Answer:
B. 1/25
Step-by-step explanation:
This question asks us to find (-2), or f(x) when x is equal to -2.
We know that f(x)=5^x. We also know that we are trying to find f(-2). Therefore, we must substitute -2 in for x.
f(x)=5^x
x= -2
f(-2)= 5^-2
Negative exponents follow the rule:
x^-b= 1/x^b
Rewrite the exponent according to this rule.
5^-2=1/5^2
f(-2)= 1/5^2
Evaluate the exponent in the denominator.
5^2=5*5=25
f(-2)=1/25
f(-2) is equal to 1/25, therefore B is the correct answer.
Answer:
12 minutes
Step-by-step explanation:
Candles can be manually packed at the rate of ...
... (400 candles)/(36 minutes) = 11 1/9 candles/minute
The two systems together pack candles at the rate of ...
... (400 candles)/(9 minutes) = 44 4/9 candles/minute
Then the machine's packing rate is the difference between the total and the manual rates:
... machine rate = (44 4/9 candles/minute) - (11 1/9 candles/minute)
... = 33 1/3 candles/minute.
At this rate, the machine working alone can pack 400 candles in ...
... (400 candles)/(33 1/3 candles/minute) = (400/(100/3)) minutes
... = 400 × 3/100 minutes = 12 minutes
Roper is not using a simple random sample. The samples are calculated to get 500 males and 500 females. This would be very unlikely or improbable to take place in a simple random sample. The design that they are using is a stratified sampling (dividing the population into groups) with 2 strata and these are the males and females.
Oh dear~This can only be simplified。
15X+18y
3(5X+6Y)
That's your answer.
Answer:
a) P(identified as containing explosives)=P(actually contains explosives and identified as containing explosives)+P(actually not contains explosives and identified as containing explosives)
=(10/(4*106))*0.95+(1-10/(4*106))*0.005 =0.005002363
hence probability that it actually contains explosives given identified as containing explosives)
=(10/(4*106))*0.95/0.005002363=0.000475
b)
let probability of correctly identifying a bag without explosives be a
hence a =0.99999763 ~ 99.999763%
c)
No as even if that becomes 1 ; proportion of true explosives will always be less than half of total explosives detected,
