An english professor determined the mean of her final exam scores to be 78 with a standard deviation of 6.3. find the minimum and maximum “usual” values of the test scores. 65.4 and 90.6 69.7 and 86.3 71.7 and 84.3 59.1 and 96.9

The minimum and maximum usual value of the test scores are:

              65.4 and 90.6

We know that:

The maximum usual value is two standard deviations above the mean and the minimum usual value is two standard deviation below the mean.

i.e.

Maximum value= Mean+2×standard deviation.

i.e.   Maximum value= 78+2×6.3

Maximum value= 90.6

and

Minimum value= Mean-2×standard deviation.

Minimum value= 78-2×6.3

i.e. Minimum value= 65.4

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places