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Mamont248 [21]
3 years ago
13

If the cost is a function depending in the number of sections, how much will 200 sections cost?

Mathematics
2 answers:
Over [174]3 years ago
6 0

Answer: So here we have a function F(x) = y. and you give some of the points (x,y).

this are (10, 409) , (50, 1889) , (125, 4664)

then, if we assume a linear relationship, we could find the slope using the next formula:  s = \frac{y2 - y1}{x2 - x1}

then for the first two points we get: s = \frac{1889 - 409}{50 - 10}  = 37

and the slope between the second and third point is:

s = \frac{4664 - 1889}{125 -50}= 37

So we have the same slope in the two segments, and it's safe to assume that our function is linear. Now we need to find the x-intercept

then F(x) = 37x + b

and F(10) = 370 + b = 409

b = 409 - 370 = 39

So our function is F(x) = 37*x + 39

then if x = 200, F(200) = 37*200 + 39 = 7439.

So the correct answer is B.

julsineya [31]3 years ago
4 0
I would say the answer is D. 7837. Since the common ratio between the x’s and y’s were are 39.5 so multiplying that times 200 gives something around 7900 so, (rounding down) the answer would be D. I really hope that helped! Also please correct me if I am wrong... thanks!
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Examine the two functions below. Use the equation and table given to answer the question below.
Natalka [10]
<span>A.
Both functions have the same y-intercept.</span>             TRUE

<span>B.
f(x) has a negative slope while g(x) has a positive slope.</span>   Tricky question.  Yes, f(x) has the negative slope m = -2.  However, g(x), being a curve, does not have a single slope; rather, it has a varying slope that increases (becomes less negative) as x increases.  Some people might argue that this statement is TRUE, while others would argue that it's FALSE, because the 2nd function does not have just one single value for the slope.

<span>C.
Both functions are linear functions.</span>  ABSOLUTELY NOT.  The first one is linear; the second is a "decaying exponential" whose graph is a curve.

D. <span> 
g(x) has a common ratio of 4.  Yes and no.  I'd say the common ratio is (1/4), because each y value is (1/4) of the previous y value

</span>
5 0
4 years ago
Bart practices his trumpet 1 1/4 hours each day. How many hours will he practice in 6 days?
horsena [70]
The equation here is 1 1/4 X 6
1X6= 6 and 1/4 X6 = 6/4 
You can add these to and the answer is 6 6/4
which can then be simplified to 7 1/2 hours in 6 days
5 0
3 years ago
Assuming that workers' salaries in your company are uniformly distributed between $15,000 and $40,000 per year, find the probabi
lord [1]

Answer:

The  probability is  P(\$20,000 <  X < \$35,000) =  0.6

Step-by-step explanation:

From the question we are told that

     Workers' salaries in your company are uniformly distributed between $15,000 and $40,000 per year

  Generally the probability that a randomly chosen worker earns an annual salary between $20,000 and $35,000 is mathematically represented as

        P(\$20,000 <  X  \$35,000) =  \frac{ 35000 -  20000}{ 40000 - 150000}

  => P(\$20,000 <  X < \$35,000) =  0.6

     

3 0
4 years ago
What is the value of the expression |a + b| + |c| when a = –5, b = 8, and c = –13?
nasty-shy [4]
|a+b| + |c| when a= -5, b= 8, c= -13:

|-5+ 8|+ |-13|
= |3|+ 13
= 3+ 13
= 16

The correct answer is D. 16~
7 0
4 years ago
The box plots below show the ages of college students in different math courses
scZoUnD [109]

Answer:

4. The mean and median age are most likely to be same for the students in Math 1.

Step-by-step explanation:

We have been given two box plots that show the ages of college students in different math courses. We are asked to choose the true statement about these box plots.

1. The median age of the students in math 1 is greater than the median age of the students in math 2.  

We can see from our given box plots that median age for both Math 1 and Math 2 is 19 that is represented by vertical middle line of box, therefore, 1st statement is not true.

2. The median age of the students in math 1 is less than the median age of the students in math 2.

Since median ages for students in Math 1 and Math 2 are same (19), therefore, 2nd statement is not true.

3. The mean and median age are most likely same for both sets of data.

We can see that box plot for ages of Math 1 students is symmetric, therefore, both mean and median will be same for Math 1 students.

The box plot representing ages of Math 2 students is skewed right as it has an outlier as 23, therefore, its mean age will be greater than median age.

Therefore, 3rd statement is not true either.

4. The mean and median age are most likely to be same for the students in Math 1.

We can see that box plot representing ages for Math 1 students is symmetric and its whiskers are equidistant from median. So both mean and median age are same that is 19, therefore, 4th statement is true.

5 0
3 years ago
Read 2 more answers
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