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Leona [35]
3 years ago
13

Based on what we know a function to be today, what did Euler mean by an analytic expression?

Mathematics
2 answers:
puteri [66]3 years ago
5 0
An expression formed from operations such as addition, multiplication, powers and roots. So, c is the answer
ludmilkaskok [199]3 years ago
3 0
<h2>Answer:</h2>

Option: C is the correct answer.

      C. an expression formed from operations such as addition, multiplication, powers, and roots.

<h2>Step-by-step explanation:</h2>

Analytic expression--

It is a mathematical expression which is composed of variables, numbers , symbols and arithmetic operators.

Also, a equation is formed with the help of two analytic expressions which are equivalent.

            Hence, the answer is:

                Option: C

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Hello there!

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-HuronGirl
5 0
3 years ago
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8 0
3 years ago
Suppose the standard deviation of a normal population is known to be 3, and H0 asserts that the mean is equal to 12. A random sa
blagie [28]

Answer:

z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9  

p_v =P(z>1.9)=0.0287  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=12.95 represent the sample mean

\sigma=3 represent the population standard deviation for the sample  

n=36 sample size  

\mu_o =12 represent the value that we want to test  

\alpha=0.05 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 12, the system of hypothesis would be:  

Null hypothesis:\mu \leq 12  

Alternative hypothesis:\mu > 12  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9  

P-value  

Since is a right tailed test the p value would be:  

p_v =P(z>1.9)=0.0287  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 5% of signficance.  

4 0
3 years ago
Read 2 more answers
(PLEASE HELP!!)
umka2103 [35]
Congruent because 2 is a larger number then one and 1 is a smalled number then 2
7 0
2 years ago
Scores on a final exam taken by 1200 students have a bell shaped distribution with mean=72 and standard deviation=9
SVETLANKA909090 [29]

Answer:

a. 72

b. 816

c. 570

d. 30

Step-by-step explanation:

Given the graph is a bell - shaped curve. So, we understand that this is a normal distribution and that the bell - shaped curve is a symmetric curve.

Please refer the figure for a better understanding.

a. In a normal distribution, Mean = Median = Mode

Therefore, Median = Mean = 72

b. We have to know that 68% of the values are within the first standard deviation of the mean.

i.e., 68% values are between Mean $ \pm $ Standard Deviation (SD).

Scores between 63 and 81 :

Note that 72 - 9 = 63 and

72 + 9 = 81

This implies scores between 63 and 81 constitute 68% of the values, 34% each, since the curve is symmetric.

Now, Scores between 63 and 81 = $ \frac{68}{100} \times 1200 $

= 68 X 12 = 816.

That means 816 students have scored between 63 and 81.

c. We have to know that 95% of the values lie between second Standard Deviation of the mean.

i.e., 95% values are between Mean $ \pm $ 2(SD).

Note that 90 = 72 + 2(9) = 72 + 18

Also, 54 = 63 - 18.

Scores between 54 and 90 totally constitute 95% of the values. So, Scores between 72 and 90 should amount to $ \frac{95}{2} \% $ of the values.

Therefore, Scores between 72 and 90 = $ \frac{95}{2(100)} \times 1200 = \frac{95}{200} \times 1200  $

$ \implies 95 \times 12 $ = 570.

That is a total of 570 students scored between 72 and 90.

d. We have to know that 5 % of the values lie on the thirst standard Deviation of the mean.

In this case, 5 % of the values lie between below 54 and above 90.

Since, we are asked to find scores below 54. It should be 2.5% of the values.

So, Scores below 54 = $ \frac{2.5}{100} \times 1200 $

= 2.5 X 12 = 30.

That is, 30 students have scored below 54.

8 0
3 years ago
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