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Sati [7]
3 years ago
15

3. Ricky buys some packs of gum (x) that cost $1.50 and some jawbreakers (y) that cost $ 50.

Mathematics
1 answer:
insens350 [35]3 years ago
5 0
Gfsretunkjkiuyretuiookkbgsad
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Which statement correctly compares the four decimals by listing them in order from least to greatest?
OverLord2011 [107]
C because it goes greater, so 0.1 is less then 03 which is greater then 01 but less the 0.6
8 0
3 years ago
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The store sells walnuts for $2.95 per pound. Corey bought 2.4 pounds of walnuts. Before tax, how much will the walnuts cost?
stepladder [879]

Answer:

$7.18

Step-by-step explanation:

Multiply the number of pounds bought by the cost of one pound.

 (2.4 pounds)($2.99 per pound) = $7.176 ≅ $7.18

7 0
2 years ago
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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 are the possible values of x in 32x + 20 = 28x − 16x2?
klemol [59]
<span>No two such factors can be found

</span>
5 0
3 years ago
How many unit tiles need to added to the expression x^2+4x+3 to form the perfect square trinomial?
Nuetrik [128]

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

So 1 need to be added.

7 0
3 years ago
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