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seropon [69]
3 years ago
5

What is the equation of the line

Mathematics
1 answer:
ohaa [14]3 years ago
6 0
The equation of the line is y = (1/2)x + 2
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Please Help! (no rush)
Anon25 [30]

Answer:

The answer to your question is below

Step-by-step explanation:

Formula  an expression that uses variables to state a rule

Constant   a number; a term containing no variables

Variable  a letter or symbol used to represent an unknown number

Term   a number or a variable, or the product of a number and variable(s)

Expression   one term or multiple terms connected by an addition or subtraction sign

3 0
3 years ago
Read 2 more answers
Y (y + 2) = y2 - 6<br> also<br> 2 [x - (1-3x)] = 3 (x + 1)
vladimir1956 [14]

Answer:

Step-by-step explanation:

y(y+2)=y²-6

y²+2y=y²-6

2y=-6

y=-3

2[x-(1-3x)]=3(x+1)

2[x-1+3x]=3x+3

2(4x-1)=3x+3

8x-2=3x+3

8x-3x=3+2

5x=5

x=1

7 0
2 years ago
What is the mean,mode,median,and range of 5,5,5,6,6,6,7,7,7,7,7​
g100num [7]

Answer:

median=6, range=2 mean=61 mode=7

the median is the middle which is 6

The range is the lowest-biggest number which would be 2

mean=average

mode is the one that appears most

7 0
3 years ago
Read 2 more answers
Solve for x: x/3 = -48
oksano4ka [1.4K]

Answer:

-144

Step-by-step explanation:

x=-48 times 3

8 0
2 years ago
Read 2 more answers
Orange M&amp;M’s: The M&amp;M’s web site says that 20% of milk chocolate M&amp;M’s are orange. Let’s assume this is true and set
SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let <em>X</em> = number of orange  milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.

The expected value of a Binomial random variable is:

E(X)=np

Compute the expected number of orange  milk chocolate M&M’s in a bag of 55 candies as follows:

E(X)=np

         =55\times 0.20\\=11

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}

                 =0.1968

The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.

This probability is quite larger than 0.05.

Thus, the correct option is (A).

4 0
3 years ago
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