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adoni [48]
3 years ago
9

Every day I swim the same number of laps in my pool. After completing a certain number of laps I have swum 20% of the total and

after one more lap I have completed 25% of the total. How many laps do I swim each day?
Please help!!
Mathematics
1 answer:
riadik2000 [5.3K]3 years ago
4 0
Each lap is 5% of the total
100/5=20
you swim 20 laps each day
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Please help!! how do i solve/answer this
nadezda [96]
It’s 1,-2!! it’s where the lines meet
3 0
2 years ago
Find the mean,median,mode and range for each set of data its 23,27,24,26,26,24,26,24
lilavasa [31]

Mean is the average of the data set, which is found by adding all values together and then dividing that sum by the number of data values.

Median is the middle number of the data set, and can be found by ordering the values from least to greatest. If there are two middle numbers, the average of the two would be the median.

Mode is the number that shows up most frequently in the data set.

Range is found by subtracting the lowest number from the highest number in the data set.

<u>Set 1</u>

Mean:

18+20+22+11+19+18+18=126

126÷7=18

Median:

11, 18, 18, 18, 19, 20, 22 →  18

Mode: 18

Range: 22-11=11

<u>Set 2</u>

Mean: 23+27+24+26+26+24+26+24=200

200÷8=25

Median:

23, 24, 24, 24, 26, 26, 26, 27 → 24+26=50 → 50÷2=25

Mode: 24 and 26

Range: 27-23=4

4 0
3 years ago
Read 2 more answers
Pls help!!! Find the measure of GPS and NPH.
Snowcat [4.5K]

Answer: gps no nph are both 140

Step-by-step explanation:

So we know that around the center is equal to 360. So (15 + 5x) = (4x + 20) to find what x equals you subtract 4x from each side to get 15 + x = 20 now subtract 15 from both sides to get x = 5. Now we know that x = 5 so the 2 angles are 40 degrees each. Now subtract those two angles from 360 and you get 280. Then divide by two and you get 140. So, now we know those two angles are equal to 140.

Good luck with your school work. Sorry it took so long.

5 0
2 years ago
F is a twice differentiable function that is defined for all reals. The value of f "(x) is given for several values of x in the
nadezda [96]

The correct answer is actually the last one.

The second derivative f''(x) gives us information about the concavity of a function: if f''(x) then the function is concave downwards in that point, whereas if f''(x)>0 then the function is concave upwards in that point.

This already shows why the first option is wrong - if the function was concave downwards for all x, then the second derivate would have been negative for all x, which isn't the case, because we have, for example, f''(8)=5

Also, the second derivative gives no information about specific points of the function. Suppose, in fact, that f(x) passes through the origin, so f(0)=0. Now translate the function upwards, for example. we have that f(x)+k doesn't pass through the origin, but the second derivative is always f''(x). So, the second option is wrong as well.

Now, about the last two. The answer you chose would be correct if the exercise was about the first derivative f'(x). In fact, the first derivative gives information about the increasing or decreasing behaviour of the function - positive and negative derivative, respectively. So, if the first derivative is negative before a certain point and positive after that point. It means that the function is decreasing before that point, and increasing after. So, that point is a relative minimum.

But in this exercise we're dealing with second derivative, so we don't have information about the increasing/decreasing behaviour. Instead, we know that the second derivative is negative before zero - which means that the function is concave downwards before zero - and positive after zero - which means that the function is concave upwards after zero.

A point where the function changes its concavity is called a point of inflection, which is the correct answer.

7 0
3 years ago
1. _____ any push or pull on an object
borishaifa [10]
1. Force any push or pull on an object.             2. Displacement is an object changing position over time.
5 0
3 years ago
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