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swat32
3 years ago
5

How do I find the interval for a frequency table

Mathematics
1 answer:
alekssr [168]3 years ago
3 0
First you need to know how to calculate the class interval;

Calculating the class interval using the following formula:

Class interval = range ÷ number of classes. If you have 15 classes of income in the distribution of income example, work out 30 ÷ 15 = $2 billion.

Anyways, back to your question how you would find the interval for a frequency table is yeah think I explained it right.
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HELP, ASAP PLEASE!!!!
algol13

Answer:Remember that the general formula of geometric sequence is 

where 

is the nth term

is the difference

is the place of the term in the sequence

Also, to find  we will use the formula: 

where 

is the current term in the sequence 

is the previous term 

a) Lets find the three first terms of our sequence to check what type of sequence we have:

We know for our problem that the initial value of the computer is $1250, so our first term is 1250. In other words .

To fin our second term , we are going to subtract 10% of the value to our original value:

and , so 

To find our third term  we are going to subtract yet again 10% to our current value:

and , so

Now that we have our sequence  lets check if we have a consistent  to prove we have a geometric sequence:

- with , and :

 

- with , and :

 

Look! our s are the same, so we can conclude that we have a geometric sequence.

b) To do this we just need to replece the values of our sequence in the general formula of a geometric sequence. We know from our previous point that  and . So lets replace those values in geometric sequence formula to find our explicit formula:

c) To find the value of the computer at the beginning of the 6th year, we just need to find the 6th therm in our geometric sequence:

We can conclude that the value of the computer at the beginning of the 6th year will be $738.1125

Step-by-step explanation:

i did some research i dont know if it helped out not but id appreciate brainliest if it did..... thank you.

3 0
3 years ago
Meg laid 15 tiles in 50 minutes.
Basile [38]

Answer:  E

Step-by-step explanation:

4 divide 50 = 8 divide 2= 4

7 0
2 years ago
Read 2 more answers
Solve for y -5-10=-60 please help
Ket [755]

Answer:

y=-45

Step-by-step explanation:

y-15=-60

y=-45

8 0
3 years ago
Read 2 more answers
Let X represent the amount of energy a city uses (in megawatt-hours) in the Kanto region. Let Y represent the amount of mismanag
Setler [38]

Answer:

Part 2: The probability of X≤2 or X≥4 is 0.5.

Part 3: The value of marginal probability of y is f_y(y)=\frac{10-y}{32} for  2\leq y\leq 10

Part 4:The value of E(y) is 4.6667.

Part 5:The value of f_{xy}(x) is \frac{2}{10-y} for 2\leq y\leq 2x\leq 10

Part 6:The value of M_{x,y}(y) is \frac{y+10}{4}

Part 7:The value of E(x) is 3.6667.

Part 8:The value of E(x,y) is 36.

Part 9:The value of Cov(x,y) is 18.8886.

Part 10:X and Y are not independent variables as f_{xy}(x,y)\neq f_x(x).f_y(y)\\

Step-by-step explanation:

As the complete question is here, however some of the values are not readable, thus the question is found online and is attached herewith.

From the given data, the joint distribution is given as

f(x,y)=\frac{1}{16} for 2\leq y\leq 2x\leq 10

Now the distribution of x is given as

f_x(x)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dy

Here the limits for y are 2\leq y\leq 2x So the equation becomes

f_x(x)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dy\\f_x(x)=\int\limits^{2x}_{2} \frac{1}{16} \, dy\\f_x(x)=\frac{1}{16} (2x-2)\\f_x(x)=\frac{x-1}{8}                        \,\,\,\,\,\,\,\,\,\,\,\, for \,\,\,\,\,\,\,\,\,\ 1\leq x\leq 5

Part 2:

The probability is given as

P(X\leq 2 U X\geq 4)=\int\limits^2_1 {f_x(x)} \, dx +\int\limits^5_4 {f_x(x)} \, dx\\P(X\leq 2 U X\geq 4)=\int\limits^2_1 {\frac{x-1}{8}} \, dx +\int\limits^5_4 {\frac{x-1}{8}} \, dx\\P(X\leq 2 U X\geq 4)=\frac{1}{16}+\frac{7}{16}\\P(X\leq 2 U X\geq 4)=0.5

So the probability of X≤2 or X≥4 is 0.5.

Part 3:

The distribution of y is given as

f_y(y)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dx

Here the limits for x are y/2\leq x\leq 5 So the equation becomes

f_y(y)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dx\\f_y(y)=\int\limits^{5}_{y/2} \frac{1}{16} \, dx\\f_y(y)=\frac{1}{16} (5-\frac{y}{2})\\f_y(y)=\frac{10-y}{32}                        \,\,\,\,\,\,\,\,\,\,\,\, for \,\,\,\,\,\,\,\,\,\ 2\leq y\leq 10

So the value of marginal probability of y is f_y(y)=\frac{10-y}{32} for  2\leq y\leq 10

Part 4

The value is given as

E(y)=\int\limits^{10}_2 {yf_y(y)} \, dy\\E(y)=\int\limits^{10}_2 {y\frac{10-y}{32}} \, dy\\E(y)=\frac{1}{32}\int\limits^{10}_2 {10y-y^2} \, dy\\E(y)=4.6667

So the value of E(y) is 4.6667.

Part 5

This is given as

f_{xy}(x)=\frac{f_{xy}(x,y)}{f_y(y)}\\f_{xy}(x)=\frac{\frac{1}{16}}{\frac{10-y}{32}}\\f_{xy}(x)=\frac{2}{10-y}

So the value of f_{xy}(x) is \frac{2}{10-y} for 2\leq y\leq 2x\leq 10

Part 6

The value is given as

\geq M_{x,y}(y)=E(f_{xy}(x))=\int\limits^5_{y/2} {x f_{xy}(x)} \, dx \\M_{x,y}(y)=\int\limits^5_{y/2} {x \frac{2}{10-y}} \, dx \\M_{x,y}(y)=\frac{2}{10-y}\left[\frac{x^2}{2}\right]^5_{\frac{y}{2}}\\M_{x,y}(y)=\frac{2}{10-y}\left(\frac{25}{2}-\frac{y^2}{8}\right)\\M_{x,y}(y)=\frac{y+10}{4}

So the value of M_{x,y}(y) is \frac{y+10}{4}

Part 7

The value is given as

E(x)=\int\limits^{5}_1 {xf_x(x)} \, dx\\E(x)=\int\limits^{5}_1 {x\frac{x-1}{8}} \, dx\\E(x)=\frac{1}{8}\left(\frac{124}{3}-12\right)\\E(x)=\frac{11}{3} =3.6667

So the value of E(x) is 3.6667.

Part 8

The value is given as

E(x,y)=\int\limits^{5}_1 \int\limits^{10}_2 {xyf_{x,y}(x,y)} \,dy\, dx\\E(x,y)=\int\limits^{5}_1 \int\limits^{10}_2 {xy\frac{1}{16}} \,dy\, dx\\E(x,y)=\int\limits^{5}_1 \frac{x}{16}\left[\frac{y^2}{2}\right]^{10}_2\, dx\\E(x,y)=\int\limits^{5}_1 3x\, dx\\\\E(x,y)=3\left[\frac{x^2}{2}\right]^5_1\\E(x,y)=36

So the value of E(x,y) is 36

Part 9

The value is given as

Cov(X,Y)=E(x,y)-E(x)E(y)\\Cov(X,Y)=36-(3.6667)(4.6667)\\Cov(X,Y)=18.8886\\

So the value of Cov(x,y) is 18.8886

Part 10

The variables X and Y are considered independent when

f_{xy}(x,y)=f_x(x).f_y(y)\\

Here

f_x(x).f_y(y)=\frac{x-1}{8}\frac{10-y}{32} \\

And

f_{xy}(x,y)=\frac{1}{16}

As these two values are not equal, this indicates that X and Y are not independent variables.

4 0
3 years ago
Can someone help please
eduard

Step-by-step explanation:

Diagonals of a rectangle are congruent and bisects each other

Therefore,BE=DE

=>2x+4=7x-1

=>2x-7x= -1-4

=> -5x= -5

=> 5x=5

=> x=1

5 0
3 years ago
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