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creativ13 [48]
2 years ago
6

What is the equivalent to 3/20 as a fraction

Mathematics
2 answers:
Ludmilka [50]2 years ago
7 0
It could be 6/40, because you multiply the numerator and denominator by 2
aalyn [17]2 years ago
5 0
6/40 is equalvilant because all you have to do is double or simplify it by 2
You might be interested in
Solve 64=V, where v is a real number. Simplify your answer as much as possible. If there is more than one solution, separate the
EastWind [94]

Answer:

no

Step-by-step explanation:

3 0
2 years ago
What is the surface area of the given figure?
morpeh [17]

Answer:

D. 2,280 cm²

Step-by-step explanation:

Surface area of triangular prism = bh + (S1 + S2 + S3)*L

Where,

b = 24 cm

h = 10 cm

S1 = 26 cm

S2 = 24 cm

S3 = 10 cm

L = 34 cm

Plug in the values

Surface area = 24*10 + (26 + 24 + 10)*34

= 240 + 2,040

= 2,280 cm²

3 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
Evaluate 4(-9) i need help lol
PIT_PIT [208]

Answer:

-36

Step-by-step explanation:

How are you in college lol

8 0
3 years ago
Read 2 more answers
(2 5/6)(6)+(-1 1/2)(1.5)-(1/16)
Jet001 [13]
Now I’m ngl my math might be wrong but I ended up with 1131/48
As a decimal I got 23.6
4 0
3 years ago
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