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Sergeu [11.5K]
3 years ago
10

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

up a simulation to mimic buying 200 small bags of milk chocolate M&M’s. Each bag contains 55 candies. We made this dotplot of the results. Normal sampling distribution of proportion of orange candies Now suppose that we buy a small bag of M&M’s. We find that 25.5% (14 of the 55) of the M&M’s are orange. What can we conclude? Group of answer choices This result is not surprising because we expect to see many samples with 14 or more orange candies. This result is surprising because we expect the orange candies to make up no more than 20% of the candies in a packet. This result is surprising because it is unlikely that we will select a random sample with 25.5% or more orange candies if 20% of milk chocolate M&M’s are orange.
Mathematics
1 answer:
SOVA2 [1]3 years ago
4 0

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).

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chubhunter [2.5K]

The correct equation is P(t) = Po*e^(kt). Please enclose "kt" inside parentheses and use the " ^ " symbol to indicate exponentiation.


"k" is the growth constant, which here is 0.017. "t" is the number of yeasrs. P(7) is the population after 7 years. Po is the initial population, which in this case is 91 million.


The initial value, when t = 0, is 91 million.


After 7 years, the population, P(7), was P(7) = (91 million)*e^(0.017*7). This evaluates to P(7) = (91 million)*(1.070) = 96 million

8 0
3 years ago
If g(x) = 4x + 1 and g(x) = -11, find x.
podryga [215]

Answer:

x = - 3

Step-by-step explanation:

g(x) = 4x + 1...... (1)

g(x) = -11.... (2)

From equations (1) & (2)

-11 = 4x + 1

-11 - 1 = 4x

-12 = 4x

-12/4 = x

-3 = x

x = - 3

3 0
2 years ago
5 ( x + 3 - 5x)<br> PLZ WHAT IS THIS IN SIMPLEIST FORM WILL GIVE BRAINLIEST
Tresset [83]

Answer:

-20x + 15

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
(3x-5)/(x-5)&gt;=0 <br> How do I get the answers for this problem?
Diano4ka-milaya [45]
\frac{3x-5}{x-5} > 0 

First, note that x is undefined at 5. / x ≠ 5
Second, replace the inequality sign with an equal sign so that we can solve it like a normal equation. / Your problem should look like: \frac{3x-5}{x-5} = 0
Third, multiply both sides by x - 5. / Your problem should look like: 3x - 5 = 0
Forth, add 5 to both sides. / Your problem should look like: 3x = 5
Fifth, divide both sides by 3. / Your problem should look like: x = \frac{5}{3}
Sixth, from the values of x above, we have these 3 intervals to test: 
x < \frac{5}{3}
\frac{5}{3} < x < 5
x > 5
Seventh, pick a test point for each interval. 

1. For the interval x < \frac{5}{3} :
Let's pick x - 0. Then, \frac{3x0-5}{0-5} > 0
After simplifying, we get 1 > 0 which is true.
Keep this interval.

2. For the interval \frac{5}{3} < x < 5:
Let's pick x = 2. Then, \frac{3x2-5}{2-5} > 0
After simplifying, we get -0.3333 > 0, which is false.
Drop this interval.

3. For the interval x > 5:
Let's pick x = 6. Then, \frac{3x6-5}{6-5} > 0
After simplifying, we get 13 > 0, which is ture.
Keep this interval.
Eighth, therefore, x < \frac{5}{3} and x > 5

Answer: x < \frac{5}{3} and x > 5

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3 years ago
Britney work 35 hours at a rate of pay of 9.55 how much did she earn before taxes ​
k0ka [10]

Answer:

3.66

Step-by-step explanation:

35÷9.55=3.66

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