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PtichkaEL [24]
3 years ago
15

PLZ HELP ON SCALE FACTOR

Mathematics
2 answers:
balu736 [363]3 years ago
7 0

56 : 7 = x : 8

x = (56 * 8) : 7

x = 448 : 7

x = 64

RS is 64

docker41 [41]3 years ago
3 0

Your answer would be C-64.

Hope this helps.

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i think

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A nurse has an income of $120,000. The income tax the nurse has to pay is 6%. What is the amount of income tax, in dollars and c
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F(x) = x2 − 8x + 15 g(x) = x − 3 h(x) = f(x) ÷g(x) The domain of h(x) is
Sphinxa [80]

Answer:

{x | x ≠ 3}

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f(x) / g(x) = x^2 - 8x + 15 / (x - 3)

= (x - 5)(x - 3) / (x - 3) = h (x)

The domain is {x | x ≠ 3)

(because h(x) has no value when x = 3:-the denominator  x-3 becomes zero when x = 3)


8 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
3 years ago
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