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kotykmax [81]
3 years ago
11

jorge took a speed reading class last saturday from 9 am to 4 pm. he learned that he can speed read 330 pages of a novel in one

hour with good comprehension. how many pages does he speed read in 225 minutes
Mathematics
1 answer:
garri49 [273]3 years ago
7 0

Answer:

1hr = 60min

225min/ 60min = 3.75 hr

330 pgs. * 3.75hr = 1237.5 pages

He read 1237.5 pages in 225 minutes.

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Plan A cost $15 a month, including 200 free texts. After 200, they cost $0.15 each
hammer [34]
If your looking how to plot this on a graph idk but I know its something like 15.00x>200
 0.15y<200


I think... I'm not sure... but it would make sense lol :)
5 0
4 years ago
How do you solve this problem <br> 7= ln(x+5)
Gnesinka [82]
Answer:

x = 1091.63315843
<span>
Setting Up:

7 = ln ( x + 5 )

ln translates to "log" with an "e" as the base or subscript ( a small "e" at the bottom right of the "g" in log).

You take the base of the log and put it to the power of "7" ( "7" is the natural log of ( x + 5 ) in this problem ).

The value of which the logarithm is calculated is set equal to the base of the logarithm to the power of the calculated logarithm of the value.

e^7 = x + 5

Solving</span>:

e = 2.71828182846

Natural logarithms are logarithms to the base of the constant 'e'.

e^7 = x + 5    ( simplify e^7 )

<span>1096.63315843 = x + 5
</span>
Subtract 5 from each side.

1091.63315843 = x


5 0
3 years ago
Find the derivative of f(x) = 6x + 2 at x = 1.
earnstyle [38]
f'(x_0)=\lim\limits_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}\\\\f(x)=6x+2;\ x_0=1\\\\f'(1)=\lim\limits_{x\to x_0}\dfrac{(6x+2)-(6\cdot1+2)}{x-1}=\lim\limits_{x\to x_0}\dfrac{6x+2-8}{x-1}\\\\=\lim\limits_{x\to x_0}\dfrac{6x-6}{x-1}=\lim\limits_{x\to x_0}\dfrac{6(x-1)}{x-1}=\lim\limits_{x\to x_0}6=6\\\\\\Answer:\boxed{f'(1)=6}
6 0
3 years ago
Read 2 more answers
In a random sample of 80 teenagers, the average number of texts handled in one day is 50. The 96% confidence interval for the me
Naily [24]

Answer:

Standard deviation of the sample = 17.421

Step-by-step explanation:

We are given that in a random sample of 80 teenagers, the average number of texts handled in one day is 50.

Also, the 96% confidence interval for the mean number of texts handled by teens daily is given as 46 to 54.

So, sample mean, xbar = 50  and   Sample size, n = 80

Let sample standard deviation be s.

96% confidence interval for the mean number of texts,\mu is given by ;

    96% confidence interval for \mu = xbar \pm 2.0537*\frac{s}{\sqrt{n} }

                              [46 , 54]           = 50 \pm 2.0537*\frac{s}{\sqrt{80} }

Since lower bound of confidence interval = 46

So,  46 = 50 - 2.0537*\frac{s}{\sqrt{80} }

       50 - 46 = 2.0537*\frac{s}{\sqrt{80} }

            s = \frac{4*\sqrt{80} }{2.0537} = 17.421

Therefore, standard deviation of the sample is 17.421 .

3 0
4 years ago
The median for the given set of six ordered data values is 26.5. what is the missing value? 7 12 21 ____ 41 50
Tpy6a [65]
I think the answer is 29.4
8 0
3 years ago
Read 2 more answers
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