Please help me solve x

Ben claims that 12 is a factor of 24. how can you check whether he is correct

AlekseyPX

Take 12 from 24 and if the the answer is 12 then it's a factor.

7 0

3 years ago

14 less than a number all divided by the same number

algol [13]

Number/number=1-14=-13

8 0

3 years ago

2. The alkalinity of a fluid (pH) can be modeled in terms of its hydrogen ion concentration (H+, measured in moles/liter) of the

nataly862011 [7]

Given the equation:

$pH=-log\mleft(H^+\mright)$

You know that a lab technician finds that a sample of fluid has this pH:

$pH=4.5$

Then, in order to find the hydrogen ion concentration of the fluid, you need to follow these steps:

1. Substitute the pH given in the exercise into the equation:

$4.5=-log(H^+)$

Multiply both sides by -1:

$-4.5=\log (H^+)$

2. Solve for:

$H^+$

Remember that, by definition:

$\log _b\mleft(c\mright)=a\Rightarrow b^a=c$

In this case, you can identify that:

$\begin{gathered} a=-4.5 \\ b=10 \\ c=H^+ \end{gathered}$

Hence, substituting values and simplifying, you get:

$\begin{gathered} b^a=c \\ 10^{-4.5}=H^+ \\ H^+=0.00003162278\text{ }\frac{moles}{liter} \end{gathered}$

Therefore, the answer is:

$H^+=0.00003162278\text{ }\frac{moles}{liter}$

4 0

1 year ago

1) Q: A: If 5 + x = 12, and you add -5 to the left side of equation, what should you add to right side of equation

san4es73 [151]

You can "add" like you say or you could just call it subtracting. If you subtract 5 from 5 you still get zero. Then on the right side, you would do the same. You would subtract 5 from 12 getting x, which is 7.

If you added you would still get the same answers but it is easier to say just subtract. Anything you do to the left, do to the right.

Hope this helps :)

3 0

3 years ago

Read 2 more answers

An example of an early application of statistics was in the year 1817. A study of chest circumference among a group of Scottish

otez555 [7]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is X: chest circumference of a Scottish man.

X≈N(μ;δ²)

μ= 40 inches

δ= 2 inches

The empirical rule states that

68% of the distribution lies within one standard deviation of the mean: μ±δ= 0.68

95% of the distribution lies within 2 standard deviations of the mean: μ±2δ= 0.95

99% of the distribution lies within 3 standard deviations of the mean: μ±3δ= 0.99

a)

The 58% that falls closest to the mean can also be referred to as the middle 58% of the distribution, assuming that both values are equally distant from the mean.

P(a≤X≤b)= 0.58

If 1-α= 0.58, then the remaining proportion α= 0.42 is divided in two equal tails α/2= 0.21.

The accumulated proportion until "a" is 0.21 and the accumulated proportion until "b" is 0.21 + 0.58= 0.79 (See attachment)

P(X≤a)= 0.21

P(X≤b)= 0.79

Using the standard normal distribution, you can find the corresponding values for the accumulated probabilities, then using the information of the original distribution:

P(Z≤zᵃ)= 0.21

zᵃ= -0.806

P(Z≤zᵇ)= 0.79

zᵇ= 0.806

Using the standard normal distribution Z= (X-μ)/δ you "transform" the values of Z to values of chest circumference (X):

zᵃ= (a-μ)/δ

zᵃ*δ= a-μ

a= (zᵃ*δ)+μ

a= (-0.806*2)+40= 38.388

and

zᵇ= (b-μ)/δ

zᵇ*δ= b-μ

b= (zᵇ*δ)+μ

b= (0.806*2)+40= 41.612

58% of the chest measurements will be within 38.388 and 41.612 inches.

b)

The measurements of the 2.5% men with the smallest chest measurements, can also be interpreted as the "bottom" 2.5% of the distribution, the value that separates the bottom 2.5% of the distribution from the 97.5%, symbolically:

P(X≤b)= 0.025 (See attachment)

Now you have to look under the standard normal distribution the value of z that accumulates 0.025 of the distribution:

P(Z≤zᵇ)= 0.025

zᵇ= -1.960

Now you reverse the standardization to find the value of chest circumference:

zᵇ= (b-μ)/δ

zᵇ*δ= b-μ

b= (zᵇ*δ)+μ

b= (-1.960*2)+40= 36.08

The chest measurement of the 2.5% smallest chest measurements is 36.08 inches.

c)

Using the empirical rule:

95% of the distribution lies within 2 standard deviations of the mean: μ±2δ= 0.95

(μ-2δ) ≤ Xc ≤ (μ+2δ)=0.95 ⇒ (40-4) ≤ Xc ≤ (40+4)= 0.95 ⇒ 36 ≤ Xc ≤ 44= 0.95

d)

The measurements of the 16% of the men with the largest chests in the population or the "top" 16% of the distribution:

P(X≥d)= 0.16

P(X≤d)= 1 - 0.16

P(X≤d)= 0.84

First, you look for the value that accumulates 0.84 of probability under the standard normal distribution:

P(Z≤zd)= 0.84

zd= 0.994

Now you reverse the standardization to find the value of chest circumference:

zd= (d-μ)/δ

zd*δ= d-μ

d= (zd*δ)+μ

d= (0.994*2)+40= 41.988

The measurements of the 16% of the men with larges chess are at least 41.988 inches.

I hope this helps!

8 0

3 years ago