PLEASE HELP!

snow_lady [41]

Answer:

HUH? What do you mean about that?

7 0

2 years ago

Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat

nirvana33 [79]

Answer:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

$\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}$

And solving for $\frac{dh}{dt}$ we got:

$\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}$

We need to convert the rate given into m^3/min and we got:

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

5 0

3 years ago

Lcm of 8 and 15 Please!

FromTheMoon [43]

Since, 8 and 5 both are prime numbers u can simply multiply 8 and 15 to get your answer..

the ans is 120..

8 0

2 years ago

Will mark brainlist:)

ella [17]

Answer:

The length of the missing side is 7 yards

Step-by-step explanation:

Simply subtract 9 from 16

16-9=7

The length of the missing side is 7 yards

Hope this helps! Plz award Brainliest : )

8 0

2 years ago

Read 2 more answers

Minor surgery on horses under field conditions requires a reliable short-term anesthetic producing good muscle relaxation, minim

andreev551 [17]

Answer:

$p_v =P(t_{74}$

If we compare the p value and a significance level for example $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean it's not significantly less than 20 min.

Step-by-step explanation:

Data given and notation

$\bar X=18.81$ represent the average lateral recumbency for the sample

$s=8.4$ represent the sample standard deviation

$n=75$ sample size

$\mu_o =20$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to apply a left tailed test.

What are H0 and Ha for this study?

Null hypothesis: $\mu \geq 20$

Alternative hypothesis : $\mu < 20$

Compute the test statistic

The statistic for this case is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{18.81-20}{\frac{8.4}{\sqrt{75}}}=-1.227$

The degrees of freedom are given by:

$df=n-1=75-1=74$

Give the appropriate conclusion for the test

Since is a one side left tailed test the p value would be:

$p_v =P(t_{74}$

Conclusion

If we compare the p value and a significance level for example $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean it's not significantly less than 20 min.

3 0

3 years ago