Given these results, does it seem that the candy company is providing consumers with the amount claimed on the label? ▼ because the probability of getting a sample mean of 0.8537 g or greater when 456 candies are selected ▼ is is not exceptionally small.

Explanation below on given result...

Step-by-step explanation:

Let´s calculate z-score and refer to the normal distribution tables.

Z = (X - μ)/σ      or       Z = (.8543 - .8605)/.0513 = -.1209

The Z-score right area translates probability of 0.5478

Given this result:

it is basically the same calculation but with a sample standard deviation:

    σx = σ/√n = .0513/√(441) = .0024

    Z = (.8543 - .8605)/.0024 = -2.5833

The Z-score right area translates probability of 0.9951

Probability;

Getting a sample is 0.8543 or greater when 441 candies are selected is not exceptionally small.

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.

boxplot