F(x) = 3x + 5 g(x) = 4x? - 2 h(x) = x² – 3x + 1 Find f(x) + g(x)

At the local grocery store, cans of beans are arranged in rows. The number of cans in each row forms an arithmetic sequence. The

Elodia [21]

In an arithmetic series, the value of the nth term is calculated using the equation,
an = ao + (n - 1)(d)
where an and ao are the nth and the 1st term, respectively. d is the common difference, and n is the number of terms.

In the given, an = 48, a0 = 93, d = -5 and n is unknown. Substituting the known values,
48 = 93 + (n - 1)(-5)
The value of n from the equation is 10. Thus, the answer is the last choice.

3 0

3 years ago

I need an answer soon. Thanks!

Anika [276]

The answer is, x-5 equals to 0. So x equals to 5. So your answer is 5.

5 0

2 years ago

The formula for the volume of w sphere is v=4\3 r3 what is the formula solved for r?

Lyrx [107]

V of w sphere = $\frac{4}{3}$ $\pi$ r³
Multiply by 3 on either sides to get rid of the fraction.
$V * 3 = \frac{4}{3} *3 \pi r^3$
3V = 4 $\pi$ r³
Now divide either sides by 4 $\pi$ to isolate r³
$\frac{3V}{4 \pi } = \frac{4 \pi }{4 \pi }$ r³
4 $\pi$ and 4 $\pi$ cancels out
$\frac{3V}{4 \pi }$ = r³
Take the cube root to isolate r.
$\sqrt[3]{ \frac{3V}{4 \pi } } = \sqrt[3]{r^3}$
the cube root cancels the cube

$\sqrt[3]{ \frac{3V}{4 \pi } }$ = r

6 0

3 years ago

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.

5 0

3 years ago

Within a computer service company, 20 percent of the male employees and 30 percent of the female employees attend night school.

Maksim231197 [3]

26% of all the employees attend night school

Step-by-step explanation:

The given is:

Within a computer service company, 20 percent of the male employees and 30 percent of the female employees attend night school

40 percent of all employees are male

We need to find what percent of all the employees attend night school

Assume that the number of all employees is x

∵ The company has x employees

∵ 40 percent of all employees are male

- Find 40% of x

∵ 40% × x = $\frac{40}{100}$ × x = 0.4 x

∴ The number of the male employees is 0.4 x

- Subtract 0.4 x from x to find the number of the female employees

∵ The number of female employees = x - 0.4 x = 0.6 x

∴ The number of the female employees is 0.6 x

∵ 20 percent of the male employees attend night school

- Multiply 20% by 0.4 x to find the number of male employees

who attend night school

∴ The number of male who attend night school = 20% × 0.4 x

∴ The number of male who attend night school = $\frac{20}{100}$ × 0.4 x

∴ The number of male who attend night school = 0.08 x

∵ 30 percent of the female employees attend night school

- Multiply 30% by 0.6 x to find the number of female employees

who attend night school

∴ The number of female who attend night school = 30% × 0.6 x

∴ The number of male who attend night school = $\frac{30}{100}$ × 0.6 x

∴ The number of male who attend night school = 0.18 x

Add the number of male and female who attend night school

∵ 0.08 x + 0.18 x = 0.26 x

∴ The number of male and female who attend night school is 0.26 x

- Find the percentage of 0.26 x of x

∵ $\frac{0.26x}{x}$ × 100% = 26%

∴ 26% of all the employees attend night school

26% of all the employees attend night school

Learn more:

You can learn more about the percentage in brainly.com/question/6871421

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5 0

3 years ago

F(x) = 3x + 5 g(x) = 4x? - 2 h(x) = x² – 3x + 1 Find f(x) + g(x) – h(x)