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3 years ago

Find K: mx=p(z/x+k + 2s/y)

Mathematics

1 answer:

ivann1987 [24]3 years ago

6 0

Answer:

$-\frac{2s}{y} -\frac{z}{x} +\frac{mx}{p} =k$

Step-by-step explanation:

$mx=p(\frac{z}{x} +k+\frac{2s}{y} )$ - Divide by p

$\frac{mx}{p} =(\frac{z}{x} +k+\frac{2s}{y} )$ - Subtract the 2 fractions

$-\frac{2s}{y} -\frac{z}{x} +\frac{mx}{p} =k$

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The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. The follow

kipiarov [429]

Answer:

The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.

Step-by-step explanation:

The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. Test if the mean cost has increased.

At the null hypothesis, we test if the mean cost is still the same, that is:

$H_0: \mu = 13252$

At the alternative hypothesis, we test if the mean cost has increased, that is:

$H_1: \mu > 13252$

The test statistic is:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

13252 is tested at the null hypothesis:

This means that $\mu = 13252$

The following year, a random sample of 20 two-year institutions had a mean of $15,560 and a standard deviation of $3500.

This means that $n = 20, X = 15560, s = 3500$

Value of the test statistic:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{15560 - 13252}{\frac{3500}{\sqrt{20}}}$

$t = 2.95$

P-value of the test and decision:

The p-value of the test is found using a t-score calculator, with a right-tailed test, with 20-1 = 19 degrees of freedom and t = 2.95. Thus, the p-value of the test is 0.0041.

The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.

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3 years ago

The 2011 super bowl had an attendance of 103,219 people. If the headline in the newspaper the next day read, about 200,000 peopl

chubhunter [2.5K]

The answer is not reasonable because even though the actual attendance for the super bowl was well over 100,000, it did not even make it to 150,000 which would make the 1 in the hundred thousands place round up to 2.

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3 years ago

Suppose the time a child spends waiting at for the bus as a school bus stop is exponentially distributed with mean 7 minutes. De

Gala2k [10]

Answer:

The probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.

Step-by-step explanation:

Let the random variable X represent the time a child spends waiting at for the bus as a school bus stop.

The random variable X is exponentially distributed with mean 7 minutes.

Then the parameter of the distribution is, $\lambda=\frac{1}{\mu}=\frac{1}{7}$ .

The probability density function of X is:

$f_{X}(x)=\lambda\cdot e^{-\lambda x};\ x>0,\ \lambda>0$

Compute the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning as follows:

$P(6\leq X\leq 9)=\int\limits^{9}_{6} {\lambda\cdot e^{-\lambda x}} \, dx$

$=\int\limits^{9}_{6} {\frac{1}{7}\cdot e^{-\frac{1}{7} \cdot x}} \, dx \\\\=\frac{1}{7}\cdot \int\limits^{9}_{6} {e^{-\frac{1}{7} \cdot x}} \, dx \\\\=[-e^{-\frac{1}{7} \cdot x}]^{9}_{6}\\\\=e^{-\frac{1}{7} \cdot 6}-e^{-\frac{1}{7} \cdot 9}\\\\=0.424373-0.276453\\\\=0.14792\\\\\approx 0.148$