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forsale [732]
3 years ago
7

Daily low temperatures in Columbus, OH in January 2014 were approximately normally distributed with a mean of 15.45 and a standa

rd deviation of 13.70. What percentage of days had a low temperature between 5 degrees and 10 degrees? (Enter a number without the percent sign, rounded to the nearest 2 decimal places)
Mathematics
1 answer:
Alona [7]3 years ago
5 0

Answer: 12.10

Step-by-step explanation:

Given : Mean : \mu = 15.45

Standard deviation : \sigma = 13.70

The formula to calculate the z-score :-

z=\dfrac{x-\mu}{\sigma}

For x= 5 degrees

z=\dfrac{5-15.45}{13.70}=-0.7627737226\approx-0.76

For x= 10 degrees

z=\dfrac{10-15.45}{13.70}=-0.397810218\approx-0.40

The P-value : P(-0.76

=0.3445783-0.2236273=0.120951\approx0.1210

In percent , 0.1210\times100=12.10\%

Hence, the percentage of days had a low temperature between 5 degrees and 10 degrees = 12.10%

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Answer: the dwarf tree grew by 3 inches.

the semi dwarf tree grew by 6 inches.

the full size tree grew by 18 inches.

Step-by-step explanation:

Let x represent how much the semi-dwarf lemon tree grew.

Last month, a dwarf lemon tree grew half as much as a semi-dwarf lemon tree. This means that the amount by which the dwarf lemon tree grew is expressed as x/2

A full-size lemon tree grew three times as much as the semi-dwarf lemon. This means that the amount by which the full-size lemon tree grew is expressed as 3x

Together, the three trees grew 27 inches. This means that

x/2 + x + 3x = 27

Cross multiplying by 2, it becomes

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x = 54/9

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3 years ago
An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 75 type K batteries and a sample of 46
lorasvet [3.4K]

The decision rule for rejecting the null hypothesis, considering the t-distribution, is of:

  • |t| < 1.9801 -> do not reject the null hypothesis.
  • |t| > 1.9801 -> reject the null hypothesis.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if there is not enough evidence to conclude that the mean voltage for these two types of batteries is different, that is, the subtraction of the sample means is of zero, hence:

H_0: \mu_1 - \mu_2 = 0

At the alternative hypothesis, it is tested if there is enough evidence to conclude that the mean voltage for these two types of batteries is different, that is, the subtraction of the sample means different of zero, hence:

H_1: \mu_1 - \mu_2 \neq 0

We have a two-tailed test, as we are testing if the mean is different of a value.

Considering the significance level of 0.05, with 75 + 46 - 2 = 119 df, the critical value for the test is given as follows:

|t| = 1.9801.

Hence the decision rule is:

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More can be learned about the t-distribution in the test of an hypothesis at brainly.com/question/13873630

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1 year ago
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Multiply the price by the 5% by turning the percent into a decimal.
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Which can be rounded to 2.30 if needed.
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