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Marrrta [24]
3 years ago
14

Last year the local dance club had 64 members. this year it has 73 members. Calculate the percentage increase in the membership.

Mathematics
1 answer:
lozanna [386]3 years ago
6 0

Answer:

14.06%

Step-by-step explanation:

To find the percent increase, find the difference between the original and new amounts, divide it by the original amount, then multiply by 100:

73 - 64 = 9, so the difference is 9.

Divide this by the original and multiply by 100:

(9 / 64) x 100

= 14.06

So, the percent increase in the membership is 14.06%

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A gas at standard temperature and pressure was measured as having a mass of 1485 grams (g) in a volume of 750 liters (L). Which
tangare [24]

Answer:

Here we will use the relation:

Density = mass/volume.

We know that:

Mass = 1485g

Volume = 750L

Density = 1485g/750L = 1.98 g/L

We do not have the options in order to select the correct one, but knowing this density, we can suppose the possible gases.

Two "common" gases with densities similar to this one are:

Carbon Dioxide, with d = 1.977 g/L

Dinitrogen Monoxide, with d = 1.977 g/L

In both cases, if we round up we will get the same density that we calculated at the beginning, so either of these can be the correct option.

6 0
3 years ago
Simplify.<br><br> 100‾‾‾‾√<br><br> 20<br><br> 10 <br><br> −1<br><br> −25
olga_2 [115]

Answer:

10

Step-by-step explanation:

\sqrt{100} =\sqrt{(10)(10)}  = 10

 

8 0
3 years ago
Not sure if any of this is correct, but it’s what I got so far
Irina18 [472]

Problem 1 is correct. You use the pythagorean theorem to find the hypotenuse.

==================================================

Problem 2 has the correct answer, but one part of the steps is a bit strange. I agree with the 132 ft/sec portion; however, I'm not sure why you wrote \frac{1 \text{ sec}}{132 \text{ ft}}=\frac{0.59\overline{09}}{78 \text{ ft}}*127 \text{ ft}

I would write it as \frac{1\text{ sec}}{132 \text{ ft}}*127 \text{ ft} = \frac{127}{132} \text{ sec} \approx 0.96 \text{ sec}

==================================================

For problem 3, we first need to convert the runner's speed from mph to feet per second.

17.5 \text{ mph} = \frac{17.5 \text{ mi}}{1 \text{ hr}}*\frac{1 \text{ hr}}{60 \text{ min}}*\frac{1 \text{ min}}{60 \text{ sec}}*\frac{5280 \text{ ft}}{1 \text{ mi}} \approx 25.667 \text{ ft per sec}

Since the runner needs to travel 90-12 = 78 ft, this means\text{time} = \frac{\text{distance}}{\text{speed}} \approx \frac{78 \text{ ft}}{25.667 \text{ ft per sec}} \approx 3.039 \text{ sec}

So the runner needs about 3.039 seconds. In problem 2, you calculated that it takes about 0.96 seconds for the ball to go from home to second base. The runner will not beat the throw. The ball gets where it needs to go well before the runner arrives there too.

-------------

The question is now: how much of a lead does the runner need in order to beat the throw?

Well the runner needs to get to second base in under 0.96 seconds.

Let's calculate the distance based on that, and based on the speed we calculated earlier above.

\text{distance} = \text{rate}*\text{time} \approx (25.667 \text{ ft per sec})*(0.96 \text{ sec}) \approx 24.64032 \text{ ft}

This is the distance the runner can travel if the runner only has 0.96 seconds. So the lead needed is 90-24.64032 = 65.35968 feet

This is probably not reasonable considering it's well over halfway (because 65.35968/90 = 0.726 = 72.6%). If the runner is leading over halfway, then the runner is probably already in the running motion and not being stationary.

As you can see, the runner is very unlikely to steal second base. Though of course such events do happen in real life. What may explain this is the reaction time of the catcher may add on just enough time for the runner to steal second base. For this problem however, we aren't considering the reaction time. Also, not all catchers can throw the ball at 90 mph which is quite fast. According to quick research, the MLB says the average catcher speed is about 81.8 mph. This slower throwing speed may account for why stealing second base isn't literally impossible, although it's still fairly difficult.

5 0
3 years ago
What is 3 times 1/3 and subtract 5
lukranit [14]
3* ( 1/3 -5)= -14 hope this helps
7 0
3 years ago
Read 2 more answers
The sum of the measures of the interior angles of a polygon with n sides is __.
melisa1 [442]

Answer: The sum of the measures of the interior angles of a polygon is always 180(n-2) degrees, where n represents the number of sides of the polygon. The sum of the measures of the exterior angles of a polygon is always 360

8 0
3 years ago
Read 2 more answers
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