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MA_775_DIABLO [31]
3 years ago
14

If Mary Alice buys the car for $35,000 and pays 20% down, how much will she finance? If she finances this balance for 6 years at

the special rate of 0.9% add-on interest, how much will her monthly payments be? What is the total cost of the car if she finances it
Mathematics
1 answer:
RUDIKE [14]3 years ago
4 0

Answer:monthly is 598.90 total is 43120

Step-by-step explanation:

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The duration of routine operations at chicago general hospital has approximately a normal distribution with an average of 130 mi
vredina [299]

Answer:

79.77%

Step-by-step explanation:

P(X≥120) = normalcdf(120,1e99,130,12) = 0.7976716754 ≈ 0.7977 = 79.77%

Therefore, about 79.77% of operations last at least 120 minutes

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Giving BRAINLIEST to correct answer
sergiy2304 [10]

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91

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There are 13 scores in total,the Middle score is the median score.

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HELP PLS HELP !!!!!!!
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19,800

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3 years ago
Stephanie had 7/8 pound of birdseed she use 3/8 pound to fill the birdfeeder how much birdseed to Stephanie have left?
marusya05 [52]

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4/8 or 1/2

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3 years ago
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The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in t
andreyandreev [35.5K]

Answer:

The test statistic is c. 2.00

The p-value is a. 0.0456

At the 5% level, you b. reject the null hypothesis

Step-by-step explanation:

We want to test to determine whether or not the mean waiting time of all customers is significantly different than 3 minutes.

This means that the mean and the alternate hypothesis are:

Null: H_{0} = 3

Alternate: H_{a} = 3

The test-statistic is given by:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

Sample of 100 customers.

This means that n = 100

3 tested at the null hypothesis

This means that \mu = 3

The average length of time it took the customers in the sample to check out was 3.1 minutes.

This means that X = 3.1

The population standard deviation is known at 0.5 minutes.

This means that \sigma = 0.5

Value of the test-statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{3.1 - 3}{\frac{0.5}{\sqrt{100}}}

z = 2

The test statistic is z = 2.

The p-value is

Mean different than 3, so the pvalue is 2 multiplied by 1 subtracted by the pvalue of Z when z = 2.

z = 2 has a pvalue of 0.9772

2*(1 - 0.9772) = 2*0.0228 = 0.0456

At the 5% level

0.0456 < 0.05, which means that the null hypothesis is rejected.

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3 years ago
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