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2 years ago

Which scenarios have a negative correlation? Check all that apply.

1 answer:

azamat2 years ago

4 0

Negative correlation is defined as an inverse relationship between two entities. If entity 1 increases, entity 2 decreases and if entity 2 increases, then entity 1 decreases.

According to this definition, option 1 and 5 are right answers.

1) the number of calories consumed and the amount of weight lost by a person - if more calories are consumed, person gains weight. If less calories are consumed, person loses weight.

5) the thickness of the ice on a lake and the likelihood of falling through the ice - if the ice sheet is thick, its less likely to fall and vice versa.

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On Tuesday, 525 people bought tickets to the county fair. Tickets cost $7 for adults and $3 for children. The total revenue from

Readme [11.4K]

You need to look at this chart The system of equations below represents the number of people and total sales for the county fair on Tuesday, where x represents the number of child tickets and y represents the number of adult tickets. you need to take the amount of money you get for adult tickets only then divid it by seven and that is you answer

8 0

2 years ago

Find dy/dx implicitly and find the largest interval of the form −a < y < a or 0 < y < a such that y is a differentia

mylen [45]

Answer:

dy/dx = -1/√(1 - x²)

For 0 < y < π

Step-by-step explanation:

Given the function cos y = x

-siny dy = dx

-siny dy/dx = 1

dy/dx = -1/siny (equation 1)

But cos²y + sin²y = 1

=> sin²y = 1 - cos²y

=> siny = √(1 - cos²y) (equation 2)

Again, we know that

cosy = x

=> cos²y = x² (equation 3)

Using (equation 3) in (equation 2), we have

siny = √(1 - x²) (equation 4)

Finally, using (equation 4) in (equation 1), we have

dy/dx = -1/√(1 - x²)

The largest interval is when

√(1 - x²) = 0

=> 1 - x² = 0

=> x² = 1

=> x = ±1

So, the interval is

-1 < x < 1

arccos(1) < y < arxcos(-1)

= 0 < y < π

6 0

2 years ago

Suppose you roll a pair of honest dice. If you roll a total of 7 you win $22, if you roll a total of 11 you win $66, if you roll

JulijaS [17]

Answer:

The expected payoff for this game is -$1.22.

Step-by-step explanation:

It is given that a pair of honest dice is rolled.

Possible outcomes for a dice = 1,2,3,4,5,6

Two dices are rolled then the total number of outcomes = 6 × 6 = 36.

$\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\\(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$

The possible ways of getting a total of 7,

{ (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) }

Number of favorable outcomes = 7

Formula for probability:

$Probability=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}$

So, the possibility of getting a total of 7 = $\frac{6}{36}=\frac{1}{6}$

The possible ways of getting a total of 11,

{(5,6), (6,5)}

So, the probability of getting a total of 11 = $\frac{2}{36}$ = $\frac{1}{18}$

Now, other possible rolls = 36 - 6 - 2 = 36 - 8 = 28,

So, the probability of getting the sum of numbers other than 7 or 11 = $\frac{28}{36}$ = $\frac{7}{9}$

Since, for the sum of 7, $ 22 will earn, for the sum of 11, $ 66 will earn while for any other total loss is $11,

Hence, the expected value for this game is

$\frac{1}{6}\times 22+\frac{1}{18}\times 66-\frac{7}{9}\times 11$

$\frac{11}{3}+\frac{11}{3}-\frac{77}{9}$

$\frac{22}{3}-\frac{77}{9}$

$\frac{66-77}{9}$

$-\frac{11}{9}$

$-1.22$

Therefore the expected payoff for this game is -$1.22.

4 0

2 years ago

A maximum can occur at the each of the following except

Vlad [161]

Answer:

im pretty sure its B correct me if im wrong

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

The first three terms of an arithmetic series are 6p+2, 4p²-10 and 4p+3 respectively. Find the possible values of p. Calculate t

Doss [256]

Answer:

First Case:

$\displaystyle p=\frac{5}{2}\text{ and } d=-2$

Second Case:

$\displaystyle p=-\frac{5}{4}\text{ and } d=\frac{7}{4}$

Step-by-step explanation:

We know that the first three terms of an arithmetic series are:

$6p+2, 4p^2-10, \text{ and } 4p+3$

Since this is an arithmetic sequence, each subsequent term is d more than the previous term, where d is our common difference.

Therefore, we can write the second term as;

$4p^2-10=(6p+2)+d$

And, likewise, for the third term:

$4p+3=(6p+2)+2d$

Let's solve for d for each of the equations.

Subtracting in the first equation yields:

$d=4p^2-6p-12$

And for the second equation:

$2d=-2p+1$

To avoid fractions, let's multiply the first equation by 2. Hence:

$2d=8p^2-12p-24$

Therefore:

$8p^2-12p-24=-2p+1$

Simplifying yields:

$8p^2-10p-25=0$

Solve for p. We can factor:

$8p^2+10p-20p-25=0$

Factor:

$2p(4p+5)-5(4p+5)=0$

Grouping:

$(2p-5)(4p+5)=0$

Zero Product Property:

$\displaystyle p_1=\frac{5}{2} \text{ or } p_2=-\frac{5}{4}$

Then, we can use the second equation to solve for d. So:

$2d_1=-2p_1+1$

Substituting:

$\begin{aligned} 2d_1&=-2(\frac{5}{2})+1 \\ 2d_1&=-5+1 \\ 2d_1&=-4 \\ d_1&=-2\end{aligned}$

So, for the first case, p is 5/2 and d is -2.

Likewise, for the second case:

$\begin{aligned} 2d_2&=-2(-\frac{5}{4})+1 \\ 2d_2&=\frac{5}{2}+1 \\ 2d_2&=\frac{7}{2} \\ d_2&=\frac{7}{4}\end{aligned}$

So, for the second case, p is -5/4, and d is 7/4.

By using the values, we can determine our series.

For Case 1, we will have:

17, 15, 13.

For Case 2, we will have:

-11/2, -15/4, -2.

8 0

1 year ago