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Vikki [24]
3 years ago
14

Insert the parenthesis to make the equality right: 30-49/42/6*8=184

Mathematics
1 answer:
Alika [10]3 years ago
5 0
(6*8) is where the parentheses should go
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Jean and Jericho who are playing in the school grounds decided to sit on a
Natalija [7]

We want to see underline the correct part in each statement.

  • 1) This situation illustrates (direct, <u>inverse</u>) variation.
  • 2) The two quantities that must vary in this situation are (<u>weight and </u>
  • <u>distance from the cente</u><u>r</u>, height and weight).
  • 3) The heavier the kid, the (<u>closer</u>, farther) he/she should be at the
  • center to balance the seesaw?
  • 4) When Jean moves farther from the center, Jericho tends to go (up,
  • <u>down</u>).
  • 5) If Jericho moves closer to the center, Jean tends to go (up, <u>down</u>).

So, Jean and Jericho are playing on a seesaw.

Jean is heavier than Jericho.

Now, notice that a seesaw is a lever. So it "amplificates" the force that you apply in one end to lift the weight that is on the other end. Depending on the form of the lever and the weights, the force that you need to do changes.

If we define:

  • W₁ = Jean's weight.
  • d₁ = Jean's distance to the center.
  • W₂ = Jericho's weight.
  • d₂ = Jericho's distance to the center.

We must have:

W₁*d₁ = W₂*d₂

Then:

1) This situation illustrates (direct, <u>inverse</u>) variation.

d₁, the position of Jean, varies inversely with respect to Jean's weight.

2) The two quantities that must vary in this situation are (<u>weight and </u>

<u>distance from the center</u>, height and weight).

(by the equation above)

3) The heavier the kid, the (<u>closer</u>, farther) he/she should be at the

center to balance the seesaw?

By the given equation, we see that d₁ must be smaller than d₂.

And the last two are trivial:

4) When Jean moves farther from the center, Jericho tends to go (up,

<u>down</u>).

5) If Jericho moves closer to the center, Jean tends to go (up, <u>down</u>).

If you want to learn more, you can read:

brainly.com/question/18320907

7 0
2 years ago
ABC is an isosceles<br> AB= 3x-4 and BC= 5x-10<br><br> what is AB?
kakasveta [241]
3x-4=5x-10
Subtract 3x from both sides
-4=2x-10
Add 10 to both sides
6= 2x
X= 3
3 times 3-4
AB=5
3 0
3 years ago
Read 2 more answers
Rewrite the equation in Ax+By=C form <br><br> use integers for A,B, and C. <br><br> Y+5=2(x+4)
Contact [7]

Answer:

2x4y=2... hope this helps.

8 0
3 years ago
Which equation represents a line that passes through (–9, –3) and has a slope of –6?
brilliants [131]
\bf \begin{array}{lllll}&#10;&x_1&y_1\\&#10;%   (a,b)&#10;&({{ -9}}\quad ,&{{ -3}})&#10;\end{array}&#10;\\\\\\&#10;% slope  = m&#10;slope = {{ m}}= \cfrac{rise}{run} \implies -6&#10;\\\\\\&#10;% point-slope intercept&#10;\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-(-3)=-6[x-(-9)]&#10;\\\\\\&#10;y+3=-6(x+9)\implies y+3=-6x-54\implies y=-6x-57
8 0
3 years ago
An insurance company examines its pool of auto insurance customers and gathers the following information: (i) All customers insu
ankoles [38]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

An insurance company examines its pool of auto insurance customers and gathers the following information: (i) All customers insure at least one car. (ii) 70% of the customers insure more than one car. (iii) 20% of the customers insure a sports car. (iv) Of those customers who insure more than one car, 15% insure a sports car. Calculate the probability that a randomly selected customer insures exactly one car, and that car is not a sports car?

Answer:

P( X' ∩ Y' ) = 0.205

Step-by-step explanation:

Let X is the event that the customer insures more than one car.

Let X' is the event that the customer insures exactly one car.

Let Y is the event that customer insures a sport car.

Let Y' is the event that customer insures not a sport car.

From the given information we have

70% of customers insure more than one car.

P(X) = 0.70

20% of customers insure a sports car.

P(Y) = 0.20

Of those customers who insure more than one car, 15% insure a sports car.

P(Y | X) = 0.15

We want to find out the probability that a randomly selected customer insures exactly one car, and that car is not a sports car.

P( X' ∩ Y' ) = ?

Which can be found by

P( X' ∩ Y' ) = 1 - P( X ∪ Y )

From the rules of probability we know that,

P( X ∪ Y ) = P(X) + P(Y) - P( X ∩ Y )    (Additive Law)

First, we have to find out P( X ∩ Y )

From the rules of probability we know that,

P( X ∩ Y ) = P(Y | X) × P(X)       (Multiplicative law)

P( X ∩ Y ) = 0.15 × 0.70

P( X ∩ Y ) = 0.105

So,

P( X ∪ Y ) = P(X) + P(Y) - P( X ∩ Y )

P( X ∪ Y ) = 0.70 + 0.20 - 0.105

P( X ∪ Y ) = 0.795

Finally,

P( X' ∩ Y' ) = 1 - P( X ∪ Y )

P( X' ∩ Y' ) = 1 - 0.795

P( X' ∩ Y' ) = 0.205

Therefore, there is 0.205 probability that a randomly selected customer insures exactly one car, and that car is not a sports car.

6 0
3 years ago
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