The data represents the daily rainfall (in inches) for one month. Construct a frequency distribution beginning with a lower class limit of and use a class width of . Does the frequency distribution appear to be roughly a normal distribution?

The frequency distribution does not appear to be normal.

Step-by-step explanation:

The data provided is as follows:

S = {0.38 , 0 , 0.22 , 0.06 , 0 , 0 , 0.21 , 0 , 0.53 , 0.18 , 0 , 0 , 0.02 , 0 , 0 , 0.24 , 0 , 0 , 0.01 , 0 , 0 , 1.28 , 0.24 , 0 , 0.19 , 0.53 , 0 , 0, 0.24 , 0}

It is provided that the first lower class limit should be 0.00 and the class width should be 0.20.

The frequency distribution table is as follows:

Class Interval  Count

 0.00 - 0.19            21

 0.20 - 0.39             6

 0.40 - 0.59             2

 0.60 - 0.79             0

 0.80 - 0.99             0

  1.00 - 1 . 19             0

  1.20 - 1. 39             1

The frequency distribution does not appear to be normal. This is because the frequencies does not start and end at almost equivalent points and the mid-distribution does not consist of the highest frequency.

Thus, the frequency distribution does not appear to be normal.

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1