Answer:
0.226, 0.374
Step-by-step explanation:
A high school is running a campaign against the over-use of technology in teens. The committee running the campaign decides to look at the difference in social media usage between teens and adults. and then Find a 90% confidence interval for the difference in proportions.
The formula for confidence interval for the difference between the proportions is given as:
p1 - p2 ± z × √p1 (1 - p1)/n1 + p2(1 - p2)/n2
From the question
We have two groups.
Group 1
They take a random sample of 200 teens in their city (Group 1) and find that 85% of them use social media,
p1 = x/n1
n1 = 200
x1 = 85% × 200 = 170
p1 = 170/200
p1 = 0.85
Group 2
Take another random sample of 180 adults in their city (Group 2) and find that 55% of them use social media.
p2 = x/n1
n2= 180
x2 = 55% × 180 = 99
p2 = 99/180
p2 = 0.55
z = z score for 90% Confidence Interval = 1.645
p1 - p2 ± z × √p1 (1 - p1)/n1 + p2(1 - p2)/n2
= 0.85 - 0.55 ± 1.645 √0.85(1 - 0.85)/200 + 0.55(1 - 0.55)/180
= 0.85 - 0.55 ± 1.645 √0.85(0.15)/200 + 0.55(0.45)/180
= 0.30 ± 1.645 × √0.0020125
= 0.30 ± 1.645 × 0.0448608961
= 0.30 ± 0.0737961741
Hence
= 0.30 - 0.0737961741
= 0.2262038259
= 0.30 + 0.0737961741
= 0.3737961741
Therefore, 90% confidence interval for the difference in proportions is (0.226, 0.374
X+4=2x+5
-1=x
hope this helps
Answer:
The ordered pair is (4,-1)
Step-by-step explanation:
Let
x ---> the first number
y ---> the second number
we know that
Five times a number minus twice another number equals twenty-two
so
-----> equation A
The sum of the numbers is three
so
isolate the variable y
----> equation B
substitute equation B in equation A
solve for x
Find the value of y (equation B)
threrefore
The ordered pair is (4,-1)
Answer:
19.3 years
Step-by-step explanation:
Given that the initial mass of a sample of Element X is 100 grams,
The formula is given as:
N(t) = No × (1/2) ^t/t½
Element X is a radioactive isotope such that every 30 years, its mass decreases by half.
N(t) = Mass after time (t)
No = Initial mass = 100 grams
t½ = Half life = 30 grams
N(t) = 100 × (1/2) ^t/30
How long would it be until the mass of the sample reached 64 grams, to the nearest tenth of a year?
This means we are to find the time
N(t) = 100 × (1/2) ^t/30
N(t) = 64 grams
64 = 100(1/2)^t/30
Divide both sides by 100
64/100 = 100(1/2)^t/30/100
0.64 = (1/2)^t/30
Take the Log of both sides
log 0.64 = log (1/2)^t/30
log 0.64 = t/30(1/2)
t = 19.315685693242 years
Approximately = 19.3 years