The lengths of a particular snake are approximately normally distributed with a given mean 15in and standard deviation 0.8in. what percentage of the snakes are longer than 16.6 in.?

We have been given that the lengths of a particular snake are approximately normally distributed with a given mean 15 in and standard deviation 0.8 in.

Let us find z-score for our given data.

Upon substituting our given values in z-score formula we will get,

We need to find . We will use the formula .

Let us find value for from normal distribution table,

Now we will subtract 0.9772 from 1 to get

Let us multiply 0.0228 by 100 to get percentage of the snakes longer than 16.6 in.

Therefore, 2.28% of the snakes are longer than 16.6 in.

3/6 chance

step-by-step explanation:

half of the nubers are odd numbers (1,3,5) which in that case if you would be told to simplify it would be 1/2

explanation:

infrared radiation has wavelength in the range between 700 nm and 1 mm. lets' convert these values into scientific notation, keeping in mind that:

let's start from the lower value, we have:

(we moved the dot two places to the left, so we increased the exponent by 2 units, from -9 to -7)

the higher value is:

so, the range of wavelength of infrared radiation is: