A study of more than 3,000 toco toucans found that their weights were approximately Normally distributed, with a mean of 20 ounces and a standard deviation of 2.3 ounces. Let Y be the weight of a randomly chosen toco toucan. Which of the following correctly interprets the value of P(Y < 16)? Find the z-table here.
P(Y < 16) = 0.50. About 50% of all toco toucans weigh less than 16 ounces.
P(Y < 16) = 0.041. About 4.1% of all toco toucans weigh less than 16 ounces.
P(Y < 16) = 0.959. About 95.9% of all toco toucans weigh less than 16 ounces.
P(Y < 16) = 0.041. About 4.1% of the toco toucans from this group weigh less than 16 ounces.
P(Y < 16) = 0.959. About 95.9% of the toco toucans from this group weigh less than 16 ounces.

they used 313 gallons of paint

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.

c = 0.429t

step-by-step explanation:

the constant rate at which carlos harvests cassava is 35 minutes per 15 cassavas 15/35 = 0.429 cassava per minute

therefore, an equation that models the number of cassava (c) that harvest carlos in a time (t) is the following:

where t is the time in minutes.

this means that when 35 minutes have elapsed 15 cassava have been harvested, at 70 minutes 30 cassava have been harvested and so on.