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Leto [7]
3 years ago
6

12 laps in 8 minutes or 16 laps in 10 minutes? Which is the greater.

Mathematics
2 answers:
nadezda [96]3 years ago
7 0

Answer:

16 laps in 10  minutes

Step-by-step explanation:if you multiply 12 and 8 you'll get 96 where when you multiply 16 and 10 you get 160

Assoli18 [71]3 years ago
4 0

Answer:

12 Laps in 8 minutes

Essentially these values are all fractions

8/12 and 10/16

If we simplify the values we get

2/3 and 5/8

In order to find which is greater we need to find the common denominator

3 and 8 have a common multiple of 24, so lets use that

16/24 and 15/24

Therefore making 2/3 larger.

Or 12 laps and 8 minutes

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Garry wants to have $750,000 in 30 years. How much does he have to invest per month in an annuity with an annual interest rate o
Nikolay [14]
I think the answer is Garry has to invest 2,035 dollars a month for 30 years to reach his goal of 750,000 dollars
4 0
3 years ago
Last Wednesday, a random sample of 24 students were surveyed to find how long it takes to walk from the Fretwell Building to the
Ray Of Light [21]

Answer:

E

Step-by-step explanation:

Solution:-

- We are to investigate the confidence interval of 95% for the population mean of walking times from Fretwell Building to the college of education building.

- The survey team took a sample of size n = 24 students and obtained the following results:

                Sample mean ( x^ ) = 12.3 mins

                Sample standard deviation ( s ) = 3.2 mins

- The sample taken was random and independent. We can assume normality of the sample.

- First we compute the critical value for the statistics.

- The z-distribution is a function of two inputs as follows:

  • Significance Level  ( α / 2 ) = ( 1 - CI ) / 2 = 0.05/2 = 0.025

Compute: z-critical = z_0.025 = +/- 1.96

- The confidence interval for the population mean ( u ) of  walking times is given below:

                      [ x^ - z-critical*s / √n  ,   x^ + z-critical*s / √n  ]

Answer:        [ 12.3 - 1.96*3.2 / √24  ,  12.3 + 1.96*3.2 / √24  ]

                   

3 0
3 years ago
Please help meeee it’s question #7
Ugo [173]

Answer:

In my own opinion i time 104 and 72and if gives me7488

3 0
3 years ago
Derivative of tan(2x+3) using first principle
kodGreya [7K]
f(x)=\tan(2x+3)

The derivative is given by the limit

f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h

You have

\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h
\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}

By this identity, you have

\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}

So in the limit you get

\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h
\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}
\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}
\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}
\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}

The first two limits are both 1, and the single term in the last limit approaches 0 as h\to0, so you're left with

f'(x)=\dfrac12\sec^2(2x+3)

which agrees with the result you get from applying the chain rule.
7 0
3 years ago
Pls help 6th grade math
Minchanka [31]

Answer:

All except 3

Hope this helps

8 0
2 years ago
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