A computer was on sale for 2/3 of the original price.If the original price was 900 dollars what was the sale price
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
1.
cot x sec⁴ x = cot x+2 tan x +tan³x
L.H.S = cot x sec⁴x
=cot x (sec²x)²
=cot x (1+tan²x)² [ ∵ sec²x=1+tan²x]
= cot x(1+ 2 tan²x +tan⁴x)
=cot x+ 2 cot x tan²x+cot x tan⁴x
=cot x +2 tan x + tan³x [ ∵cot x tan x =1]
=R.H.S
2.
(sin x)(tan x cos x - cot x cos x)=1-2 cos²x
L.H.S =(sin x)(tan x cos x - cot x cos x)
= sin x tan x cos x - sin x cot x cos x
= sin²x -cos²x
=1-cos²x-cos²x
=1-2 cos²x
=R.H.S
3.
1+ sec²x sin²x =sec²x
L.H.S =1+ sec²x sin²x
= [
]
=1+tan²x
=sec²x
=R.H.S
4.
L.H.S=
= 2 csc x
= R.H.S
5.
-tan²x + sec²x=1
L.H.S=-tan²x + sec²x
= sec²x-tan²x
=
=1
Answer:
The probability that the sample proportion is between 0.35 and 0.5 is 0.7895
Step-by-step explanation:
To calculate the probability that the sample proportion is between 0.35 and 0.5 we need to know the z-scores of the sample proportions 0.35 and 0.5.
z-score of the sample proportion is calculated as
z= where
- p(s) is the sample proportion of first time customers
- p is the proportion of first time customers based on historical data
- N is the sample size
For the sample proportion 0.35:
z(0.35)= ≈ -1.035
For the sample proportion 0.5:
z(0.5)= ≈ 1.553
The probabilities for z of being smaller than these z-scores are:
P(z<z(0.35))= 0.1503
P(z<z(0.5))= 0.9398
Then the probability that the sample proportion is between 0.35 and 0.5 is
P(z(0.35)<z<z(0.5))= 0.9398 - 0.1503 =0.7895