9695

$MoE = t_{\alpha/2}(\frac{\sigma}{\sqrt{n} } ) \\\\MoE = 1.9695\cdot \frac{12.6}{\sqrt{250} } \\\\MoE = 1.9695\cdot 0.7969\\\\MoE = 1.569\\\\$

So the required 95% confidence interval is

$\text {confidence interval} = \mu \pm MoE\\\\\text {confidence interval} = 9 \pm 1.569\\\\\text {LCI } = 9 - 1.569 = 7.431\\\\\text {UCI } = 9 + 1.569 = 10.569$

The 95% interval estimate of the population mean $\mu$ is

LCL = 7.431 hours to UCL = 10.569 hours

8 0

3 years ago

The segments shown below could form a triangle.OA. TrueOB. False​

The segments shown below could form a triangle.OA. TrueOB. False