These are the two equation
10-2
13-5
Answer:
33+12t−21t^2
Step-by-step explanation:
(2t-7)²-(5t-4)²
Use binomial theorem (a−b)^2 = a^2−2ab+b^2 to expand (2t-7)².
4t^2−28t+49−(5t-4)²
Use binomial theorem (a−b)^2 = a^2−2ab+b^2 to expand (5t-4)².
4t^2−28t+49−(25t^2−40t+16)
To find the opposite of 25t^2
−40t+16, find the opposite of each term.
4t^2−28t+49−25t^2−40t+16
Combine 4t^2 and −25t^2 to get −21t^2.
−21t^2−28t+49+40t−16
Combine −28t and 40t to get 12t.
−21t^2+12t+49−16
Subtract 16 from 49 to get 33.
−21t^2+12t+33
Swap terms to the left side.
33+12t−21t^2
I hope this helped!
__Brainliest if helped!
5 Parts Yellow with 3 Parts blue
5Yellow with 3 Blue
1Yellow would be with 3/5 Blue
Hence ,
2 Yellow = (3/5)*2 Blue =6/5 Cans of blue or 1.2Cans of blue.
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
Answer:
ans -------8/5 pansnhshshsh
