The calibration of a scale is to be checked by weighing a 10-kg test specimen 25 times. suppose that the results of different weighings are independent of one another and that the weight on each trial is normally distributed with σ = .200 kg. let µ denote the true average weight reading on the scale. (a) what hypotheses should be tested? (b) with the sample mean itself as the test statistic, what is the p-value when x = 9.85, and what would you conclude at significance level .01? (c) for a test with α = .01, what is the probability that recalibration is judged unnecessary when in fact µ = 10.2?

a:  That the mean weight of the trials is 10 kg

b:  See attached photo for work

Step-by-step explanation:

We want to see if the scale is weighing properly and are using a 10 kg weight to calibrate it.  That means our hypothesis test is that the mean weight of the trails (in this case 25) is 10 kg.

The hypothesis we will use are

H0:  µ = 10

Ha:  µ ≠ 10

The alternate hypothesis has a not equals to sign because if the scale weighs too much or to little, then it needs to be better calibrated, so it's a two tailed test.

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.

answer: apothems is your answer

step-by-step explanation: hope u get it right