The estimate of the number of students studying abroad in 2003 is 169 and the estimate of the number of students studying abroad in 2018 is 433
<h3>a. Estimate the number of students studying abroad in 2003.</h3>
The function is given as:
y = 123(1.065)^x
Where x represents years from 1998 to 2013
2003 is 5 years from 1998.
This means that
x = 5
Substitute the known values in the above equation
y = 123(1.065)^5
Evaluate the exponent
y = 123 * 1.37008666342
Evaluate the product
y = 168.520659601
Approximate
y = 169
Hence, the estimate of the number of students studying abroad in 2003 is 169
<h3>b. Assuming this equation continues to be valid in the future, use this equation to predict the number of students studying abroad in 2018.</h3>
2018 is 20 years from 1998.
This means that
x = 20
Substitute the known values in the above equation
y = 123(1.065)^20
Evaluate the exponent
y = 123 * 3.52364506352
Evaluate the product
y = 433.408342813
Approximate
y = 433
Hence, the estimate of the number of students studying abroad in 2018 is 433
Read more about exponential functions at:
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Answer:
it does not move.
Step-by-step explanation:
....
If it is 40% off, it is selling at 60% of it original price.
The sales price is $48
60% = 58
1% = 58 ÷ 60 = 0.93
100% = 0.93 x 100 = $93
The original price is $93
Answer:
Step-by-step explanation:
Given that 10000 times a fair coin is flipped. Since fair coin prob for head of tail =0.5
If tail comes it will be taken as 1. Hence for the sum to be less than 5000, no of heads <5000
Since trials are large we approximate to normal with mean = np = 5000 and variance = np(1-p) = 2500
Thus no of heads x is normal with (5000, 50)
Since this prob is less than 0.5 only minimum amount to be bet i.e. 1 dollar
If sum of coins <5100, then Z = 
P(Z<2) = 0.9500
This time we can bet maximum of 100 dollars as probability is very high nearer to 1.
