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Gnom [1K]
3 years ago
13

How do you write algebraic expressions to model quantities

Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
8 0
To write algebraic expressions<span> to model quantities is you base the algebraic expressions or mathematical values declared in words by the manner of how the sentence illustrates the said problem. For example, one minus a number is the difference of zero shall be </span>
<span>1. 1 – n = 0. </span>

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What is 6 1/4 written as a decimal? Please help me I am very confused.
MrRa [10]

Answer: 6.2

Step-by-step explanation:

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2 years ago
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Consider the squence following a minus 8 pattern 9,1,-7,-15 write an explict formula for the squence
Andrew [12]
A(x) = 17 - 8x. Hope this helps!
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3 years ago
He logical statement represents the inverse of the conditional statement “if you are human, then you were born on earth.”
Artyom0805 [142]

The conditional statement “if you are human, then you were born on earth.”

The logical statement is the inverse of the conditional statement.

So, the logical statement is If you are born on earth then you are a human.

Now we can say that our conditional statement is absolutely true in all condition

 which states If i am a human⇒  I am born on earth

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The answer can be depicted by venn diagram also.


8 0
3 years ago
Bad gums may mean a bad heart. Researchers discovered that 79% of people who have suffered a heart attack had periodontal diseas
zysi [14]

Answer:

(A) 0.297

(B) 0.595

Step-by-step explanation:

Let,

H = a person who suffered from a heart attack

G = a person has the periodontal disease.

Given:

P (G|H) = 0.79, P(G|H') = 0.33 and P (H) = 0.15

Compute the probability that a person has the periodontal disease as follows:

P(G)=P(G|H)P(H)+P(G|H')P(H')\\=P(G|H)P(H)+P(G|H')(1-P(H))\\=(0.79\times0.15)+(0.33\times(1-0.15))\\=0.399

(A)

The probability that a person had periodontal disease, what is the probability that he or she will have a heart attack is:

P(H|G)=\frac{P(G|H)P(H)}{P(G)}\\=\frac{0.79\times0.15}{0.399} \\=0.29699\\\approx0.297

Thus, the probability that a person had periodontal disease, what is the probability that he or she will have a heart attack is 0.297.

(B)

Now if the probability of a person having a heart attack is, P (H) = 0.38.

Compute the probability that a person has the periodontal disease as follows:

P(G)=P(G|H)P(H)+P(G|H')P(H')\\=P(G|H)P(H)+P(G|H')(1-P(H))\\=(0.79\times0.38)+(0.33\times(1-0.38))\\=0.5048

Compute the probability of a person having a heart attack given that he or she has the disease:

P(H|G)=\frac{P(G|H)P(H)}{P(G)}\\=\frac{0.79\times0.38}{0.5048}\\ =0.59469\\\approx0.595

The probability of a person having a heart attack given that he or she has the disease is 0.595.

4 0
3 years ago
Pls answer this for me, this is the remainder theorem for grade 9 so help me.
professor190 [17]

Answer:

1.  3

2. 2

Step-by-step explanation:

| |  

x | + | 1 | | x^2 | + | 4 x | - | 2

x^3 | + | 5 x^2 | + | 2 x | + | 1

x^3 | + | x^2 | | | |  

| | 4 x^2 | + | 2 x | |  

| | 4 x^2 | + | 4 x | |  

| | | | -2 x | + | 1

| | | | -2 x | - | 2

| | | | | | 3

__________________________________________

| |  

x | - | 5 | | x^2 | - | x | + | 0

x^3 | - | 6 x^2 | + | 5 x | + | 2

x^3 | - | 5 x^2 | | | |  

| | -x^2 | + | 5 x | |  

| | -x^2 | + | 5 x | |  

| | | | | | 2

| | | | | | 0

| | | | | | 2

5 0
3 years ago
Read 2 more answers
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