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Artist 52 [7]
3 years ago
11

A company employee has to print 3145 copies of a 1-page document to be mailed to clients. The company's printer prints 85 pages

per minute. The function f(x)=−85x+3145 represents the number of unprinted pages x minutes after the printing begins.
What is the practical domain of function f?

A all whole numbers from 0 to 37
B all multiples of 85 between 0 and 3145
C all real numbers from 0 to 37
D all real numbers from 0 to 3145
Mathematics
2 answers:
MakcuM [25]3 years ago
8 0
The domain is going to be ur x values....the number of minutes
therefore, ur domain is : I know it is either A or C.....I am thinking C because it does not have to be a whole minute...it can be 1.5 minutes...so that would eliminate A. final answer    is gonna be C
Alex Ar [27]3 years ago
4 0
I think it's 
<span>
C.all real numbers from 0 to 3145</span>
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Answer:

First numerator and second denominator

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A 6-ounce tube of toothpaste costs $2.00. What is the approximate difference in the unit price from the original price if a $1.2
LenKa [72]
I believe it is D, because if you divide 2/6 its .333 minus 1.25/6, which is .124.
4 0
3 years ago
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PLEASE HELP I WILL MARK BRAINLEIST
icang [17]

Answer:

25 mm

Step-by-step explanation:

Set up a proportion where x is the number of millimeters for 160 kilometers

\frac{5}{32} = \frac{x}{160}

Cross multiply and solve for x:

32x = 800

x = 25

So, 25 millimeters represent 160 kilometers

7 0
3 years ago
Find the mean, variance &amp;a standard deviation of the binomial distribution with the given values of n and p.
MrMuchimi
A random variable following a binomial distribution over n trials with success probability p has PMF

f_X(x)=\dbinom nxp^x(1-p)^{n-x}

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.

\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1

The mean is given by the expected value of the distribution,

\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}
\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}
\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}
\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}
\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}
\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}
\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}

The remaining sum has a summand which is the PMF of yet another binomial distribution with n-1 trials and the same success probability, so the sum is 1 and you're left with

\mathbb E(x)=np=126\times0.27=34.02

You can similarly derive the variance by computing \mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2, but I'll leave that as an exercise for you. You would find that \mathbb V(X)=np(1-p), so the variance here would be

\mathbb V(X)=125\times0.27\times0.73=24.8346

The standard deviation is just the square root of the variance, which is

\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834
7 0
3 years ago
Help me on this one too
never [62]

Answer:

C) -2.5<3

Hope this helps!

Hope this helps!

6 0
3 years ago
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