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konstantin123 [22]
3 years ago
8

Expression is a trinonmial

Mathematics
2 answers:
Leno4ka [110]3 years ago
6 0

Answer:

A: The expression is a trinomial with a degree of 4.

Step-by-step explanation:

The expression has three parts, so it's a tronomial.

The largest exponent is 4, so it has a degree of 4.  

Karo-lina-s [1.5K]3 years ago
5 0

Answer:

Yea he is right it is a

Step-by-step explanation:

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Analyze the diagram below and complete the instructions that follow.
Jlenok [28]

Step-by-step explanation:

scale \: factor =  \frac{7}{28}  =  \frac{1}{4}  \\

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3 0
3 years ago
13 13 13 13 = 43. Which multiplication expression shows another way to make 43?
nikdorinn [45]
13+13+13+13=52, not 43 .
<span>43 is a prime number therefore the only two factors are 1 and itself, Technically, you can say that 21.5 multiplied by 2 would be 43.</span>
6 0
4 years ago
Simplify.<br> -3(w+ 3) + 4w
densk [106]

Hello! :)

To solve you equation, we will first Distribute

(-3) (w) + (-3) (3) + 4w

= -3w + -9 +4w

Now, we will combine like terms

-3w + -9 + 4w

= (-3w + 4w) + (-9)

= w + 9

Hope this helped you!

THEDIPER

4 0
4 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
4 years ago
What are the coordinates of point S?
mote1985 [20]

Answer:

I think you forgot to attach the notes/picture

6 0
4 years ago
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