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seropon [69]
3 years ago
7

A high school track is shaped as a rectangle with a half circle on either side.Jake plans on running four laps. How many meters

will Jake run? Use 3.14 for Pi.
Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
3 0

Answer:

if you go around a track one time thats 400 meters but if you go around 4 times thats 1600 meters, you dont need to use 3.14 pi for this " no offense", i do track and field myself and i do the short distance, which is 100 meters and 200 meter but for for long distance runners they go around the track 4- 8 times so it is 1600 meters is ur answer. and when u go around 8 times, thats 3200 meters.

Step-by-step explanation:

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Therefore, ur equation is : y = -1
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What is the focus of the parabola? <img src="https://tex.z-dn.net/?f=y%3D%20%5Cfrac%7B1%7D%7B4%7D%20x%5E%7B2%7D%20-x%2B3" id="Te
OlgaM077 [116]
We know that
the equation of the parabola is of the form
y=ax²+bx+c
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3 0
2 years ago
Read 2 more answers
When a distribution is mound-shaped symmetrical, what is the general relationship among the values of the mean, median, and mode
yuradex [85]

Answer:

The mean, median, and mode are approximately equal.

Step-by-step explanation:

The mean, median, and mode are <em>central tendency measures</em> in a distribution. That is, they are measures that correspond to a value that represents, roughly speaking, "the center" of the data distribution.

In the case of a <em>normal distribution</em>, these measures are located at the same point (i.e., mean = median = mode) and the values for this type of distribution are symmetrically distributed above and below the mean (mean = median = mode).

When a <em>distribution is not symmetrical</em>, we say it is <em>skewed</em>. The skewness is a measure of the <em>asymmetry</em> of the distribution. In this case, <em>the mean, median and mode are not the same</em>, and we have different possibilities as the mentioned in the question: the mean is less than the median and the mode (<em>negative skew</em>), or greater than them (<em>positive skew</em>), or approximately equal than the median but much greater than the mode (a variation of a <em>positive skew</em> case).  

In the case of the normal distribution, the skewness is 0 (zero).

Therefore, in the case of a <em>mound-shaped symmetrical distribution</em>, it resembles the <em>normal distribution</em> and, as a result, it has similar characteristics for the mean, the median, and the mode, that is, <em>they are all approximately equal</em>. So, <em>the </em><em>general</em><em> relationship among the values for these central tendency measures is that they are all approximately equal for mound-shaped symmetrical distributions, </em>considering they have similar characteristics of the <em>normal distribution</em>, which is also a mound-shaped symmetrical distribution (as well as the t-student distribution).

5 0
2 years ago
The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
marusya05 [52]

Answer:

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

\hat p =\frac{955-812}{955}= 0.150

0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131

0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169

And the 90% confidence interval would be given (0.131;0.169).

Step-by-step explanation:

We have the following info given:

n= 955 represent the sampel size slected

x = 812 number of students who read above the eighth grade level

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

\hat p =\frac{955-812}{955}= 0.150

The confidence interval for the proportion  would be given by this formula

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

For the 90% confidence interval the significance is \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution and we got.

z_{\alpha/2}=1.64

And replacing into the confidence interval formula we got:

0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131

0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169

And the 90% confidence interval would be given (0.131;0.169).

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