We measure the diameters for a random sample of 25 oak trees in a neighbourhood. Diameters of oak trees in the neighbourhood follow a normal distribution with standard deviation 8.25 cm. A confidence interval for the true mean diameter of all oak trees in the neighbourhood is calculated to be (36.191, 42.969). What is the confidence level of this interval

The required confidence inteval = 94.9%.

Step-by-step explanation:

Confidence interval: Mean ± Margin of error

Given: A confidence interval for the true mean diameter of all oak trees in the neighbourhood is calculated to be (36.191, 42.969).

i.e.  Mean + Margin of error = 42.969               (i)

Mean - Margin of error = 36.191                    (ii)

Adding (i) and (ii), we get

Margin of error = 42.969-39.58             [from (i)]

= 3.389

Margin of error =

here n= 25

i.e.

Using  excel function 1-TDIST.2T(2.054,24)

The required confidence inteval = 94.9%.

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.

7

step-by-step explanation:

yuo ned the activity or vertical angle of the form of 1,7 but 2,2 has a c4,2 as it verticie so i not have a car