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ddd [48]
3 years ago
6

Which number is irrational? A.0.14, .14 is repeating B. 1/3 C.the square root of 4 D. The square root of 6

Mathematics
2 answers:
Svetradugi [14.3K]3 years ago
7 0
I think it is A in this statment
ss7ja [257]3 years ago
3 0
The answer would be A.
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Which of the following tables represents a ratio which is greater than the ratio in the table above?
finlep [7]

complete question please.

4 0
3 years ago
I am confused. please help!
wel

Answer:

Step-by-step explanation:

D=90°

E=147°

F= 33°

4 0
3 years ago
Got confused?? Any help??
Leviafan [203]
Length x width = area
5 1/2 x 3 1/2 = 19 1/4

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8 0
3 years ago
Read 2 more answers
A sample of size 100 is selected from a population with p = .40.
zhuklara [117]

Answer:

(a) 0.40

(b) 0.049

(c) \bar p\sim N(0.40,0.049^{2})

(d) Explained below

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

 E(\bar p)=p

The standard deviation of this sampling distribution of sample proportion is:

 SE(\bar p)=\sqrt{\frac{p(1-p)}{n}}

Given:

n = 100

p = 0.40

As <em>n</em> = 100 > 30 the Central limit theorem is applicable.

(a)

Compute the expected value of \bar p as follows:

E(\bar p)=p=0.40

The expected value of \bar p is 0.40.

(b)

Compute the standard error of \bar p as follows:

SE(\bar p)=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.40(1-0.40)}{100}}=0.049

The standard error of \bar p is 0.049.

(c)

The sampling distribution of \bar p is:

\bar p\sim N(0.40,0.049^{2})

(d)

The sampling distribution of p show that as the sample size is increasing the distribution is approximated by the normal distribution.

3 0
3 years ago
Prove the following with Euler's formula to write e^itheta and e^-itheta in terms of sine and cosine. Then subtract them.
kirill115 [55]
Euler's formula tells us that

e^{i\theta}=\cos\theta+i\sin\theta
e^{-i\theta}=\cos\theta-i\sin\theta

Suppose we subtract the two. This eliminates the cosine terms.

e^{i\theta}-e^{-i\theta}=i\sin\theta-(-i\sin\theta)=2i\sin\theta

Divide both sides by 2i and you're done.
8 0
3 years ago
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