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Stells [14]
3 years ago
8

How many solutions does this linear equation system have y=2/3x+2

Mathematics
1 answer:
Neko [114]3 years ago
3 0

Answer:

infinite many solutions

Step-by-step explanation:

You might be interested in
The first term of an arithmetic sequence is -3 and the fifteenth term is 53 what is the common difference of the sequence
saul85 [17]

Solution

The 1st term = -3

15th term = 53

15th term = a + d(n-1)

53 = -3 + d(15- 1)                       [Here n = 15)

53 + 3 = d(14)

14d = 56

d = 4

The common difference is 4.

7 0
4 years ago
If two different people are randomly selected from the 864 subjects, find the probability that they are both heavy smokers. Roun
AlekseyPX

Answer:

P = 0.008908

Step-by-step explanation:

The complete question is:

The table below describes the smoking habits of a group of asthma sufferers

               Nonsmokers      Light Smoker     Heavy smoker      Total

Men                        303                     35                           37        375

Women                   413                     31                            45        489

Total                        716                     66                           82        864

If two different people are randomly selected from the 864 subjects, find the probability that they are both heavy smokers.

The number of ways in which we can select x subjects from a group of n subject is given by the combination and it is calculated as:

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

Now, there are 82C2 ways to select subjects that are both heavy smokers. Because we are going to select 2 subjects from a group of 82 heavy smokers. So, it is calculated as:

82C2=\frac{82!}{2!(82-2)!}=3321

At the same way, there are 864C2 ways to select 2 different people from the 864 subjects. It is equal to:

864C2=\frac{864!}{2!(864-2)!}=372816

Then, the probability P that two different people from the 864 subjects are both heavy smokers is:

P=\frac{82C2}{864C2}=\frac{3321}{372816}=0.008908

4 0
3 years ago
SAT scores (out of 1600) are distributed normally with a mean of 1100 and a standard deviation of 200. Suppose a school council
fgiga [73]

Answer:

0.91517

Step-by-step explanation:

Given that SAT scores (out of 1600) are distributed normally with a mean of 1100 and a standard deviation of 200. Suppose a school council awards a certificate of excellence to all students who score at least 1350 on the SAT, and suppose we pick one of the recognized students at random.

Let A - the event passing in SAT with atleast 1500

B - getting award i.e getting atleast 1350

Required probability = P(B/A)

= P(X>1500)/P(X>1350)

X is N (1100, 200)

Corresponding Z score = \frac{x-1100}{200}

P(X>1500)/P(X>1350)\\= \frac{P(Z>2)}{P(Z>1.25} \\=\frac{0.89435}{0.97725} \\=0.91517

4 0
3 years ago
How many zero's would 10 to the power of 3 have?
Tom [10]
I’m pretty sure it’s 3 cause 10 to the power of 3 is 1000
7 0
3 years ago
Read 2 more answers
4 (3x + 2)-6<br> I neeeed help
yuradex [85]

Answer:

12x - 2

Step-by-step explanation:

Step 1: Write expression

4(3x + 2) - 6

Step 2: Distribute 4

12x + 8 - 6

Step 3: Combine like terms

12x - 2

5 0
3 years ago
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