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3 years ago

A sound wave is traveling at 6,420m/s and has a frequency of 600Hz. What is the wavelength?

Mathematics

1 answer:

GaryK [48]3 years ago

4 0

Answer:

10.7 m

Step-by-step explanation:

Given that,

The speed of a sound wave, v = 6,420 m/s

The frequency of the wave, f = 600 Hz

We need to find the wavelength of the wave. We can find it using the below relation as follows :

$v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{6420}{600}\\\\\lambda=10.7\ m$

So, the wavelength of the sound wave is equal to 10.7 m.

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Answer:

3/5(-8+5/9x-3)

3/5(-8+5x/9-3)

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4 years ago

The population of a certain town was 10,000 in 1990. The rate of change of the population, measured in people per year, is model

Katena32 [7]

The question is incomplete. The complete question is :

The population of a certain town was 10,000 in 1990. The rate of change of a population, measured in hundreds of people per year, is modeled by P prime of t equals two-hundred times e to the 0.02t power, where t is measured in years since 1990. Discuss the meaning of the integral from zero to twenty of P prime of t, d t. Calculate the change in population between 1995 and 2000. Do we have enough information to calculate the population in 2020? If so, what is the population in 2020?

Solution :

According to the question,

The rate of change of population is given as :

$\frac{dP(t)}{dt}=200e^{0.02t}$ in 1990.

Now integrating,

$\int_0^{20}\frac{dP(t)}{dt}dt=\int_0^{20}200e^{0.02t} \ dt$

$=\frac{200}{0.02}\left[e^{0.02(20)}-1\right]$

$=10,000[e^{0.4}-1]$

$=10,000[0.49]$

=4900

$\frac{dP(t)}{dt}=200e^{0.02t}$

$\int1.dP(t)=200e^{0.02t}dt$

$P=\frac{200}{0.02}e^{0.02t}$

$P=10,000e^{0.02t}$

$P=P_0e^{kt}$

This is initial population.

k is change in population.

So in 1995,

$P=P_0e^{kt}$

$=10,000e^{0.02(5)}$

$=11051$

In 2000,

$P=10,000e^{0.02(10)}$

$=12,214$

Therefore, the change in the population between 1995 and 2000 = 1,163.

8 0

3 years ago

The manager of a local nightclub has recently surveyed a random sample of 280 customers of the club. She would now like to deter

Lunna [17]

Answer:

$z = \frac{35.6-35}{\frac{5}{\sqrt{280}}}= 2.007$

$p_v = P(z>2.007) = 0.0224$

Since the p value is lower than the significance level given of 0.05 we have enough evidence to reject the null hypothesis on this case. And the best conclusion for this case is:

We (reject) the null hypothesis. That means that we (found) evidence to support the alternative.

Step-by-step explanation:

We have the following info given:

$\bar X = 35.6$ represent the sampel mean for the age of customers

$\sigma = 5$ represent the population standard deviation

$n = 280$ represent the sample size selected

We want to test if the mean age of her customers is over 35 so then the system of hypothesis for this case are:

Null hypothesis: $\mu \leq 35$

Alternative hypothesis $\mu >35$

The statistic for this case is given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing the data given we got:

$z = \frac{35.6-35}{\frac{5}{\sqrt{280}}}= 2.007$

We can calculate the p value since we are conducting a right tailed test like this:

$p_v = P(z>2.007) = 0.0224$

Since the p value is lower than the significance level given of 0.05 we have enough evidence to reject the null hypothesis on this case. And the best conclusion for this case is:

We (reject) the null hypothesis. That means that we (found) evidence to support the alternative.

5 0

3 years ago