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Dmitry [639]
3 years ago
14

The bar graph shows the number of school days Jalen had homework and did not have homework during the first six months of

Mathematics
1 answer:
Oksanka [162]3 years ago
4 0

Given:

<u>In July:</u>

Number of days he has done the homework = 6

Number of days he has not done the homework = 4

<u>In August:</u>

Number of days he has done the homework = 16

Number of days he has not done the homework = 7

<u>In September:</u>

Number of days he has done the homework = 14

Number of days he has not done the homework = 4

<u>In October:</u>

Number of days he has done the homework = 14

Number of days he has not done the homework = 8

<u>In November:</u>

Number of days he has done the homework = 16

Number of days he has not done the homework = 4

<u>In December:</u>

Number of days he has done the homework = 5

Number of days he has not done the homework = 10

To find the months in which the difference between the number of days with homework and with no homework greater than 6.

Now,

<u>In July,</u>

The difference between the number of days with homework and with no homework = 6-4 = 2

<u>In August,</u>

The difference between the number of days with homework and with no homework = 16-7 = 9

<u>In September,</u>

The difference between the number of days with homework and with no homework = 14-4 = 10

<u>In October,</u>

The difference between the number of days with homework and with no homework = 14-8 = 6

<u>In November,</u>

The difference between the number of days with homework and with no homework = 16-4 = 12

<u>In December,</u>

The difference between the number of days with homework and with no homework = 10-5 = 5

Hence,

In 3 months the difference between the number of days with homework and with no homework greater than 6 and these are: August, September, November.

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Pizza is served in the school cafeteria every fourth school day. Tacos are served every third school day. Both Pisa and tacos we
Oliga [24]

<u>Answer: </u>

Tacos and Pizza both will be served on 12th day of school

<u>Explanation:</u>

Given, pizza is served in the school cafeteria every fourth school day.

And, on every 3rd school day Tacos are being served.

If both are served today then we have to find the number of days after which both will be served on the same day again.

Taking LCM  of 4 and 3 to solve the question,

LCM of 4 and 3 = 4×3 = 12

Therefore, after 12 days both will be served on the same day.

5 0
3 years ago
How many millimeters are in 15 fluid ounces
Helen [10]

There are 443,603 cubic millimeters in 15 fluid ounces.

Because there are 29,573.5 in 1 fluid ounces, so you would just multiply that answer by 15.

Good luck and have a great day Monkeybutt8103!

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.

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3 years ago
I need the inequality answer.florence wants to buy onions and doesn't want to spend no more than $3. Can she buy 2.5 pounds ? Th
Lapatulllka [165]
No because if you divide 1.25 by two to find out what the half is in 2.5, it would be .62… and since she wants to buy 2.5 pounds, you've already figured out the .5, so 1.25x2 + .62, it would be 3.12. 
Illusion103
3 0
3 years ago
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Please I need to get this done by tomorrow
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Answer:

i think it is 8

4 0
3 years ago
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