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2 years ago

The three sides of AJKL

Mathematics

1 answer:

BaLLatris [955]2 years ago

7 0

Answer:

right triangle

Step-by-step explanation:

I find it convenient to answer questions like this by computing a "form factor" that subtracts the square of the long side from the sum of the squares of the shorter sides. The sign of this value is the sign of the cosine of the largest angle in the triangle.

f = a² +b² -c²

f = 11.2² +21² -23.8²

f = 125.44 +441 -566.44 = 0

The cosine of the largest angle is 0, so that angle is 90° and the triangle is a right triangle.

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Scores on the SAT Mathematics test are believed to be normally distributed. The scores of a simple random sample of five student

AysviL [449]

Answer:

The mean calculated for this case is $\bar X=584$

And the 95% confidence interval is given by:

$584-2.776\frac{86.776}{\sqrt{5}}=476.271$

$584+2.776\frac{86.776}{\sqrt{5}}=691.729$

So on this case the 95% confidence interval would be given by (476.271;691.729)

Step-by-step explanation:

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 for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

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 $\bar X=584$

The sample deviation calculated $s=86.776$

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 tabel to find the critical value. The excel command would be: "=-T.INV(0.025,4)".And we see that $t_{\alpha/2}=2.776$

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

$584-2.776\frac{86.776}{\sqrt{5}}=476.271$

$584+2.776\frac{86.776}{\sqrt{5}}=691.729$

So on this case the 95% confidence interval would be given by (476.271;691.729)

3 0

2 years ago

-3x+10y=-60<br> 3x+5y=15

Verdich [7]

Answer:

x=10, y=-3

Step-by-step explanation:

solve by addition/elimination

6 0

2 years ago

What is the answer? Please help me!!! PLEASE!!!!

sladkih [1.3K]

Answer:

a.) y=1/3x +3

Step-by-step explanation:

8 0

2 years ago

What is the length of line segment FE?

Jobisdone [24]

Answer:

(7,-1)because it is problem of vector lesson so that we have to use formula of vector

6 0

3 years ago

Read 2 more answers

It is assumed that the test results for a class follow a normal distribution with a mean of 78 and a standard deviation of 36. I

HACTEHA [7]

Answer:

0.7143

- You know the minimum possible grade of the student

- You want to find out the probability that that grade which is greater than 72 is also greater than 84

- Expect the answer to be 0.7143

- The explanation/steps is given below

- The answer seems (and is) reasonable. You only created a new lower limit.

Step-by-step explanation:

Distribution Type: Normal

Mean: 78

Standard deviation: 36

Range of marks: [78 - 36], [78 + 36] = [42 to 114]

If you know that a student's grade is greater than 72, what is the probability that it's greater than 84?

In this case, the full probability level is between 72 and the upper limit of 114 (instead of between 42 and 114). First, find the fraction of this probability.

[114 - 72] ÷ [114 - 42] = 42/72 = 0.5833

So, the probability of a student's mark falling between 72 and 114 is 0.5833.

Now making this a whole interval, 114 - 72 = 42

What fraction of this interval of 42 will bear marks between 84 and 114?

[114 - 84] = 30

30/42 = 0.7143

Because of the first part of the question "If you know (are sure) that a student's grade is greater than 72...", your answer stops at 0.7143, since 72 was used as the lower limit.

8 0

2 years ago