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Bond [772]
3 years ago
12

A random sample of CO2 levels in a school has a sample mean of x¯=598.4 ppm and sample standard deviation of s=86.7 ppm. Use the

Empirical Rule to determine the approximate percentage of CO2 levels that lie between 338.3 and 858.5 ppm. Round your answer to the nearest tenth.
Mathematics
1 answer:
astraxan [27]3 years ago
6 0

Answer:

99.7%

Step-by-step explanation:

we start by getting the z-scores for the values given

We can get the z-scores using the formula below;

z-scores = (x-mean)/SD

for 338.5;

z = (338.5-598.4)/86.7 = -2.99 approximately-3

for 858.5;

z = (858.5-598.4)/86.7 = 3

so we want to get the percentages that lie within 3 standard deviations from the mean

According to the empirical rule, the percentages within 3 SD of the mean is;

100-0.15-0.15 = 99.7%

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Suppose a principal amount of $2079 were invested into a simple interest account that pays an annual rate of 2.31% then the amou
Anna35 [415]

Answer:

The amount of interest earned at the end of 14 years would be $672.3486

Step-by-step explanation:

This is a simple interest problem.

The simple interest formula is given by:

E = P*I*t

In which E are the earnings, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time.

In this problem, we have that:

P = 2079, I = 0.0231, t = 14

So

E = 2079*0.0231*14 = 672.3486

The amount of interest earned at the end of 14 years would be $672.3486

4 0
3 years ago
3,480
Serjik [45]

Answer

B

explanation

3480ones = 3480x1=3480

8 0
2 years ago
Explain how to multiply the following whole numbers 21 x 14
Lesechka [4]

Answer:

\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}

________

\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}

Step-by-step explanation:

Given

21\:\times \:14

Line up the numbers

\begin{matrix}\space\space&2&1\\ \times \:&1&4\end{matrix}

Multiply the top number by the bottom number one digit at a time starting with the ones digit left(from right to left right)

Multiply the top number by the bolded digit of the bottom number

\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}

Multiply the bold numbers:    1×4=4

\frac{\begin{matrix}\space\space&2&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&\space\space&4\end{matrix}}

Multiply the bold numbers:    2×4=8

\frac{\begin{matrix}\space\space&\textbf{2}&1\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}

Multiply the top number by the bolded digit of the bottom number

\frac{\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}

Multiply the bold numbers:    1×1=1

\frac{\begin{matrix}\space\space&\space\space&2&\textbf{1}\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&\space\space&1&\space\space\end{matrix}}

Multiply the bold numbers:    2×1=2

\frac{\begin{matrix}\space\space&\space\space&\textbf{2}&1\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&2&1&\space\space\end{matrix}}

Add the rows to get the answer. For simplicity, fill in trailing zeros.

\frac{\begin{matrix}\space\space&\space\space&2&1\\ \space\space&\times \:&1&4\end{matrix}}{\begin{matrix}\space\space&0&8&4\\ \space\space&2&1&0\end{matrix}}

adding portion

\begin{matrix}\space\space&0&8&4\\ +&2&1&0\end{matrix}

Add the digits of the right-most column: 4+0=4

\frac{\begin{matrix}\space\space&0&8&\textbf{4}\\ +&2&1&\textbf{0}\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&\textbf{4}\end{matrix}}

Add the digits of the right-most column: 8+1=9

\frac{\begin{matrix}\space\space&0&\textbf{8}&4\\ +&2&\textbf{1}&0\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{9}&4\end{matrix}}

Add the digits of the right-most column: 0+2=2

\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}

Therefore,

\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}

________

\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}

6 0
3 years ago
1.2x + 3 =4x + 2
Dmitry_Shevchenko [17]
1. 2x+3=4x+2
  ⇒ 4x-2x=3-2
2x=1
x=1/2
There is one solution.

2. 18p-10p=5+6/2
8p=5+3
8p=8
p=8:8
p=1

3. 6a/3=2a
2a-a= -2 -3
a= - 5

4. 5+r-2=9,
 5-2+r=9
3+r=9
r=9 - 3
    r=6

5. -0,9 m= 9
     m= - 10

6. -2 (x+1/4)=5-1=4
     x+1/4= -2
     x= -2 - 1/4
     x= -8/4 - 1/4= - 9/4

7. 6f+4f= 6 + 12
    10 f = 18
      f=18/10=1,8

8. 7n - 3n= 16
    4n=16
    n=16:4
    n=4

9. 1/3m-5/6m= - 15 - 3
     2/6 m-5/6=-18
     - 3/6m=-18  ⇒ 1/2m=18
       m=2*18=36
     
7 0
3 years ago
What is the value of x in units
nikdorinn [45]

\triangle RQT\ \text{and}\ \triangle SRT\ \text{are similar. Therefore the sides are in proportion.}\\\\\dfrac{RT}{TQ}=\dfrac{TS}{RT}\\\\\text{We have}\ RT=x,\ TQ=16,\ TS=9.\ \text{Substitute:}\\\\\dfrac{x}{16}=\dfrac{9}{x}\qquad\text{cross multiply}\\\\x^2=(9)(16)\\\\x^2=144\to x=\sqrt{144}\\\\\boxed{x=12}

6 0
3 years ago
Read 2 more answers
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