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TiliK225 [7]
2 years ago
6

Simplify. (8 + 14m) + m

Mathematics
2 answers:
Neko [114]2 years ago
5 0
8 + 14m + 8
Add 8 and 8
So 16
16 + 14m would be your answer
Ronch [10]2 years ago
4 0
8 + 14m + m
=8 + 15m
adding number with common variable.
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A curious student asks his math teacher: "I heard you have three daughters. How old are they?" The math teacher says: "Well, the
scZoUnD [109]

Answer:

There are 7 solutions.

The possible answers are:

The oldest daughter is 72 and the others are 1 and 1 and the room number is 74.

The oldest daughter is 36 and the others are 1 and 2 and the room number is 39.

The oldest daughter is 24 and the others are 1 and 3 and the room number is 28.

The oldest daughter is 18 and the others are 1 and 4 and the room number is 23.

The oldest daughter is 12 and the others are 2 and 3 and the room number is 17.

The oldest daughter is 12 and the others are 1 and 6 and the room number is 19.

The oldest daughter is 9 and the others are 1 and 8 and the room number is 18.

The first few solutions are laughable and unlikely, but all are possible.  

OR..

The clue that the sums of the daughters' ages is the room number is the sleeper.  The student's first response is that two possible age combinations meet the necessary conditions.  The teacher's reference to the "oldest daughter" resolves the ambiguity, so one of the solutions must consist of a set of ages in which at least two of the oldest daughters are the same age.

The prime factors of 72 are 2^3 *3^2.  Because one of the solutions requires the two oldest daughters to be the same age, only combination 6, 6, 2 would be admissible.  There are insufficient factors for any other combination in which the duplicate ages represent the oldest daughter.  The ages of these daughters sum to 14, which is the room number.  The other combination must therefore add to 14 and have an unique oldest age.  The combination 8, 3, 3 meets all conditions and is the answer.

4 0
2 years ago
Malik randomly picked two numbers from 1 to 9 (Includin 1 and 9). the same number could be picked more than once. The first of t
Tju [1.3M]

Answer:

2/9 = 0.22

Step-by-step explanation:

There are two ways to pick the first number odd and less than 5:  1 and 3.

With each of these, the second number drawn can be 1, 2, 3, 4, 5, 6, 7, 8 or 9.

This makes 18 total possibilities.

Out of these, the only ways to have a sum less than 5 are 1 and 1, 1  and 2, 1 and 3, and 3 and 1.  This is 4 ways out of 18:

4/18 = 2/9 = 0.22

8 0
2 years ago
Paul needs to find 310% of 72. Which expression should he use
Nataliya [291]

Answer:

3.10 * 72

Step-by-step explanation:

To find 310% of 72

Of means multiply and is means equal

310% * 72 = answer

Change to decimal form

3.10 * 72 = answer

223.20

3 0
2 years ago
In the diagram, AB←→ || CD←→. Which pair of angles can be proven congruent?
slega [8]
For the answer to the question above, the opposite angles at the intersection of two straight lines as shown in the graph are congruent. 
So the answer is simply∡ EIA  and ∡ KIJ are congruent.

I hope my answer helped you. Have a nice day!
7 0
2 years ago
Read 2 more answers
A supermarket has two customers waiting to pay for their purchases at counter I and one customer waiting to pay at counter II. L
Pachacha [2.7K]

Answer:

b. 0.864

Step-by-step explanation:

Let's start defining the random variables.

Y1 : ''Number of customers who spend more than $50 on groceries at counter 1''

Y2 : ''Number of customers who spend more than $50 on groceries at counter 2''

If X is a binomial random variable, the probability function for X is :

P(X=x)=(nCx)p^{x}(1-p)^{n-x}

Where P(X=x) is the probability of the random variable X to assume the value x

nCx is the combinatorial number define as :

nCx=\frac{n!}{x!(n-x)!}

n is the number of independent Bernoulli experiments taking place

And p is the success probability.

In counter I :

Y1 ~ Bi (n,p)

Y1 ~ Bi(2,0.2)

P(Y1=y1)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}

With y1 ∈ {0,1,2}

And P( Y1 = y1 ) = 0 with y1 ∉ {0,1,2}

In counter II :

Y2 ~ Bi (n,p)

Y2 ~ Bi (1,0.3)

P(Y2=y2)=(1Cy2)(0.3)^{y2}(0.7)^{1-y2}

With y2 ∈ {0,1}

And P( Y2 = y2 ) = 0 with y2 ∉ {0,1}

(1Cy2) with y2 = 0 and y2 = 1 is equal to 1 so the probability function for Y2 is :

P(Y2=y2)=(0.3)^{y2}(0.7)^{1-y2}

Y1 and Y2 are independent so the joint probability distribution is the product of the Y1 probability function and the Y2 probability function.

P(Y1=y1,Y2=y2)=P(Y1=y1).P(Y2=y2)

P(Y1=y1,Y2=y2)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}(0.3)^{y2}(0.7)^{1-y2}

With y1 ∈ {0,1,2} and y2 ∈ {0,1}

P( Y1 = y1 , Y2 = y2) = 0 when y1 ∉ {0,1,2} or y2 ∉ {0,1}

b. Not more than one of three customers will spend more than $50 can mathematically be expressed as :

Y1 + Y2 \leq 1

Y1 + Y2\leq 1 when Y1 = 0 and Y2 = 0 , when Y1 = 1 and Y2 = 0 and finally when Y1 = 0 and Y2 = 1

To calculate P(Y1+Y2\leq 1) we must sume all the probabilities that satisfy the equation :

P(Y1+Y2\leq 1)=P(Y1=0,Y2=0)+P(Y1=1,Y2=0)+P(Y1=0,Y2=1)

P(Y1=0,Y2=0)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{0}(0.7)^{1-0}=(0.8)^{2}(0.7)=0.448

P(Y1=1,Y2=0)=(2C1)(0.2)^{1}(0.8)^{2-1}(0.3)^{0}(0.7)^{1-0}=2(0.2)(0.8)(0.7)=0.224

P(Y1=0,Y2=1)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{1}(0.7)^{1-1}=(0.8)^{2}(0.3)=0.192

P(Y1+Y2\leq 1)=0.448+0.224+0.192=0.864\\P(Y1+Y2\leq 1)=0.864

7 0
2 years ago
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