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Inessa [10]
3 years ago
12

Wanna check out my question? :) Brainliest will be marked soooo

Mathematics
1 answer:
lara [203]3 years ago
6 0

Answer:

144

Step-by-step explanation:

We have 240 7th graders

60% check out fantasy books

240 *60%

240 *.60

144 will check out fantasy books

You might be interested in
5 plus 2 minus 6 times 4​
nevsk [136]

Answer:

714

Step-by-step explanation:

5+2=7

7-6=1

1x4=4

3 0
3 years ago
Read 2 more answers
A water tank is in the shape of a cone.Its diameter is 50 meter and slant edge is also 50 meter.How much water it can store In i
Aneli [31]
To get the most accurate answer possible, we're going to have to go into some unsightly calculation, but bear with me here:

Assessing the situation:

Let's get a feel for the shape of the problem here: what step should we be aiming to get to by the end? We want to find out how long it will take, in minutes, for the tank to drain completely, given a drainage rate of 400 L/s. Let's name a few key variables we'll need to keep track of here:

V - the storage volume of our tank (in liters)
t - the amount of time it will take for the tank to drain (in minutes)

We're about ready to set up an expression using those variables, but first, we should address a subtlety: the question provides us with the drainage rate in liters per second. We want the answer expressed in liters per minute, so we'll have to make that conversion beforehand. Since one second is 1/60 of a minute, a drainage rate of 400 L/s becomes 400 · 60 = 24,000 L/min.

From here, we can set up our expression. We want to find out when the tank is completely drained - when the water volume is equal to 0. If we assume that it starts full with a water volume of V L, and we know that 24,000 L is drained - or subtracted - from that volume every minute, we can model our problem with the equation

V-24000t=0

To isolate t, we can take the following steps:

V-24000t=0\\ V=24000t\\ \frac{V}{24000}=t

So, all we need to do now to find t is find V. As it turns out, this is a pretty tall order. Let's begin:

Solving for V:

About units: all of our measurements for the cone-shaped tank have been provided for us in meters, which means that our calculations will produce a value for the volume in cubic meters. This is a problem, since our drainage rate is given to us in liters per second. To account for this, we should find the conversion rate between cubic meters and liters so we can use it to convert at the end.

It turns out that 1 cubic meter is equal to 1000 liters, which means that we'll need to multiply our result by 1000 to switch them to the correct units.

Down to business: We begin with the formula for the area of a cone,

V= \frac{1}{3}\pi r^2h

which is to say, 1/3 multiplied by the area of the circular base and the height of the cone. We don't know h yet, but we are given the diameter of the base: 50 m. To find the radius r, we divide that diameter in half to obtain r = 50/2 = 25 m. All that's left now is to find the height.

To find that, we'll use another piece of information we've been given: a slant edge of 50 m. Together with the height and the radius of the cone, we have a right triangle, with the slant edge as the hypotenuse and the height and radius as legs. Since we've been given the slant edge (50 m) and the radius (25 m), we can use the Pythagorean Theorem to solve for the height h:

h^2+25^2=50^2\\ h^2+625=2500\\ h^2=1875\\ h=\sqrt{1875}=\sqrt{625\cdot3}=25\sqrt{3}

With h=25\sqrt{3} and r=25, we're ready to solve for V:

V= \frac{1}{3} \pi(25)^2\cdot25\sqrt{3}\\ V= \frac{1}{3} \pi\cdot625\cdot25\sqrt{3}\\ V= \frac{1}{3} \pi\cdot15625\sqrt{3}\\\\ V= \frac{15625\sqrt{3}\pi}{3}

This gives us our volume in cubic meters. To convert it to liters, we multiply this monstrosity by 1000 to obtain:

\frac{15625\sqrt{3}\pi}{3}\cdot1000= \frac{15625000\sqrt{3}\pi}{3}

We're almost there.

Bringing it home:

Remember that formula for t we derived at the beginning? Let's revisit that. The number of minutes t that it will take for this tank to drain completely is:

t= \frac{V}{24000}

We have our V now, so let's do this:

t= \frac{\frac{15625000\sqrt{3}\pi}{3}}{24000} \\ t= \frac{15625000\sqrt{3}\pi}{3}\cdot \frac{1}{24000} \\ t=\frac{15625000\sqrt{3}\pi}{3\cdot24000}\\ t=\frac{15625\sqrt{3}\pi}{3\cdot24}\\ t=\frac{15625\sqrt{3}\pi}{72}\\ t\approx1180.86

So, it will take approximately 1180.86 minutes to completely drain the tank, which can hold approximately V= \frac{15625000\sqrt{3}\pi}{3}\approx 28340615.06 L of fluid.
5 0
3 years ago
Greg builds a new pond which has a volume of 7.35 m3. What is the width of the pond?
arsen [322]
The pong will be at 3.5 m
5 0
4 years ago
83 random samples were selected from a normally distributed population and were found to have a mean of 32.1 and a standard devi
arlik [135]

Answer:

\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}

4.525 \leq \sigma^2 \leq 7.602

Now we just take square root on both sides of the interval and we got:

2.127 \leq \sigma \leq 2.757

Step-by-step explanation:

Information given

\bar X=32.1 represent the sample mean

\mu population mean (variable of interest)

s=2.4 represent the sample standard deviation

n=83 represent the sample size  

Confidence interval

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

The degrees of freedom given by:

df=n-1=8-1=7

The confidence level is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,82)" "=CHISQ.INV(0.95,82)". so for this case the critical values are:

\chi^2_{\alpha/2}=104.139

\chi^2_{1- \alpha/2}=62.132

The confidence interval is given by:

\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}

4.525 \leq \sigma^2 \leq 7.602

Now we just take square root on both sides of the interval and we got:

2.127 \leq \sigma \leq 2.757

5 0
3 years ago
What is the measure of IL?
svetoff [14.1K]

Answer:

L-6 test results measure the amount of IL-6 circulating in the blood, and are used as one sign of systemic inflammation. A normal level of IL-6 may represent people who do not have an infection or inflammation. A low level of IL-6 may be expected for most patients with a less severe inflammatory response.

Step-by-step explanation:

7 0
3 years ago
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