Given a minimum usual value of 135.8 and a maximum usual value of 155.9, determine which (1 point) of the following values would be considered unusual. a. 137 b. 134 c. 146 d. 155

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3

answer: give more info

step-by-step explanation:

 b. 134

Step-by-step explanation:

Given : A minimum usual value of 135.8 and a maximum usual value of 155.9.

Let x denotes a usual value.

i.e.  135.8< x < 155.9

Therefore , the interval for the usual values is [135.8, 155.9] .

If interval for any usual value is [135.8, 155.9] , then any value should lie in this otherwise we call it unusual.

Let's check all options

a. 137  ,

since  135.8< 137 < 155.9

So , it is usual.

b. 134

since 134<135.8 (Minimum value)

So , it is unusual.

c. 146  

since  135.8< 146 < 155.9

So , it is usual.

d. 155  

since 135.8< 1155 < 155.9

So , it is usual.

Hence, the correct answer is b. 134 .