Answer:
a
b

c

Step-by-step explanation:
From the question we are told that
The population proportion is p = 0.53
The sample size is n = 10
Generally the distribution of the confidence of US adults in newspapers follows a binomial distribution
i.e
and the probability distribution function for binomial distribution is
Here C stands for combination hence we are going to be making use of the combination function in our calculators
Generally the probability that the number of U.S. adults who have very little confidence in newspapers is exactly five is mathematically represented as
=>
=>
Generally the probability that the number of U.S. adults who have very little confidence in newspapers is at least six is mathematically represented as

=> ![P( X \ge 6 )= [^{10}C_6 * [0.53]^6 * (1- 0.53)^{10-6}] + [^{10}C_7 * [0.53]^7 * (1- 0.53)^{10-7}] + [^{10}C_8 * [0.53]^8 * (1- 0.53)^{10-8}] + [^{10}C_9 * [0.53]^9 * (1- 0.53)^{10-9}] + [^{10}C_{10} * [0.53]^{10} * (1- 0.53)^{10-10}]](https://tex.z-dn.net/?f=P%28%20X%20%5Cge%20%206%20%29%3D%20%20%5B%5E%7B10%7DC_6%20%2A%20%20%5B0.53%5D%5E6%20%2A%20%20%281-%200.53%29%5E%7B10-6%7D%5D%20%2B%20%20%5B%5E%7B10%7DC_7%20%2A%20%20%5B0.53%5D%5E7%20%2A%20%20%281-%200.53%29%5E%7B10-7%7D%5D%20%2B%20%20%5B%5E%7B10%7DC_8%20%2A%20%20%5B0.53%5D%5E8%20%2A%20%20%281-%200.53%29%5E%7B10-8%7D%5D%20%2B%20%20%5B%5E%7B10%7DC_9%20%2A%20%20%5B0.53%5D%5E9%20%2A%20%20%281-%200.53%29%5E%7B10-9%7D%5D%20%2B%20%5B%5E%7B10%7DC_%7B10%7D%20%2A%20%20%5B0.53%5D%5E%7B10%7D%20%2A%20%20%281-%200.53%29%5E%7B10-10%7D%5D)
=> ![P( X \ge 6 )= [0.227] + [0.1464] + [0.0619] + [0.0155] + [0.00082]](https://tex.z-dn.net/?f=P%28%20X%20%5Cge%20%206%20%29%3D%20%20%5B0.227%5D%20%2B%20%20%5B0.1464%5D%20%2B%20%20%5B0.0619%5D%20%2B%20%20%5B0.0155%5D%20%2B%20%5B0.00082%5D)
=> 
Generally the probability that the number of U.S. adults who have very little confidence in newspapers is less than four is mathematically represented as

=> ![P( X < 4 )= [^{10}C_3 * 0.53^3 * (1- 0.53)^{10-3}] + [^{10}C_2 * 0.53^2 * (1- 0.53)^{10-2}] + [^{10}C_1 * 0.53^1 * (1- 0.53)^{10-1}] + [^{10}C_0 * 0.53^0 * (1- 0.53)^{10-0}]](https://tex.z-dn.net/?f=P%28%20X%20%3C%20%204%20%29%3D%20%20%5B%5E%7B10%7DC_3%20%2A%20%200.53%5E3%20%2A%20%20%281-%200.53%29%5E%7B10-3%7D%5D%20%2B%20%20%5B%5E%7B10%7DC_2%20%2A%20%200.53%5E2%20%2A%20%20%281-%200.53%29%5E%7B10-2%7D%5D%20%2B%20%20%5B%5E%7B10%7DC_1%20%2A%20%200.53%5E1%20%2A%20%20%281-%200.53%29%5E%7B10-1%7D%5D%20%2B%20%20%5B%5E%7B10%7DC_0%20%2A%20%200.53%5E0%20%2A%20%20%281-%200.53%29%5E%7B10-0%7D%5D%20)
=> 
=> 
Hello!
log₃(x) + log₃(x - 6) = log₃(7) <=>
<=> log₃(x * (x - 6)) = log₃(7) <=>
<=> log₃(x² - 6x) = log₃(7) <=>
<=> x² - 6x = 7 <=>
<=> x² - 6x - 7 = 0 <=>
<=> x² + x - 7x - 7 = 0 <=>
<=> x * (x + 1) - 7 * (x + 1) = 0 <=>
<=> (x + 1) * (x - 7) = 0 <=>
<=> x + 1 = 0 and x - 7 = 0 <=>
<=> x = -1 and x = 7, x ∈ { 6; +∞ } <=>
<=> x = 7
Good luck! :)
Answer:
Mean track length for this rock specimen is between 10.463 and 13.537
Step-by-step explanation:
99% confidence interval for the mean track length for rock specimen can be calculated using the formula:
M±
where
- M is the average track length (12 μm) in the report
- t is the two tailed t-score in 99% confidence interval (2.977)
- s is the standard deviation of track lengths in the report (2 μm)
- N is the total number of tracks (15)
putting these numbers in the formula, we get confidence interval in 99% confidence as:
12±
=12±1.537
Therefore, mean track length for this rock specimen is between 10.463 and 13.537
What about it? What is the question?
well and 300% increase means you multiply by 3 and then add the original number
in this case 300% of 25 is 75 and its an increase so 25+75=100
the answer is 100