Answered Essay: Using R language answer the following problem

Using R language answer the following problem Exercise 2.1 a. Using R, verify that 6a + 42 34.2-3.62 29.50556 when a = 2.3 b. Which of the following squares negative 4 and adds 2 to the result? i. (-4)*2+2 ii. -4*2+2 iv. -4*(2+2) Using R, how would you calculate the square root of half of the average of the numbers 25.2, 15, 16.44, 15.3, and 18.6? Find logc 0.3. Compute the exponential transform of your answer to (d) Identify Rs representation of-0.00000000423546322 when printing this number to the console c. d. e.

a. Using R, verify that 6a + 42/3^4.2-3.62 = 29.50556 when a = 2.3. b. Which of the following squares negative 4 and adds 2 to the result? i. (-4)^2+2 ii. -4^2+2 iii. (-4)^(2 + 2) iv. -4^(2+2) c. Using R, how would you calculate the squareroot of half of the average of the numbers 25.2, 15, 16.44, 15.3, and 18.6? d. Find log_e 0.3. e. Compute the exponential transform of your answer to (d). f. Identify R’s representation of -0.00000000423546322 when printing this number to the console.

Expert Answer

 

#*******(a) solution*****************

a = 2.3
numerator = 6*a+42
denomenoterspower = 4.2-3.62
denomenoter = 3^denomenoterspower
print(numerator/denomenoter)
if(numerator/denomenoter==29.50556)
print(“Verified”)
#***********(b) solution************
negative4 = -4
squarenegative4 =negative4^2
squarenegative4add2 =squarenegative4+2
print(squarenegative4add2)
#**********check for option i)
a = (-4)^2+2
if(a == squarenegative4add2)
print(“Option a is correct”)
#**********check for option ii)
b = -4^2+2
if(b == squarenegative4add2)
print(“Option b is correct”)
#**********check for option iii)
c = (-4)^(2+2)
if(c == squarenegative4add2)
print(“Option a is correct”)
#**********check for option iv)
d = -4^(2+2)
if(d == squarenegative4add2)
print(“Option a is correct”)

#*******(c) solution
avg = (25.2+15+16.44+15.3+18.6)/5
print(sqrt(avg/2.0))

#****(d) soution****
#log(x, base = exp(1))
#x is number and base is base of the log
#exp(1) = e ie base of the log is e
print(log(0.3, base = exp(1)))

#*****(e) solution
#exp function find exponential
print(exp(log(0.3, base = exp(1))))

#*****(f)solution
num = -0.00000000423546322
print(‘R representation when printing on console’)
print(num)

#*******Output Screenshot***************

#*******R code part (1)Screenshot**************

#*******R code part 2 screenshot***********

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