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sertanlavr [38]
3 years ago
6

Have students write the answer to the essential question and draw examples to explain their answer. Essential Question is how ca

n you use models to find factors?

Mathematics
1 answer:
OLga [1]3 years ago
7 0
Objective:

Understand what an essential question is & is not

Write one effectively

Definition: An essential question lies at the heart of a subject & promotes the asking of more questions to uncover more about the subject.

Example:

Modeling factors of Numbers:

A factor is a number multiplied by another number that gives a product.

every whole number greater than 1 has at least two factors, 1 and itself.


Use 20 tiles to make as many different arrays as you can.

factors are 4*5=20

2*10=20

1*20=20

factors of 20: 1,2,4,5,10 and 20.



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Number 3 I need some help
Nadusha1986 [10]

Answer:

15+10x>_75

Step-by-step explanation:

so last one

6 0
3 years ago
A researcher is exploring the notion that there is an economy of scale in raising children. While having one child might add to
trasher [3.6K]

Answer:

The probability is   G  =  85%

Step-by-step explanation:

From the question we are told that

   The percentage of parents that have one child under 18 in their homes is  k = 15%

  The percentage of parents that have two children is  T = 65%

    The percentage of parents that have three or more children is  Y = 20%

Generally the probability that any given family who is selected have two or more children at home is mathematically evaluated as

     G  =  T  +  Y

=>  G  =  65% +  20%

=>  G  =  85%

4 0
3 years ago
what positive number satisfies the condition that twice the number minus three times the reciprocal of the number is equal to 1?
marshall27 [118]
Solve:

"<span>twice the number minus three times the reciprocal of the number is equal to 1."
                                                     3(1)
Let the number be n.  Then 2n - ------- = 1
                                                       n

Mult all 3 terms by n to elim. the fractions:

2n^2 - 3 = n.  Rearranging this, we get 2n^2 - n - 3 = 0.

We need to find the roots (zeros or solutions) of this quadratic equation.

Here a=2, b= -1 and c= -3.  Let's find the discriminant b^2-4ac first:

disc. = (-1)^2 - 4(2)(-3) = 1 + 24 = 25.

That's good, because 25 is a perfect square.
                -(-1) plus or minus 5         1 plus or minus 5
Then x = ------------------------------ = --------------------------
                            2(2)                                  4

x could be 6/4 = 3/2, or -5/4.

You must check both answers in the original equation.  If the equation is true for one or the other or for both, then you have found one or more solutions.</span>
8 0
3 years ago
Simplify Negative 15.6 divided by negative 4.
masha68 [24]
Simplify:

-15.6/-4

( \frac{-15.6}{-4} )

Divide to solve:

You get 3.9

3 0
3 years ago
An alarming number of U.S. adults are either overweight or obese. The distinction between overweight and obese is made on the ba
madreJ [45]

Answer:

(A) The probability that a randomly selected adult is either overweight or obese is 0.688.

(B) The probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C) The events "overweight" and "obese" exhaustive.

(D) The events "overweight" and "obese" mutually exclusive.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = a person is overweight

<em>Y</em> = a person is obese.

The information provided is:

A person is overweight if they have BMI 25 or more but below 30.

A person is obese if they have BMI 30 or more.

P (X) = 0.331

P (Y) = 0.357

(A)

The events of a person being overweight or obese cannot occur together.

Since if a person is overweight they have (25 ≤ BMI < 30) and if they are obese they have BMI ≥ 30.

So, P (X ∩ Y) = 0.

Compute the probability that a randomly selected adult is either overweight or obese as follows:

P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\=0.331+0.357-0\\=0.688

Thus, the probability that a randomly selected adult is either overweight or obese is 0.688.

(B)

Commute the probability that a randomly selected adult is neither overweight nor obese as follows:

P(X^{c}\cup Y^{c})=1-P(X\cup Y)\\=1-0.688\\=0.312

Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C)

If two events cannot occur together, but they form a sample space when combined are known as exhaustive events.

For example, flip of coin. On a flip of a coin, the flip turns as either Heads or Tails but never both. But together the event of getting a Heads and Tails form a sample space of a single flip of a coin.

In this case also, together the event of a person being overweight or obese forms a sample space of people who are heavier in general.

Thus, the events "overweight" and "obese" exhaustive.

(D)

Mutually exclusive events are those events that cannot occur at the same time.

The events of a person being overweight and obese are mutually exclusive.

5 0
2 years ago
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