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xz_007 [3.2K]
2 years ago
6

Which will help you build a good credit history

Mathematics
1 answer:
tresset_1 [31]2 years ago
5 0
Make 100% of your payments on time and keep your credit card debt low.
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Increase 386.3 by 4%
astraxan [27]
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multiply 386.3 by 4% (.04) then add the total to 386.3
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3 years ago
What role do individuals play in a capitalist system?
Vitek1552 [10]
The very rich control the market and decide what industries to invest in or lend money too. They influence the government so that they don't have to pay very much in taxes, which would support the nation's infrastructure, & provide jobs. The middle class works hard to produce products, and also spends money on products, which helps to make the rich investors even more wealthy. The poor are typically unable to compete in a capitalist economy (for a number of reasons), & require public assistance to survive.


6 0
3 years ago
CCN and ActMedia provided a television channel targeted to individuals waiting in supermarket checkout lines. The channel showed
tatyana61 [14]

Answer:

(a) Null Hypothesis, H_0 : \mu = 8 minutes    

    Alternate Hypothesis, H_A : \mu \neq 8 minutes

(b) The P-value of the test statistics is 0.0436.

(c) We conclude that the actual mean waiting time equals the standard at 0.05 significance level.

(d) 95% confidence interval for the population mean is [7.93 minutes , 9.07 minutes].

Step-by-step explanation:

We are given that the length of the program was based on the assumption that the population mean time a shopper stands in a supermarket checkout line is 8 minutes.

A sample of 120 shoppers showed a sample mean waiting time of 8.5 minutes. Assume a population standard deviation of 3.2 minutes.

<u><em>Let </em></u>\mu<u><em> = actual mean waiting time.</em></u>

(a) So, Null Hypothesis, H_0 : \mu = 8 minutes     {means that the actual mean waiting time does not differs from the standard}

Alternate Hypothesis, H_A : \mu \neq 8 minutes     {means that the actual mean waiting time differs from the standard}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;

                        T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean waiting time = 8.5 minutes

            \sigma = population standard deviation = 3.2 minutes

            n = sample of shoppers = 120

So, <u><em>test statistics</em></u>  =  \frac{8.5-8}{\frac{3.2}{\sqrt{120} } }  

                               =  1.71

The value of t test statistics is 1.71.

(b) Now, the P-value of the test statistics is given by;

                    P-value = P(Z > 1.71) = 1 - P(Z \leq 1.71)

                                  = 1 - 0.9564 = 0.0436

(c) Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

<em>Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which </em><u><em>we fail to reject our null hypothesis</em></u><em>.</em>

Therefore, we conclude that the actual mean waiting time equals the standard.

(d) Now, the pivotal quantity for 95% confidence interval for the population mean is given by;

                        P.Q. = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean waiting time = 8.5 minutes

            \sigma = population standard deviation = 3.2 minutes

            n = sample of shoppers = 120

            \mu = population mean

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.</em>

<u>So, 95% confidence interval for the population proportion, </u>\mu<u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                             of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times{\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.96 \times{\frac{\sigma}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X-1.96 \times{\frac{\sigma}{\sqrt{n} } } , \bar X+1.96 \times{\frac{\sigma}{\sqrt{n} } } ]

  = [ 8.5-1.96 \times{\frac{3.2}{\sqrt{120} } } , 8.5+1.96 \times{\frac{3.2}{\sqrt{120} } } ]

  = [7.93 minutes , 9.07 minutes]

Therefore, 95% confidence interval for the population mean is [7.93 minutes , 9.07 minutes].

Yes, it support our above conclusion.

3 0
3 years ago
The top of the table can be modeled with a cylinder, and the legs can be modeled with rectangular prisms. Estimate the amount of
faltersainse [42]

Answer:

A. 3,703 inches squared

Step-by-step explanation:

I took the test and got it right. #platofam

4 0
1 year ago
(Please please, actually type the answer and dont send it in a file because it wont work)
NNADVOKAT [17]

Answer: like 36 i think

Step-by-step explanation:

6 0
2 years ago
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