Answer: (0.076, 0.140)
Step-by-step explanation:
Confidence interval for population proportion (p) is given by :-

, where
= sample proportion.
n= sample size.
= significance level .
= critical z-value (Two tailed)
As per given , we have
sample size : n= 500
The number of Independents.: x= 54
Sample proportion of Independents
Significance level 98% confidence level :
By using z-table , Critical value :
The 98% confidence interval for the true percentage of Independents among Haywards 50,000 registered voters will be :-

Hence, the 98% confidence interval for the true percentage of Independents among Haywards 50,000 registered voters.= (0.076, 0.140)
Answer:
Step-by-step explanation:
(7,-6) (13,2)
2--6. 8. 4
------= ------= ------ (2--6 turns to a +)
13-7. 6. 3
4
Slope:------
3
Points=(7,-6)
4
y--6=----(x-7) (y--6 turns to a +)
3
4. 28
y+6= ---- x - ----
3. 3
-6. -6
-------------------------
4. 46
y=----- x - -------
3. 3
Well, remember we can't take the square root of a negative
so we see that we have

so find those values that take sqrt of a negative and restrict hem from the domain
anny value greater than 1 and less than -1
so domain is from -1 to 1, including those numbers
D=[-1,1]
a. D=[-1,1] or from -1 to 1 is domain
b. for a TI-84, go to y-editor then input

for y1
c. for a TI-84, click 2nd then window (gets to tbset) scrol down to set Δx to 0.1, then cilick 2nd again then click graph (to select table) and scroll down till you see that value of y that is the biggest, that value is x=0.7
A. domain is from -1 to 1
B. use your brain or google the instructions for your calulator
C. at x=0.7
Answer:
x>13
Step-by-step explanation:
Answer:
- P(t) = 100·2.3^t
- 529 after 2 hours
- 441 per hour, rate of growth at 2 hours
- 5.5 hours to reach 10,000
Step-by-step explanation:
It often works well to write an exponential expression as ...
value = (initial value)×(growth factor)^(t/(growth period))
(a) Here, the growth factor for the bacteria is given as 230/100 = 2.3 in a period of 1 hour. The initial number is 100, so we can write the pupulation function as ...
P(t) = 100·2.3^t
__
(b) P(2) = 100·2.3^2 = 529 . . . number after 2 hours
__
(c) P'(t) = ln(2.3)P(t) ≈ 83.2909·2.3^t
P'(2) = 83.2909·2.3^2 ≈ 441 . . . bacteria per hour
__
(d) We want to find t such that ...
P(t) = 10000
100·2.3^t = 10000 . . . substitute for P(t)
2.3^t = 100 . . . . . . . . divide by 100
t·log(2.3) = log(100)
t = 2/log(2.3) ≈ 5.5 . . . hours until the population reaches 10,000