Answer:
brainliest plzzzzz
Step-by-step explanation:
2(x-2) - 3(x-4)
Simplify:
2(x-2)-3(x-4)
2x-4-3(x-4)
Distribute:
2x-4-3(x-4)
2x-4-3x+12
Add the numbers:
2x-4-3x+12
2x+8-3x
Combine like terms:
2x+8-3x
-x+8
Hence Solution = -x+8
The pH of the weak acid is 3.21
Butyric acid is known as a weak acid, we need the concentration of [H+] formula of weak acid which is given by this equation :
![[H^{+}]=\sqrt{Ka . Ma}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%3D%5Csqrt%7BKa%20.%20Ma%7D)
where [H+] is the concentration of ion H+, Ka is the weak acid ionization constant, and Ma is the acid concentration.
Since we know the concentration of H+, the pH can be calculated by using
pH = -log[H+]
From question above, we know that :
Ma = 0.0250M
Ka = 1.5 x 10¯⁵
By using the equation, we can determine the concentration of [H+]
[H+] = √(Ka . Ma)
[H+] = √(1.5 x 10¯⁵ . 0.0250)
[H+] = 6.12 x 10¯⁴ M
Substituting the value of [H+] to get the pH
pH = -log[H+]
pH = -log(6.12 x 10¯⁴)
pH = 3.21
Hence, the pH of the weak acid c3h7cooh is 3.21
Find more on pH at: brainly.com/question/14466719
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Answer: 0.84
Step-by-step explanation:
let p be the population proportion of adults who smoked a cigarette in the past week.
As per given , we have

Sample size : n= 1491
The sample proportion of adults smoked a cigarette= 
The test statistic for proportion is given by :-

Substitute all the values , we get

Hence, the value of the test statistic = 0.84
Answer:
Step-by-step explanation:
Hello!
The definition of the Central Limi Theorem states that:
Be a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.
As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.
X[bar]≈N(μ;σ²/n)
If the variable of interest is X: the number of accidents per week at a hazardous intersection.
There is no information about the distribution of this variable, but a sample of n= 52 weeks was taken, and since the sample is large enough you can approximate the distribution of the sample mean to normal. With population mean μ= 2.2 and standard deviation σ/√n= 1.1/√52= 0.15
I hope it helps!