Answer:
d. Approximate the standard normal distribution with the Student's t distribution
(0.2199 ; 0.2327)
Step-by-step explanation:
Given that :
Sample size, n = 31
Sample mean, xbar = 0.2258
Sample standard deviation, s = 0.0188
Confidence interval (C. I) :
xbar ± margin of error
Margin of Error : Tcritical * s/sqrt(n)
Degree of freedom, df = n - 1 = 31 - 1 = 30
Tcritical value :
T0.05/2, 30 = 2.042
Margin of Error = 2.042 * 0.0188/sqrt(31)
Margin of Error = 0.0068949
C. I = 0.2258 ± 0.0068949
Lower boundary : (0.2258 - 0.006895) = 0.2189
Upper boundary : (0.2258 - 0.006895) = 0.2327
(0.2199 ; 0.2327)
Answer:
6cm
Step-by-step explanation:
Volume of cuboid = length × breadth × height
Length = x
Breadth = 4cm
Height = 5cm
Volume = 120cm³
Volume = 4 × 5 × x
120 = 20x
Divide both sides by 20
20x/20 = 120/20
x = 6cm
I hope this was helpful, please mark as brainliest
✴<u>Given:</u>
↪Purchase price = $190,000
↪Appraisal fee = $450
↪Processing fee = $575
↪Title fee = $600
✴<u>Lets , find total closing costs :</u>
![\small\bold\red{[Total \: closing \: costs = 2 \% \: of \: the \: purchase \: price + appraisal \: fee + processing \: fee + title \: fee]}](https://tex.z-dn.net/?f=%5Csmall%5Cbold%5Cred%7B%5BTotal%20%5C%3A%20%20closing%20%5C%3A%20%20costs%20%3D%202%20%5C%25%20%5C%3A%20of%20%20%5C%3A%20the%20%5C%3A%20%20purchase%20%5C%3A%20%20price%20%2B%20appraisal%20%20%5C%3A%20fee%20%2B%20processing%20%20%5C%3A%20fee%20%2B%20title%20%20%5C%3A%20fee%5D%7D)
↪ $190000 × 0.02 + $450 + $575 + $600
↪ $3800 + $450 + $575 + $600
↪ $5425
Answer :- 
Answer: the answer is polar bear and Hippo
Step-by-step explanation:
1,750+3,865
Which equals 5,615
<h2>x = 6°</h2>
Step-by-step explanation:
<h3>^bac = ^dac</h3><h3>3x + 12° = 5x</h3><h3>12° = 5x - 3x</h3><h3>12° = 2x</h3><h3>6° = x</h3>
<h2>MARK ME AS BRAINLIST </h2><h2>PLZ FOLLOW ME</h2>
