MATHEMATICS COLLEGE

A cell site is a site where electronic communications equipment is placed in a cellular network for the use of mobile phones. The numbers y of cell sites from 1985 through 2011 can be modeled byy = 269573/1+985e^-0.308t where t represents the year, with t = 5 corresponding to 1985. Use the model to find the numbers of cell sites in the years 1998, 2003, and 2006.

Answers

Answer 1

Answer:

Answer:

(a) 3178

(b) 14231

Step-by-step explanation:

Given

$y = (269573)/(1+985e^(-0.308t))$

Solving (a): Year = 1998

1998 means t = 8 i.e. 1998 - 1990

So:

$y = (269573)/(1+985e^(-0.308*8))$

$y = (269573)/(1+985e^(-2.464))$

$y = (269573)/(1+985*0.08509)$

$y = (269573)/(84.81365)$

$y = 3178$ --- approximated

Solving (b): Year = 2003

2003 means t = 13 i.e. 2003 - 1990

So:

$y = (269573)/(1+985e^(-0.308*13))$

$y = (269573)/(1+985e^(-4.004))$

$y = (269573)/(1+985*0.01824)$

$y = (269573)/(18.9664)$

$y = 14213$ --- approximated

Solving (c): Year = 2006

2006 means t = 16 i.e. 2006 - 1990

So:

$y = (269573)/(1+985e^(-0.308*16))$

$y = (269573)/(1+985e^(-4.928))$

$y = (269573)/(1+985*0.00724)$

$y = (269573)/(8.1314)$

$y = 33152$ --- approximated

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Write the slope-intercept form of the equation of the line
2x + 3y = 18

Answers

Answer:

y=-2/3x+18

Step-by-step explanation:

Slope intercept form equals y=mx+b

Step 1 move 2x to the other side. 3y=-2x+18

Step 2 divide by 3. y=-2/3x+18

To save money you set aside $50. For each following month you set aside 10% more than the previous month.How much money will you save in a year? Explain each step.

Answers

A lot :) hahdinshsiskvshsknsvsudkdnhdjdjdd

Which conclusion can be drawn from the data?Question 9 options:

For ten weeks, City A received less rainfall, on average, than City B.

The range between the maximum and minimum values for City B is greater than the range between maximum and minimum values for City A.

During the 10 wk period, the rainfall amount recorded most often for City B was 1 in.

The median for City A is less than the median for City B.

Answers

Answer:

During the 10 wk period, the rainfall amount recorded most often for City B was 1 in.

Step-by-step explanation:

Given the data :

City A :

Reordered data:

0, 0.2, 0.2, 0.3, 0.4, 1, 1.3, 1.5, 2.5, 3

City B :

Reordered data:

0, 0, 0.1, 0.1, 0.2, 0.3, 0.4, 1, 1, 1

Using a calculator :

Mean Rainfall for City A = 1.04

Mean rainfall for city B = 0.41

Range : maximum - minimum

City A = 3 - 0 = 3

City B = 1 - 0 = 1

Mode (most occurring) :

City A = 0.2

City B = 1

Median :

City A = 0.7

City B = 0.25

The only true conclusion in the options given that can be drawn from the data is that ;During the 10 wk period, the rainfall amount recorded most often for City B was 1 in.

Answer:

the Answer is C.

Step-by-step explanation:

I just took the test

P(x)=3x² + 4x³-8+x⁴-7x Degree; Type; Leading coefficent;

Answers

Answer:

Degree: 4; Type: quartic; Leading coefficient: 1

Step-by-step explanation:

Help please!!!!! asap

Answers

Answer:

$w \leq \: 26$

Step-by-step explanation:

$(9w - 4)/(5) \leq (7w + 2)/(4)$

First of all cross multiply

That's

$4(9w - 4) \leq5(7w + 2)$

Expand the terms

$36w - 16 \leq \: 35w + 10$

Add 16 to both sides

That's

$36w + 16 - 16 \leq \: 35w + 10 + 16 \n 36w \leq35w + 26$

Subtract 35w from both sides

We have

$36w - 35w \leq35w - 35w + 26 \n$

We have the final answer as

$w \leq \: 26$

Hope this helps you

Consider a disease whose presence can be identified by carrying out a blood test. Let p denote the probability that a randomly selected individual has the disease. Suppose n individuals are independently selected for testing. One way to proceed is to carry out a separate test on each of the n blood samples. A potentially more economical approach, group testing, was introduced during World War II to identify syphilitic men among army inductees. First, take a part of each blood sample, combine these specimens, and carry out a single test. If no one has the disease, the result will be negative, and only the one test is required. If at least one individual is diseased, the test on the combined sample will yield a positive result, in which case the n individual tests are then carried out. [The article "Random Multiple-Access Communication and Group Testing"† applied these ideas to a communication system in which the dichotomy was active/idle user rather than diseased/nondiseased.] If p = 0.15 and n = 5, what is the expected number of tests using this procedure? (Round your answer to three decimal places.)