The heights of flowering cherry trees are normally distributed with a mean of 11.5 feet, and a standard deviation of 1.7 ft. find the probability that a randomly selected tree is less than 13.5 feet tall.

Mean (μ) = 11.5 feet

Standard deviation (σ) = 1.7 feet

First we need to find the z-score for less than 13.5 feet.

The formula of z-score is : z = (X - μ)/σ

Here X= 13.5, so z =

z = = 1.18

P(X < 13.5) = P(z< 1.18) =P(z< (1.1 + .08)) = 0.8810 (from z-score table)

P(X< 13.5) = 88.1% (for making percentage from decimal, we need to multiply by 100)

So, the probability that a randomly selected tree is less than 13.5 feet tall = 88.1%

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.