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Liono4ka [1.6K]
3 years ago
12

Which of the scatterplots to the right show ​a) no​ association? ​b) a negative​ association? ​c) a linear​ association? ​d) a w

eak or moderately strong​ association? ​e) a very strong​ association?
Mathematics
1 answer:
Len [333]3 years ago
4 0

Answer:

A scatter plot to the right shows  a very strong​ association.

Step-by-step explanation:

Because  a scatter plot to the right shows that both variables are positive and they both increase, so it shows a strong association between those variables.

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The petrol tank at the local service station has a capacity of 6000L. It is 66.06 percent full. How many litres of petrol are ne
Rina8888 [55]


Since the tank is 66.06% full, it is (100 - 66.06) = 33.94% empty.

Its capacity is 6,000L.  So, in order to fill it to capacity, the service station
must order  33.94% of 6,000L .

               33.94% = 0.3394
               "of" means "times"

0.3394 x 6,000L = <em>2,036.4 liters</em> needed, to fill the tank to the rim.


7 0
3 years ago
What is the average distance from Rose Hill Tennesse to Miami Florida?
OLEGan [10]
The average distance from Rose Hill Tennessee to Miami Florida is 832 miles.
3 0
3 years ago
Read 2 more answers
Solve the triangle m&lt;N= 118°, m&lt;P= 33° and m=15. Round to the nearest tenth. Please show work.​
kirill [66]

Answer:

M = 29

27.31838095 = n

16.8511209 = p

Step-by-step explanation:

M = 180 -33-118 = 29

We can use the rule of sines

sin A          sin B           sin C

------------ = ---------- = ------------

a                  b               c

sin 118          sin 29        

------------ = ----------

n                  15              

Using cross products

15 sin 118 = n sin 29

Divide by sin 29

15 sin 118 / sin 29 = n

27.31838095 = n

sin 33          sin 29        

------------ = ----------

p                  15              

Using cross products

15 sin 33 = p sin 29

Divide by sin 29

15 sin 33 / sin 29 = p

16.8511209 = p

3 0
2 years ago
A school is organizing a weekend trip to a nature preserve. For each student, there is a $50 charge, which covers food and lodgi
k0ka [10]

The number of students going on the trip is 50.

Given,

A school is organizing a weekend trip to a nature preserve.

For each student, there is a $50 charge, which covers food and lodging. There is also a $25 charge per student for the bus.

The school must also pay a $15 cleaning fee for the bus.

The total cost of the weekend is $3,765.

We need to find out how many students will be going on the trip.

<h3>How do we find the number of items from the cost of each item and the total amount?</h3>

We simply divide the total cost by the cost of each item.

Example:

Total amount = 10

Each item cost = 2

Number of items = 10/2 = 5

Find the charge of each student.

= Food and lodging charge + bus charge

= $50 + $25

= $75

Find the additional charge for the trip.

= Cleaning charge

= $15

Find the total cost of the trip.

= $3,765

Find the total charge from the students.

= $3,765 - $15

= $3,750

Find the number of students who went on the trip.

= Total charge from the student's ÷ cost of each student

= 3750 ÷ 75

= 3750 / 75

= 50

This means 50 students.

Thus the number of students who went on the trip is 50.

Learn more about finding the number of students who like both games with a ratio given here:

brainly.com/question/24125573

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8 0
1 year ago
Read 2 more answers
Let X and Y be discrete random variables. Let E[X] and var[X] be the expected value and variance, respectively, of a random vari
Ulleksa [173]

Answer:

(a)E[X+Y]=E[X]+E[Y]

(b)Var(X+Y)=Var(X)+Var(Y)

Step-by-step explanation:

Let X and Y be discrete random variables and E(X) and Var(X) are the Expected Values and Variance of X respectively.

(a)We want to show that E[X + Y ] = E[X] + E[Y ].

When we have two random variables instead of one, we consider their joint distribution function.

For a function f(X,Y) of discrete variables X and Y, we can define

E[f(X,Y)]=\sum_{x,y}f(x,y)\cdot P(X=x, Y=y).

Since f(X,Y)=X+Y

E[X+Y]=\sum_{x,y}(x+y)P(X=x,Y=y)\\=\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y).

Let us look at the first of these sums.

\sum_{x,y}xP(X=x,Y=y)\\=\sum_{x}x\sum_{y}P(X=x,Y=y)\\\text{Taking Marginal distribution of x}\\=\sum_{x}xP(X=x)=E[X].

Similarly,

\sum_{x,y}yP(X=x,Y=y)\\=\sum_{y}y\sum_{x}P(X=x,Y=y)\\\text{Taking Marginal distribution of y}\\=\sum_{y}yP(Y=y)=E[Y].

Combining these two gives the formula:

\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y) =E(X)+E(Y)

Therefore:

E[X+Y]=E[X]+E[Y] \text{  as required.}

(b)We  want to show that if X and Y are independent random variables, then:

Var(X+Y)=Var(X)+Var(Y)

By definition of Variance, we have that:

Var(X+Y)=E(X+Y-E[X+Y]^2)

=E[(X-\mu_X  +Y- \mu_Y)^2]\\=E[(X-\mu_X)^2  +(Y- \mu_Y)^2+2(X-\mu_X)(Y- \mu_Y)]\\$Since we have shown that expectation is linear$\\=E(X-\mu_X)^2  +E(Y- \mu_Y)^2+2E(X-\mu_X)(Y- \mu_Y)]\\=E[(X-E(X)]^2  +E[Y- E(Y)]^2+2Cov (X,Y)

Since X and Y are independent, Cov(X,Y)=0

=Var(X)+Var(Y)

Therefore as required:

Var(X+Y)=Var(X)+Var(Y)

7 0
3 years ago
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