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Lilit [14]
2 years ago
6

I really need major help .

Mathematics
1 answer:
frozen [14]2 years ago
6 0

Answer:

Step-by-step explanation:

A)

exponential ,  b/c it curves upwards

B)

a linear graph with have a constant number of growth, and an exponential, needs that number to change, so that's why the exponent is required.

C)

Maybe a temperature function that could follow the outdoor temp over a few hour period , say in the morning.. as the temp goes up in a linear fashion.

maybe the temp at 8 am would be 50  then at 9 it would be 55 and at 10 it would be 60 and at 11 it could be 65  , something like that seems like it would be a good working example of a linear function that was not chosen

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2. Sam can make 15 pizzas in 2.5 hours. At this
Verizon [17]

Answer:

48

Step-by-step explanation:

You can do it with cross multiplication

6 0
2 years ago
Read 2 more answers
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 the function f(p) = p^2 + 3p + 1 for p =- 2.
Anon25 [30]
(-2)^2 + 3(-2) + 1

4-6+1

-1
4 0
2 years ago
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Create a model to show that 2(2x+3y)=4x+6y
Tom [10]
You would first do 2 multiplied by 2x and that would give you 4x, then you would do 2 multiplied by 3y which would give you 6y, you would put 4x under 2x and put 6y under 3y and bring down your addition symbol 
4 0
3 years ago
Paloma ran 3 3/4 miles around the school track. If each lap is 1/2 mile, how many laps did she run?
Ratling [72]

Answer:

9

Step-by-step explanation:

It is 9 because 1/2 goes into 3, 6 times, and then 3/4 divided by 1/2 is 1 1/2 so 1/2 goes into 3/4 1/12 times, so then figure out how mnay 1/2 is in 1 1/2 and youll get 3 so 6 plus 3 equals 9 :)

7 0
2 years ago
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