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3 years ago

If p-1/p=2, find the value of p^3+1/p^3. Help me fast.

Mathematics

1 answer:

tekilochka [14]3 years ago

4 0

Answer:

$\\\\\frac{P^3 \ + \ 1}{P^3} = 0$

Step-by-step explanation:

Given;

$\frac{P-1 }{P} = 2\\\\From \ the \ equation \ above \ we \ determine \ the \ value \ of \ p \ as \ follows;\\\\P-1 = 2P\\\\P-2P = 1\\\\-P = 1\\\\P = -1$

$Now\ solving \ the \ given \ question;\\\\\frac{P^3 \ + \ 1}{P^3} \\\\Substitute \ the \ value \ of \ "P" \ into \ the \ equation;\\\\\frac{P^3 \ + \ 1}{P^3} = \frac{(-1)^3 \ + \ 1}{(-1)^3} = \frac{-1 \ + \ 1}{-1} = \frac{0}{-1} = 0$

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Men are believed to have a slightly longer average length of short hospital stays than women; 5.2 days versus 4.5 days respectiv

Georgia [21]

Answer:

Since our calculated value is lower than our critical value, $z_{calc}=2.02$ , we have enough evidence to FAIL to reject the null hypothesis at 1% of singificance. So there is not nough evidence to conclude that mean for men it's significantly higher than the mean for female.

Step-by-step explanation:

1) Data given and notation

$\bar X_{1}=5.2$ represent the mean for group men

$\bar X_{2}=4.5$ represent the mean for group women

Assuming these values for the remaining data:

$\sigma_{1}=1.2$ represent the population standard deviation for the sample men

$\sigma_{2}=1.5$ represent the population standard deviation for the sample women

$n_{1}=32$ sample size for the group men

$n_{2}=30$ sample size for the group women

z would represent the statistic (variable of interest)

$p_v$ represent the p value

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for men it's higher than the mean for women, the system of hypothesis would be:

H0: $\mu_{1} \leq \mu_{2}$

H1: $\mu_{1} > \mu_{2}$

If we analyze the size for the samples both are higher than 30, and we know the population deviation's, so for this case is better apply a z test to compare means, and the statistic is given by:

$z=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

3) Calculate the statistic

We have all in order to replace in formula (1) like this:

$z=\frac{5.2-4.5}{\sqrt{\frac{1.2^2}{32}+\frac{1.5^2}{30}}}=2.02$

4) Find the critical value

In order to find the critical value we need to take in count that we are conducting a right tailed test, so we are looking on the normal standard distribution a value that accumulats 0.01 of the area on the right and 0.99 of the area on the left. We can us excel or a table to find it, for example the code in Excel is:

"=NORM.INV(1-0.01,0,1)", and we got $z_{critical}=2.33$

5) Statistical decision

Since our calculated value is lower than our critical value, $z_{calc}=2.02$ , we have enough evidence to FAIL to reject the null hypothesis at 1% of significance. So there is not nough evidence to conclude that mean for men it's significantly higher than the mean for female.

6 0

4 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

0.719 rounded to the nearest hundredth

lapo4ka [179]

Answer:

0.72

Step-by-step explanation:

hope i helped

pls mark brainliest im trying to level up

4 0

4 years ago

Read 2 more answers

Determine how many zeros, how many real or complex, and find the roots for f(x) = x3 − 5x2 − 25x + 125.

miskamm [114]

We can write this as:-

P(x) = + x^3 - 5x^2 - 25x + 125
There are 2 changes of real sign so by Descartes Rule of signs there are either 2 positive real roots or 0 positive roots.

P(-x) = - x^3 - 5x^2 + 25x + 125
There is just one change of sign so there is exactly 1 real negative root.

125 is a multiple of 5 so By rational root theorem 5 could be a positive root.

P(5) = 125 - 125 - 125 + 125 = 0 so one zero is 5
if we divide the polynomial by (x - 5) we get the quadratic
x^2 - 25
(x + 5)(x - 5) = 0

x = 5,-5

so the roots are 5 (multiplicity 2) and -5.
2 real positive zeroes and one real negative zero

7 0

3 years ago

Rewrite each expression using each base only once 5^4*5^-6*5

erastovalidia [21]

5^4 * 5^(-6) * 5

= 5^4 * 5^(-6) * 5^1

= 5^[ 4 + (-6) + 1 ]

= 5^[ 4 - 6 + 1 ]

= 5^(-1) <----- this is the answer.

I hope this helps. =)

7 0

4 years ago

If p-1/p=2, find the value of p^3+1/p^3. ​Help me fast.

If p-1/p=2, find the value of p^3+1/p^3. Help me fast.