The ability of lizards to recognize their predators via tongue flicks can often mean life or death for lizards. Seventeen juvenile common lizards were exposed to the chemical cues of the viper snake. Their responses, in number of tongue flicks per 20 minutes, are presented below. 727,217, 268, 438, 625, 319, 200, 591, 574, 727, 693, 336, 302, 761, 268, 353, 370 Assuming a normal distribution, find a 90% confidence interval for the mean number of tongue flicks per 20 minutes for all juvenile common lizards. Confidence interval:

Confidence level is (380.7133, 533.2867)

Step-by-step explanation:

Responses in number of tongue flicks per 20 minutes of lizards, are: 727,217, 268, 438, 625, 319, 200, 591, 574, 727, 693, 336, 302, 761, 268, 353, 370

n = 17

Mean (μ) is:

Standard deviation (σ) is:

The confidence interval (c) = 90% = 0.9

Margin of error (e) =

Confidence level = μ ± e = 457 ± 76.2867 = (380.7133, 533.2867)

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

b

step-by-step explanation: