Answer:
Step-by-step explanation:
Volume = 198 ft
length * width * height = 198
(x - 1) * (x + 4) * 3 = 198
(x - 1) * (x + 4) = 198/3
(x - 1) * (x + 4) = 66
x*x + x*4 -1*x - 1*4 = 66 {Use FOIL method to multiply}
x² + 4x - x - 4 = 66
x² + 3x - 4 =66
x² + 3x - 4 - 66 = 0
x² + 3x - 70 = 0
Now factorize.
x² + 10x - 7x - 7*10 = 0
x(x +10) - 7(x + 10 ) = 0
(x + 10)(x - 7) = 0
x - 7 = 0 {x + 10 is ignored because measurement cannot be in negative}
x = 7
length = x - 1 = 7 - 1 = 6ft
Width = x + 4 = 7 + 4 = 11 ft
Answer:
I believe the answer is b
Step-by-step explanation: I think the answer is b because he gets $5 every week and he already has $25 the only thing we dont know is the weeks and you multiply 5 by the weeks so, 5(how much he gets each week) times x (The amount of weeks) plus 25(the amount he already has) equals y(the amount of everything put together)
Answer: provided in the explanation segment
Step-by-step explanation:
(a). from the question, we can see that since that б is known, we can use standard normal, z.
we are asked to find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error?
⇒ 80% confidence interval for the average weight of Allen's hummingbirds is given thus;
x ± z * б / √m
which is
3.15 ± 1.28 * 0.32/√10
= 3.15 ± 0.1295 = 3.0205 or 3.2795
(b). normal distribution of weight (c) б is known
(c). option (a) and (e) are correct
(d). from the question, let sample size be given as S
this gives';
1.28 * 0.32/√S = 0.15
√S = (1.28 * 0.32) / 0.15 = 2.73
S = 7.4529
cheers i hope this helps
you do 47%/100% and X/896 and cross multiply . so 47 times 896 = 42112 and divide that by hundred so the answer should be 421 photos.
Terminal digit filing is a filing system that
works by means of the last two numbers on files to consolidate them. While
middle digit filling uses a tripartite number, using the first part of the
number for the second sort, middle part for the initial sort, and the third portion
for the final sort. One way they are similar is that the two evade working with
large numbers and overpowers the difficulties of crowding at the end of the
storage.
