Let's find the unknown variables in the following cases. ) (a) If C.P. Rs 150 and profit = Rs 12, find profit percent

Explain the circumstances for which the interquartile range is the preferred measure of dispersion. What is an advantage that th

KatRina [158]

Answer:

Explain the circumstances for which the interquartile range is the preferred measure of dispersion

Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.

What is an advantage that the standard deviation has over the interquartile range?

The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.

Step-by-step explanation:

Previous concepts

The interquartile range is defined as the difference between the upper quartile and the first quartile and is a measure of dispersion for a dataset.

$IQR= Q_3 -Q_1$

The standard deviation is a measure of dispersion obatined from the sample variance and is given by:

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

Solution to the problem

Explain the circumstances for which the interquartile range is the preferred measure of dispersion

Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.

What is an advantage that the standard deviation has over the interquartile range?

The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.

8 0

3 years ago

4. A triangle has angle measures of 72º and 3°. What is the measure of the third angle?

8090 [49]

Answer:

105

the whole triangle adds up to 180

72+3=75

180-75=105

7 0

3 years ago

Find the area of the shaded region. Round the nearest hundredth if necessary. YZ=14.2m

andreyandreev [35.5K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

1

$A_1 = 67.58 \ in^2$

2

$A_2 =415.4 \ ft^2$

3

$A_3 = 8.48 \ cm^2$

4

$A_4 = 480.38 \ m^2$

Step-by-step explanation:

Generally the area of a sector is mathematically represented as

$A = \frac{\theta}{360} * \pi r^2$

Now at $r_1 = 11 in$ and $\theta_1 = 64^o$

$A_1 = \frac{64}{360} * 3.142 * 11^2$

$A_1 = 67.58 \ in^2$

Now at $r_2 = 20 ft$ in and $\theta_2 = 119 ^o$

$A_2 = \frac{119}{360} * 3.142 * 20^2$

$A_2 =415.4 \ ft^2$

Now at $r_3 = 6.5 cm$ and $\theta_3 = 23 ^o$

$A_3 = \frac{23}{360} * 3.142 * 6.5 ^2$

$A_3 = 8.48 \ cm^2$

Now at $r_4 = 14.2 m$ and $\theta_4 = 360 -87 = 273 ^o$

$A_4 = \frac{273}{360} * 3.142 * 14.2^2$

$A_4 = 480.38 \ m^2$

6 0

3 years ago

Please help :(,,,,,,???,,,,,,

nadya68 [22]

Step-by-step explanation:

The first one (f×g)(x) is asking you what is f of x times g of x. This means you multiply the function. You can multiply by the first term (5x) then the second (3) and combine like terms.

${15x}^{2} (5x) = {75x}^{3} \\ 19x(5x) = 95 {x}^{2} \\ 6(5x) = 30x$