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2 years ago

9

Assume this is a positive ordered set of data which statistics show. values that would most likely describe the following set of

data? X,9,11,13,y,20

1 answer:

andrew11 [14]2 years ago

7 0

I think it is 20 because you have to multiply and then divide them add 11 then subtract 9 idk what I am saying

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Solve the system of linear equations y= 2x -3Y= x + 6

Alja [10]

Answer:

x=9, y=15. (9, 15).

Step-by-step explanation:

y=2x-3

y=x+6

----------

2x-3=x+6

2x-x-3=6

x-3=6

x=6+3

x=9

y=9+6=15

7 0

2 years ago

PLEASE HELP ILL GIVE BRAINLIEST

dusya [7]

1)

$(-2+\sqrt{-5})^2\implies (-2+\sqrt{-1\cdot 5})^2\implies (-2+\sqrt{-1}\sqrt{5})^2\implies (-2+i\sqrt{5})^2 \\\\\\ (-2+i\sqrt{5})(-2+i\sqrt{5})\implies +4-2i\sqrt{5}-2i\sqrt{5}+(i\sqrt{5})^2 \\\\\\ 4-4i\sqrt{5}+[i^2(\sqrt{5})^2]\implies 4-4i\sqrt{5}+[-1\cdot 5] \\\\\\ 4-4i\sqrt{5}-5\implies -1-4i\sqrt{5}$

3)

let's recall that the conjugate of any pair a + b is simply the same pair with a different sign, namely a - b and the reverse is also true, let's also recall that i² = -1.

$\cfrac{6-7i}{1-2i}\implies \stackrel{\textit{multiplying both sides by the denominator's conjugate}}{\cfrac{6-7i}{1-2i}\cdot \cfrac{1+2i}{1+2i}\implies \cfrac{(6-7i)(1+2i)}{\underset{\textit{difference of squares}}{(1-2i)(1+2i)}}} \\\\\\ \cfrac{(6-7i)(1+2i)}{1^2-(2i)^2}\implies \cfrac{6-12i-7i-14i^2}{1-(2^2i^2)}\implies \cfrac{6-19i-14(-1)}{1-[4(-1)]} \\\\\\ \cfrac{6-19i+14}{1-(-4)}\implies \cfrac{20-19i}{1+4}\implies \cfrac{20-19i}{5}\implies \cfrac{20}{5}-\cfrac{19i}{5}\implies 4-\cfrac{19i}{5}$

7 0

2 years ago

Read 2 more answers

Answer Is not C<br>please help

Katyanochek1 [597]

Answer:

B

Step-by-step explanation:

Well since it is arthimetic sequence we know the formula

We know d is difference between second term and first so it will be -5- (-12)= -5+12= 7

$a_{n}= a _{1}+ d (n-1) \\a_{n}= -12+7(n-1)\\a_{n} = -12+7n-7\\a_{n}= 7n-19$

the domain of this is 1,2,3,4.......

range is -12,-5,2,9.......

Therefore answer is B.

5 0

3 years ago

Laura sees a horse pulling a buggy. She wonders how it can accelerate if the action of the horse pulling the cart would causes a

KatRina [158]

Answer:

The net forces exerted on the horse and cart are not the same, so they are not balanced forces.

Step-by-step explanation:

Please see the Newton's 2nd Law which states that an object accelerates if there is a net or unbalanced force on it. In this scenario there is just one force exerted on the wagon i.e: the force that the horse exerts on it. The wagon accelerates because the horse pulls on it. And the amount of acceleration equals the net force on the wagon divided by its mass.

As there are two forces the push and pull the horse; the wagon pulls the horse backwards, and the ground pushes the horse forward. The net force is determined by the relative sizes of these two forces.

If the ground pushes harder on the horse than the wagon pulls, there is a net force in the forward direction, and the horse accelerates forward, and if the wagon pulls harder on the horse than the ground pushes, there is a net force in the backward direction, and the horse accelerates backward.

If the force that the wagon exerts on the horse is the same size as the force that the ground exerts, the net force on the horse is zero, and the horse does not accelerate.

7 0

2 years ago

Read 2 more answers

HELP I NEED THIS DONE QUICKK CAN ANNYONE HELP???????????????????????????????????????????????????????????????????????????????????

natulia [17]

For the first question, it would be D: $\frac{1}{3}$ ÷ 9. This is because to solve a division problem with fractions, you would need to multiply the first fraction by the reciprocal of the second fraction. For the second question, the answer would be D: 32. This can be solved by dividing 8 by $\frac{1}{4}$ or 0.25. For the third question, it would be D. $\frac{1}{12}$ . This can be solved by dividing $\frac{1}{4}$ by 3. For the fourth question, the answer would be $\frac{1}{12}$ . This can be found by flipping the 4, which would make it look like $\frac{1}{4}$ , and multiplying it by $\frac{1}{3}$ . For the last question, 1 would be B, 2 would be C, 3 would be C, and 4 would be A.

Hope this helps ^-^

6 0

2 years ago