What is 9.66 rounded to the nearest tenth

(1/16)^x+3=(1/4)^x+1

Anika [276]

Answer:

x=1

....................

6 0

2 years ago

6 multiplication problems that ar involved in the solving of 325×89

gogolik [260]

325*89.1

Let's decompose this into:

325 = 300 + 25

89.1 = 80 + 9 + 0.1

Then we have the multiplication:

(300 + 25)*(80 + 9 + 0.1)

Let's distribute the multiplication:

300*80 + 300*9 + 300*0.1 + 25*80 + 25*9 + 25*0.1

So now we have 6 multiplications that are a lot easier to solve than the initial one that we had.

Then the list of six multiplications involved in solving this problem are:

300*80 = 24,400

300*9 = 2,700

300*0.1 = 30

25*80 = 2,000

25*9 = 225

25*0.1 = 2.5

Now we add all of those and get:

325*89.1 = 24,400 + 2,700 + 30 + 2,000 + 225 + 2.5 = 28,957.5

4 0

2 years ago

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

Find the surface area S of the following cylinder.

Tanya [424]

Answer:

384π

Step-by-step explanation:

From the diagram,

Diameter ( d ) = 16

Height ( h ) = 16

To find : Surface area of Cylinder = ?

Radius ( r ) = ?

Formula : -

r = d / 2

r = 16 / 2

r = 8

Formula : -

Surface area of Cylinder = 2πr ( r + h )

= 2π x 8 ( 8 + 16 )

= 16π x 24

Surface area of Cylinder = 384π

4 0

2 years ago

Hours worked / pay

Nadusha1986 [10]

It is c because 8x2=16 and 8x4=32 and so on

7 0

3 years ago