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lions [1.4K]
2 years ago
9

80 PTS!!!

Mathematics
1 answer:
timama [110]2 years ago
6 0
It’s c

8 squared is 64
If you multiply that by 4 you will get 256 pi
You will not have to approximate it.
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Recently, a random sample of 13–18 year olds was asked, "How much do you currently have in savings?" The data in the table repre
Evgesh-ka [11]

The estimate for the mean is of: $227.15.

The estimate for the standard deviation is of: $219.4.

-------------------------------

  • First, we estimate the mean, which is the <u>sums of the multiplications of the relative frequencies and the halfway point of each interval.</u>
  • The size of the sample was of 339 + 93 + 53 + 17 + 9 + 8 + 1 = 520

The halfway points and relative frequencies are given as follows, and will be used to calculate both the mean and the standard deviation:

  • (0 + 199)/2 = 99.5, with a relative frequency of 339/520 = 0.6519.
  • (200 + 399)/2 = 299.5, with a relative frequency of 93/520 = 0.1788.
  • (400 + 599)/2 = 499.5, with a relative frequency of 53/520 = 0.1019.
  • (600 + 799)/2 = 699.5, with a relative frequency of 17/520 = 0.0327.
  • (800 + 999)/2 = 899.5, with a relative frequency of 9/520 = 0.0173.
  • (1000 + 1199)/2 = 1099.5, with a relative frequency of 8/520 = 0.0154.
  • (1200 + 1399)/2 = 1299.5, with a relative frequency of 1/520 = 0.0019.  

-------------------------------

Thus, the mean is of:

M = 0.6519(99.5) + 0.1788(299.5) + 0.1019(499.5) + 0.0327(699.5) + 0.0173(899.5) + 0.0154(1099.5) + 0.0019(1299.5) = 227.15

The mean of the amount of savings is of $227.15.

-------------------------------

The standard deviation is the <u>square root of the sum of the difference squared between each observation and the mean</u>, thus:

S = \sqrt{0.6519(99.5 - 227.15)^2 + 0.1788(299.5 - 227.15)^2 + 0.1019(499.5 - 227.15)^2 + ...} = 219.4

The standard deviation is of $219.4.

A similar problem is given at brainly.com/question/24651197

8 0
3 years ago
Help plz i reeeaaalllyyy dont get it :(
Irina-Kira [14]

Answer:

4 = b

5 = a

6 = a

6 0
3 years ago
There is a ratio of 2 boys to 3 girls in a class. Which of the following are correct ratios for the class?
denis23 [38]
All but the second one are correct
4 0
3 years ago
Read 2 more answers
If solving a system of linear equations by elimination or substitution I get that 0=0, as final or intermediate step depending o
aniked [119]

Answer:

If I get 0=0 then it means:

  • If the system of equations is 2 linear equations in 2 variables, there is infinity number of solutions
  • If the system of equations is 3 linear equations in 3 variables, there might be infinite  number of solutions

Step-by-step explanation:

Linear equation systems can consist of two or three equations with two or three unknowns respectively.

A system of linear equations in two variables has infinite solutions when the lines made by them overlap each other and similarly a system with three variables has infinite solutions when two lines overlap each other and third plane is parallel to them

Hence,

If I get 0=0 then it means:

  • If the system of equations is 2 linear equations in 2 variables, there is infinity number of solutions
  • If the system of equations is 3 linear equations in 3 variables, there might be infinite  number of solutions
4 0
3 years ago
A family has four children. If the genders of these children are listed in the order they are born, there are sixteen possible o
Agata [3.3K]

Answer:

\begin{array}{cccccc}X&0&1&2&3&4\\Pr&\dfrac{1}{16}&\dfrac{1}{4}&\dfrac{3}{8}&\dfrac{1}{4}&\dfrac{1}{16}\end{array}

Step-by-step explanation:

A family has four children. If the genders of these children are listed in the order they are born, there are sixteen possible outcomes: BBBB, BBBG, BBGB, BGBB, GBBB, BGBG, GBGB, BGGB, GBBG, BBGG, GGBB, BGGG, GBGG, GGBG, GGGB, and GGGG.

Let X represent the number of children that are girls. Then

1. When X=0, there is one possible outcome BBBB. So

Pr(X=0)=\dfrac{1}{16}

2. When X=1, then there are 4 possible outcomes GBBB, BGBB, BBGB, BBBG, so

Pr(X=1)=\dfrac{4}{16}=\dfrac{1}{4}

3. When X=2, then there are 6 possible outcomes BGBG, GBGB, BGGB, GBBG, BBGG, GGBB, so

Pr(X=2)=\dfrac{6}{16}=\dfrac{3}{8}

4. When X=3, then there are 4 possible outcomes GGGB, GGBG, GBGG, BGGG, so

Pr(X=3)=\dfrac{4}{16}=\dfrac{1}{4}

5. When X=4, then there is one possible outcome GGGG, so

Pr(X=4)=\dfrac{1}{16}

Now, the probability distribution table is

\begin{array}{cccccc}X&0&1&2&3&4\\Pr&\dfrac{1}{16}&\dfrac{1}{4}&\dfrac{3}{8}&\dfrac{1}{4}&\dfrac{1}{16}\end{array}

5 0
4 years ago
Read 2 more answers
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