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1 year ago

11

The fact that the speed of a vehicle is lower than the prescribed limits shall relieve the driver from the duty to decrease spee

d when approaching and crossing an intersection. True or false

Computers and Technology

2 answers:

blsea [12.9K]1 year ago

4 0

Explanation:

the answer is false ........

Harrizon [31]1 year ago

4 0

Answer:

Of course. ......false

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Import java.util.scanner; public class sumofmax { public double findmax(double num1, double num2) { double maxval; // note: if-e

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Here you go,

Import java.util.scanner

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// Note: if-else statements need not be understood to

// complete this activity

if (num1 > num2) { // if num1 is greater than num2,

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return maxVal;

}

public static void main(String[] args) {

double numA = 5.0;

double numB = 10.0;

double numY = 3.0;

double numZ = 7.0;

double maxSum = 0.0;

/* Your solution goes here */

maxSum = findMax(numA, numB); // first call of findMax

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System.out.print("maxSum is: " + maxSum);

return;

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/*

Output:

maxSum is: 17.0

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George is working with an image in Photoshop. He added a lot of effects to the image. He saved the first draft of the file. The

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Find a 95% confidence interval for the mean failure pressure for this type of roof panel.The article "Wind-Uplift Capacity of Re

aliya0001 [1]

Answer:

The 95% confidence interval would be given by (2.351;3.525)

Explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

2) Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=2.938$

The sample deviation calculated $s=0.472$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=5-1=4$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,4)".And we see that $t_{\alpha/2}=2.78$

Now we have everything in order to replace into formula (1):

$2.938-2.78\frac{0.472}{\sqrt{5}}=2.351$

$2.938+2.78\frac{0.472}{\sqrt{5}}=3.525$

So on this case the 95% confidence interval would be given by (2.351;3.525)

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