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3 years ago

1. A right circular cylinder has a height of 5 in. and a base area of 20 in2. What is the volume of the cylinder? Use π = 3.14.

Mathematics

2 answers:

Natalka [10]3 years ago

5 0

1. Volume of a cylinder is πr^2h

Area of the base would be πr^2

Volume would be the height x the area of the base:

Volume = 5 x 20 = 100 cubic in.

2. Divide the volume by the area:

207 / 20.7 = 10

The height is 10 inches.

Kruka [31]3 years ago

4 0

Answer:

Step-by-step explanation:

The formula of a volume of a cylinder:

$V=Bh$

B - base area

h - height

1. We have

h = 5in, B = 20 in².

Substitute:

$V=(5)(20)=100\ in^3$

2. We have

B = 20.7 in², V = 207 in³.

Substitute:

$207=20.7h$ divide both sides by 20.7

$\dfrac{207}{20.7}=\dfrac{20.7h}{20.7}\\\\10=h\to h=10\ in$

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If g(x) = 4x^2 - 16 were shifted 7 units to the right and 3 down, what would the

Kitty [74]

Answer:

The new equation is $h(x)=4(x-7)^2-19$

Step-by-step explanation:

Given : Function $g(x) = 4x^2 - 16$ were shifted 7 units to the right and 3 down.

To find : What would the new equation be?

Solution :

Shifting to the right with 'a' unit is

f(x)→f(x-a)

So, shifting g(x) 7 units to the right is

$h(x) = 4(x-7)^2-16$

Shifting to the down with 'b' unit is

f(x)→f(x)-b

So, shifting g(x) 3 units down is

$h(x)=(4(x-7)^2-16)-3$

$h(x)=4(x-7)^2-19$

The new equation is $h(x)=4(x-7)^2-19$

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3 years ago

ill give brainliest n a recent survey, 3 out of every 5 students said Math is their favorite class.If 200 students were surveyed

anyanavicka [17]

Answer: D. 120 students

Step-by-step explanation:

This is an easy solution

Take 300 and divide it by 5

300/5

This will result in 40

Now Multiply 40 by 3

Like so: 40x3

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6 0

2 years ago

Read 2 more answers

The histogram below shows the distribution of times, in minutes, required for 25 rats in an animal behavior experiment to naviga

tino4ka555 [31]

Answer:

c. The median and the IQR (Interquartile range).

Step-by-step explanation:

Considering the histogram for the distribution of times for this experiment (see the graph below), we can notice that this distribution is skewed positively because of "a few scores creating an elongated tail at the higher end of the distribution" (Urdan, 2005). There is also a probable outlier (an animal that navigates around nine minutes). An outlier is an extreme value that is more than two standard deviations below or above the mean.

In these cases, when we have skewed and extreme values in a distribution, it is better to avoid using the mean and standard deviations as measures of central tendency and dispersion, respectively. Instead, we can use the median and the interquartile range for those measures.

With skewed distributions, the mean is more "sensible" to extreme data than the median, that is, it tends to not represent the most appropriate measure for central position (central tendency) in a distribution since in a positively skewed distribution like the one of the question, the mean is greater than the median, that is, the extreme values tend to pull the mean to them, so the mean tends to not represent a "reliable" measure for the central tendency of all the values of the experiments.

We have to remember that the dispersion measures such as the standard deviation and interquartile range, as well as the variance and range, provide us of measures that tell us how spread the values are in a distribution.

Because the standard deviation depends upon the mean, i.e., to calculate it we need to subtract each value or score from the mean, square the result, divide it by the total number of scores and finally take the square root for it, we have to conclude that with an inappropriate mean, the standard deviation is not a good measure for the dispersion of the data, in this case (a positively skewed distribution).

Since the median represents a central tendency measure in which 50% of the values for distribution falls below and above this value, no matter if the distribution is skewed, the median is the best measure to describe the center of the distribution in this case.

Likewise, the quartiles are not affected by skewness, since they represent values of the distribution for which there is a percentage of data below and above it. For example, the first quartile (which is also the 25th percentile) splits the lowest 25% of the data from the highest 75% of them, and the third quartile, the highest 25%, and the lowest 75%. In other words, those values do not change, no matter the extreme values or skewness.

For these reasons, we can say that the median and the interquartile range (IQR) describe the center and the spread for the distribution presented, and not the most usual measures such as the mean and standard deviation.

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3 years ago

The perimeter of a rectangle is 150 cm. the length is 4 cm greater than the width find the width and length

Fudgin [204]

Let's call the width of our rectangle $w$ and the length $l$ . We can say $l = w + 4$ , since the length is equal to 4 cm greater than the width.

Also remember that the perimeter of a rectangle is the sum of two times the width and two times the length, or $P = 2l + 2w$ . To solve this problem, we can substitute in the information we know, as shown below: