Luis colors 1/4 of a spinner using a red crayon. write the fraction of the spinner Luis colored red in twelfths.

Find the rate of change given the following solutions: (6, 2) and (1, 12).

blagie [28]

Answer:

The rate of change is -2

Step-by-step explanation:

f(b)-f(a) 12-2 10

---------- = ------------ = -------- = -2

b-a 1-6 -5

7 0

3 years ago

Music streaming services are the most popular way to listen to music. Data gathered over the last 12 months show Apple Music was

marysya [2.9K]

Answer:

There is enough statistical evidence to suggest that there is no difference between the mean number of households picking either service

Step-by-step explanation:

The parameters of the music streaming service are;

The average number of households using Apple Music, $\overline {X}_1$ = 1.73 million

The standard deviation, σ₁² = 0.46 million

The average number of households using Spotify, $\overline {X}_2$ = 2.17 million

The standard deviation, σ₂² = 0.33 million

The significance level = 0.01

Hypothesis testing with different population standard deviation

$Z = \dfrac{\overline {X}_1 - \overline {X}_2}{\sqrt{\dfrac{\sigma_1^2}{n_1} +\dfrac{\sigma_2^2}{n_2}} }$

H₀: $\overline {X}_1$ ≠ $\overline {X}_2$

H₁: $\overline {X}_1$ = $\overline {X}_2$

$Z = \dfrac{1.73- 2.17}{\sqrt{\dfrac{0.46^2}{12} +\dfrac{0.33^2}{12}} } = -2.6923$

p-value for the test statistic = 2 × P(Z <-2.6923) = 2 × 0.00357 = 0.00714

Therefore, the p-value is less than the significance level, and it is therefore unlikely that the result will be observed under the null hypothesis

We therefore reject the null hypothesis and there is enough statistical evidence to suggest that there is no difference between the mean number of households picking either service

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 X = number of orange milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, p = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, n = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable X follows a Binomial distribution with parameter n = 55 and p = 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:

$P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}$