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Kisachek [45]
2 years ago
14

The populations of two cities after t years can be modeled by

Mathematics
2 answers:
anyanavicka [17]2 years ago
6 0

Answer:

Step-by-step explanation:

Mkey [24]2 years ago
5 0
<h3>Hello There!!</h3>

<h3><u>Given</u></h3>

Value of 't' = 4

<h3><u>To </u><u>Find</u></h3>

Difference between the population of the two cities

<h3><u>Solution</u></h3>

\text{{Assuming the population of two cities as}}  \: \text{c}_1 \text{ and}  \: \text{c}_2

\text{ In Case of }\text{c}_1

\text{c}_1 = -150\text{t} + 50,000 \\  \implies \text{c}_1  =  - 150  \times 4 +  50,000 \\  \implies  \text{c}_1 =  - 600 + 50,000 \\ \implies  \text{c}_1 = 49,400

\text{ In Case of }\text{c}_2

\text{c}_2 = 50\text{t} + 75,000 \\ \implies \text{c}_2 = 50\times 4 + 75,000  \\\implies \text{c}_2 = 200+ 75,000 \\ \implies \text{c}_2 = 75,200

Now, We need to find the difference in their population

\text{c}_2 - \text{c}_1 \\  = 75,200 - 49,400 \\  = 25,800

\therefore  \text{Difference between their Population=} \red{25,800}

<h3>Hope This Helps!!</h3>
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3 years ago
A certain tennis player makes a successful first serve 6969​% of the time. Suppose the tennis player serves 9090 times in a matc
Wewaii [24]

Answer:

a) 4.387

b) Yes, because np & npq are greater than 10.

c) = 0.017          

Step-by-step explanation:

Give data:

p = 0.69

n = 90

a) a

E(X) = np = 62.1

SD(X) = \sqrt{(np(1-p))}

          =\sqrt{90\times 0.69(1- 0.69)}

          = 4.387

b)

np = 62.1  

q = 1 - p  = 1 - 0.69 = 0.31

npq = 19.251

Yes, because np & npq are greater than 10.

c.

P(X \geq 72   ) = P(X > 71.5) [continuity correction]

=    P(Z> \frac{((71.5-62.1)}{ 4.387})

= P(Z> 2.14 )      

= 1 - P(Z<2.14)              

= 1 - 0.983   (using table)          

= 0.017          

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