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3 years ago

What is the distance between the pivot point and the marking point of a compass when drawing a circle?

1 answer:

3 0

I'm sure you have seen or used a compass already. The pivot point is the needle part. It is called as such because it is the point where the compass turns so that the movie end of the other leg of the compass can make arcs.

As you create a circle using a compass, that same distance between the two legs of the compass is the radius of the circle. By definition, a circle is a round two-dimensional figure that is composed of a series of points that is equidistant to a fixed point inside the circle. That point is called the center of the circle represented by the pivot or needle part. The equal distance of all the points along the circumference is the radius.

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A salesman for a shoe company claimed runners would run faster races if they wore the company's brand of running shoe. A track c

gregori [183]

Answer:

Correct option: (c) Paired samples t confidence interval.

Step-by-step explanation:

A paired data is collected from an experiment that is conducted twice; once before applying a treatment and second time after the treatment has been applied.

For example, scores of students before the coaching classes and after the coaching classes.

In this case a paired data is collected.

The runners were asked to run 100-yard wearing their usual shoes one day and do same thing the next day wearing the company's shoes.

The two data collected from both the days is a paired data.

To determine whether the salesman's claim that the runners would run faster races if they wore the company's brand of running shoe is correct or not a confidence interval can be constructed for the paired the data.

If the true mean difference is contained in the interval then the salesman's claim can be considered true otherwise false.

To construct this confidence interval use a paired sample t confidence interval.

Because the population standard deviations are not known.

The (1 - <em>α</em>)% confidence interval for the paired difference is:

$CI=\bar X_{diff}\pm t_{\alpha/2, (n-1)} \frac{\hat \sigma_{diff}}{\sqrt{n}}$

3 0

3 years ago

Correct Answers only!

SOVA2 [1]

Answer:

$128000

Step-by-step explanation:

Using the given formula :

I = P × R × T

I = 80000 × 15/100 × 4

I = $48000

He saves = 80000 + 48000 = $128000

5 0

3 years ago

How many times does 2 go into4

Oduvanchick [21]

Twice, two times. Good luck with it all.

7 0

4 years ago

Read 2 more answers

Finn changes his mind and, from now on, decides to take the normal route to work everyday. On any given day, the time (in minute

Margaret [11]

Answer:

The 33rd percentile of the time it takes Finn to get to work on any given day is 31.04 minutes.

There is a 61.92% probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.

Step-by-step explanation:

This can be solved by the the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

$Z = \frac{X - \mu}{\sigma}$

Each z-score value has an equivalent p-value, that represents the percentile that the value X is:

The problem states that:

Mean = 35, so $\mu = 35$

Variance = 81. The standard deviation is the square root of the variance, so $\sigma = \sqrt{81} = 9$ .

Find the 33rd percentile of the time it takes Finn to get to work on any given day. Do not include any units in your answer.

Looking at the z-score table, $z = -0.44$ has a pvalue of 0.333. So what is the value of X when $z = -0.44$ .

$Z = \frac{X - \mu}{\sigma}$

$-0.44 = \frac{X - 35}{9}$

$X - 35 = -3.96$

$X = 31.04$

The 33rd percentile of the time it takes Finn to get to work on any given day is 31.04 minutes.

Over the next 2 days, find the probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.

$P = P_{1} + P_{2}$

$P_{1}$ is the probability that Finn took more than 40.5 minutes to get to work on the first day. The first step to solve this problem is finding the z-value of $X = 40.5$ .

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{40.5 - 35}{9}$

$Z = 0.61$

$Z = 0.61$ has a pvalue of 0.7291. This means that the probability that it took LESS than 40.5 minutes for Finn to get to work is 72.91%. The probability that it took more than 40.5 minutes if $P_{1} = 100% - 72.91% = 27.09% = 0.2709$

$P_{2}$ is the probability that Finn took more than 38.5 minutes to get to work on the second day. Sine the probabilities are independent, we can solve it the same way we did for the first day, we find the z-score of

$X = 38.5$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{38.5 - 35}{9}$

$Z = 0.39$

$Z = 0.39$ has a pvalue of 0.6517. This means that the probability that it took LESS than 38.5 minutes for Finn to get to work is 65.17%. The probability that it took more than 38 minutes if $P_{1} = 100% - 65.17% = 34.83% = 0.3483$

So:

$P = P_{1} + P_{2} = 0.2709 + 0.3483 = 0.6192$

There is a 61.92% probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.

7 0

4 years ago

Help me answer this please!!! will mark brainliest

Anni [7]

Answer:WX=65

Step-by-step explanation:as both traiangles are homologous,

WX/XZ=VW/ ZY

WX= 52×10/8

5 0

3 years ago