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GrogVix [38]
3 years ago
15

Which of the following are valid names for the triangle below? Check all that apply.

Mathematics
1 answer:
Sholpan [36]3 years ago
3 0

Answer:

the answer i think is DMR

Step-by-step explanation:

because u use the letters from the corners

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What is the measure of JM?
Ray Of Light [21]

Answer:

10???

Step-by-step explanation:

8 0
3 years ago
What is 1/4x+1+3/4x-2/3-1/2x
Aleks04 [339]

Answer: 1/2x + 1/3

Step-by-step explanation:

Given:

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

Step 1: Combine like terms

1/4(x) and 3/4(x) have a common denominator of 4.  This means that you can add them together.

1/4(x) + 3/4(x) = 4/4(x) = x

Step 2: Find the common denominator of x in step 1 and combine like terms  

x - 1/2(x) = 2/2(x) - 1/2(x)

Now that we have the common denominator of x, we can combine like terms.  Its the same as adding or subtracting fractions without a variable.  In this case, you must subtract 1/2(x) from 2/2(x).

2/2(x) - 1/2(x) = 1/2(x)

Step 3: Find the common denominator of the constants and combine like terms

1 - 2/3 = 3/3 - 2/3

Now combine like terms.  Simply subtract 2/3 from 3/3.

3/3 - 2/3 = 1/3

Step 4: Write the simplified equation

1/2(x) + 1/3

This is the answer

4 0
3 years ago
Orange M&M’s: The M&M’s web site says that 20% of milk chocolate M&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
What does he mean on question one
Katarina [22]
You plug -2 in for the function k(p) and add it to the function g(w), getting 
(-2+3)*(-2-7)+(-2-5)^2=1*-9+49=40 for a - I challenge you to do B on your own!
6 0
3 years ago
If a 11 inch wheel cost $21.35 what is the cost per square inch
Natali [406]

Answer:

$1.94 per square inch

8 0
3 years ago
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