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lions [1.4K]
3 years ago
8

3. Find the equation of the line parallel to taxis

Mathematics
1 answer:
Gre4nikov [31]3 years ago
6 0

Answer:

here the answer you need

(i)passing through the point (3 ,-4)

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I need a answer right now I’ll give you points!
Kipish [7]

Answer:

<h2>G. 118⁰</h2>

100% right answer

8 0
3 years ago
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Help Would Be Greatly Appreciated :)
aliina [53]

Answer:

n is y=\frac{1}{2}x+b, p is y=\frac{1}{2}x-5, and r has infinitely many different lines that have this property.

Step-by-step explanation:

If two lines have no solutions, that means that they are parallel. The slopes of two lines are the same if they are parallel. So n's slope is \frac{1}{2}. The y-intercept can be anything except -5, or +b. If two lines have infinitely many solutions, those two lines are exactly the same. So p is just m. If two lines have one and only one solution, that just means the slopes and y-intercepts are different. Since we don't know anything else about line r, we an't model an equation for it.

4 0
3 years ago
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
Help me please... I have a lot more things to do and I wanna get this done.
lianna [129]

Answer:

-6

Step-by-step explanation:

count

7 0
3 years ago
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Solve for the value of z: 90/15 = 6/z
lara31 [8.8K]

Answer:

z = 1

Step-by-step explanation:

90/15 = 6/z

Cross Multiply

15 * 6 = 90 * z

90 = 90z

Divide both sides of the equation by 90

z = 1

Hope this was useful to you!

3 0
3 years ago
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