Answer:
14.4 × 10^13
Step-by-step explanation:
.........
Answer:
In 1981 was the building worth double it’s value.
Step-by-step explanation:
Given : A townhouse in San Francisco was purchased for $80,000 in 1975. The appreciation of the building is modeled by the equation : 
, where t represents time in years.
To find : In what year was the building worth double it’s value in 1975?
Solution :
The amount is $80,000.
The building worth double it’s value in 1975.
i.e. amount became A=2(80000).
Substitute in the model,
Taking log both side,
i.e. Approx in 6 years.
So, 1975+6=1981
Therefore, in 1981 was the building worth double it’s value.
(4b+3) (2b−5)
(4b+3) (2b+−5)
(4b) (2b) + (4b) (−5) +(3) (2b) +(3) (−5)
8b²−20b+6b−15.
= 8b²−14b−15.
Answer:
The answer to the question is;
Yes, it is very significant as the number of of observed vaccinated children is below the number of actually vaccinated children by 78.
Step-by-step explanation:
The result of the survey of more than 13,000 children indicate that only 89.4 % had actually been and the P-value indicate that the chance of having a sample proportion of 89.4 % vaccinated is 1.1 %.
P is low at 0.011 for which however the proportion of those vaccinated is between 0.889 and 0.899 using a 95% confidence interval, whereby the decrease from 90 % believed to 89.9 % is small, albeit it depends on the size of the population.
At 89.4 %, in a sample of 13,000, the number of children expected to have been vaccinated but were missed is equal to 90 - 89.4 = 0.6 % = 0.006
Therefore the children missed = 78 children which is significant.
50 percent because 30 plus 20 is 50 so 100-50 which is 50 percent of retail price
