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REY [17]
3 years ago
15

The number of days (X) after mailout it takes a utility company to receive payment for a customer’s bill is a normal random vari

able with a mean of 31 days and a standard deviation of 6 days.The latest 2.5% of bills will be received more than how many days after mailout?
Mathematics
1 answer:
Monica [59]3 years ago
4 0

Answer:

The latest 2.5% of bills will be received in 20 more days after mail-out.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 31 days

Standard Deviation, σ = 6 days

We are given that the distribution of The number of days (X) is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P(X \leq x) = 0.025

We have to find the value of x such that the probability is 0.025

P(X \leq x) = 0.025

P( X \leq x) = P( z \leq \displaystyle\frac{x - 31}{6})=0.025  

Calculation the value from standard normal z table, we have,  

P( z \leq -1.960) = 0.025

\displaystyle\frac{x - 31}{6} = -1.960\\x = 19.24 \approx 20  

Hence, the latest 2.5% of bills will be received in 20 more days after mail-out.

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Answer:

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What is the image of (1, -6) for a 180° counterclockwise rotation about the origin?
a_sh-v [17]

Answer: option b

Step-by-step explanation:

It is important to know that, by definition, the rule for 180 degrees counterclockwise rotation is:

(x,y)→(-x,-y)

Therefore, given the point (1, -6) we can find the image obtained for  for a 180° counterclockwise rotation about the origin by applying the rule mentioned before.

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Determine whether the sequence is arithmetic or geometric.
katen-ka-za [31]

Answer:

A. Sequence 2 is arithmetic and Sequence 1 is geometric

Step-by-step explanation:

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3 years ago
I need to know how to do the whole thing and understand it.
patriot [66]

We are given the data on the number of candies handed by neighborhood A and neighborhood B.

Let us first find the mean and variance of each neighborhood.

Mean:

\bar{x}_A=\frac{\sum x}{N_1}=\frac{12}{6}=2\bar{x}_B=\frac{\sum x}{N_2}=\frac{20}{6}=3.33

Variance:

s_A^2=\frac{\sum x^2}{N_1}-\bar{x}_A^2=\frac{28}{6}-2^2=0.667s_B^2=\frac{\sum x^2}{N_2}-\bar{x}_B^2=\frac{80}{6}-3.33^2=2.244

A. Null hypothesis:

The null hypothesis is that there is no difference in the mean number of candies handed out by neighborhoods A and B.

H_0:\;\mu_A=\mu_B

Research hypothesis:

The research hypothesis is that the mean number of candies handed out by neighborhood A is more than neighborhood B.

H_a:\;\mu_A>\mu_B

Test statistic (t):

The test statistic of a two-sample t-test is given by

t=\frac{\bar{x}_A-\bar{x}_B}{s_p}

Where sp is the pooled standard deviation given by

\begin{gathered} s_p=\sqrt{\frac{N_1s_1^2+N_2s_2^2}{N_1+N_2-2}(\frac{N_1+N_2}{N_1\cdot N_2}}) \\ s_p=\sqrt{\frac{6\cdot0.667+6\cdot2.244}{6+6-2}(\frac{6+6}{6\cdot6})} \\ s_p=0.763 \end{gathered}t=\frac{2-3.33}{0.763}=-1.74

So, the test statistic is -1.74

Critical t:

Degree of freedom = N1 + N2 - 2 = 6+6-2 = 10

Level of significance = 0.05

The right-tailed critical value for α = 0.05 and df = 10 is found to be 1.81

Critical t = 1.81

We will reject the null hypothesis because the calculated t-value is less than the critical value.

Interpretation:

This means that we do not have enough evidence to conclude that neighborhood A gives out more candies than neighborhood B.

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