A population of insect weights is normal with a mean of μ = 16. If the probability of a weight falling between 10 and 22 is 95.44%, what is the standard deviation for this distribution a. 1.5
b. 3
c. 6
d. 12

6

Step-by-step explanation:

This is calculated as using z score

z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

Hence,

We find the z score of 95.44% confidence interval

95.44% = 1.9991

Hence.

1.9991 = ( X = 22) - (X = 10)

1.9991 = (22 - 16/ σ) - (10 - 16/ σ)

1.9991 = (6/σ - (-6/σ)

1.9991 = ( 6/σ + 6/σ)

1.9991 = (6 + 6/σ)

1.9991 = 12/σ

Cross Multiply

= 1.9991 × σ = 12

σ = 12/1.9991

= 6.00270121555

Approximately 6

Hence, the standard deviation is 6

An obtuse triangle.

Step-by-step explanation:

We have given the different sides of a triangle that is Side A = 2, Side B = 8 and Side C = 7.

According to the question, we can tell the triangle is acute, obtuse or a right triangle.

We know that,    

For given sides x,y,z of a triangle,

If = , then its a right triangle.

If < , then its an acute triangle.

If > , then its an obtuse triangle.

We have a = x = 2, b = z = 8 and c = y = 7.

Note: We always take z as the greatest side length.

Now,

We can see that, 4 + 49 < 64.

So, we can apply rule

if > , then its an obtuse triangle.

Therefore,

Given sides are of an obtuse triangle.