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Otrada [13]
3 years ago
7

Dan made 85% of his free throws. If he took 45 shots how many did he make?

Mathematics
2 answers:
likoan [24]3 years ago
6 0

Answer:

38

Step-by-step explanation:

85%=.85

So 45 × .85 = 38.25

38 shots he made.

Hope this helps

Cloud [144]3 years ago
6 0
<h3>Answer:  38</h3>

Work Shown:

85% = 85/100 = 0.85

85% of 45 = 0.85*45 = 38.25

Rounding to the nearest whole number, we get the result 38

As a check, 38/45 = 0.8444 approximately which converts to 84.44% and that's fairly close to 85%. If we tried 39, then 39/45 = 0.8667 = 86.67% which is a bit too high.

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So one key thing to remember here is that the direction of the correlation is irrelevant, that is it does not matter if your correlation is + or - what matters is how close that number is to 1.0.

To help you out here are the ranges of correlation strength

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So to start off with 0.26 and 0.18 are very small correlations so you'd call those weak correlations.

Let me know if you need help doing the other ones? It should be simple enough with the data I gave you :)


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Find the domain of the following piecewise function
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The domain of the function will be [-4, 6). Then the correct option is B.

The complete question is attached below.

<h3>What are domain and range?</h3>

The domain means all the possible values of x and the range means all the possible values of y.

The function is given below.

f(x) = \left\{\begin{matrix}x+4, & if & -4\leq x < 3 \\\\2x-1, & if & 3 \leq x < 6 \\\end{matrix}\right.

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Step-by-step explanation:

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the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?​
madam [21]

Answer:

About 0.6548 grams will be remaining.  

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

f(t)=a(r)^t

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}

One year has 365 days.

Therefore, the amount remaining after one year will be:

\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548

About 0.6548 grams will be remaining.  

Alternatively, we can use the standard exponential growth/decay function modeled by:

f(t)=Ce^{kt}

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

255=510e^{38k}

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\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)

Thus, our function is:

f(t)=510e^{t\ln(.5)/38}

Then after one year or 365 days, the amount remaining will be about:

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3 years ago
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