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2 years ago

Solve for e. T=e29 e=±3T√T e=±3T e=±3T√ e=±9T√

Mathematics

1 answer:

Anon25 [30]2 years ago

4 0

Answer:

option (3) is correct.

$e= \pm 3\sqrt{T}$

Step-by-step explanation:

Given $T=\frac{e^2}{9}$

We have to solve for e.

Consider the given statement,

$T=\frac{e^2}{9}$

Cross multiply, we get,

$\Rightarrow 9T={e^2}$

Taking square root both sides , we get,

$\Rightarrow \sqrt{9T}=e$

We know square root of 9 is 3.

$\Rightarrow \pm 3\sqrt{T}=e$

Thus, option (3) is correct.

$e= \pm 3\sqrt{T}$

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A) Write the sequence of natural numbers which are multiplied by 3 ?

Volgvan

Answer:

a) 3, 6, 9, 12, 15,..., $3\cdot n$ , b) 4, 7, 10, 13, 16,..., $3\cdot n +1$ , c) Both sequences are arithmetic.

Step-by-step explanation:

a) The sequence of natural numbers which are multiplied by 3 are represented by the function $f(n) = 3\cdot n$ , $n\in \mathbb{N}$ . Let see the first five elements of the sequence: 3, 6, 9, 12, 15,...

b) The sequence of natural numbers which are multiplied by 3 and added to 1 is represented by the function $f(n) = 3\cdot n + 1$ , $n\in \mathbb{N}$ . Let see the first five elements of the sequence: 4, 7, 10, 13, 16,...

c) Both sequences since differences between consecutive elements is constant. Let prove this statement:

(i) $f(n) = 3\cdot n$

$\Delta f = f(n+1) -f(n)$

$\Delta f = 3\cdot (n+1) -3\cdot n$

$\Delta f = 3$

(ii) $f(n) = 3\cdot n +1$

$\Delta f = f(n+1)-f(n)$

$\Delta f = [3\cdot (n+1)+1]-(3\cdot n+1)$

$\Delta f = 3$

Both sequences are arithmetic.

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2 years ago

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2 years ago

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Setler79 [48]

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3 years ago

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In college basketball, a shot made from beyond a designated arc radiating about 20 ft from the basket is worth three points ins

denis-greek [22]

Answer:

a) By the Central Limit Theorem, the distribution would be approximately normal, with mean $\mu = 0.45$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.45*0.55}{100}} = 0.0497$

b) 4.02 standard deviations below the mean.

c) Making 25 or less shots has a pvalue of -4.02. Z-scores lower than -2 are considered surprising, so yes, it would be surprising for him to make only 25 of these shots.

Step-by-step explanation:

To solve this question, we are going to need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For proportions, we have that the mean is $\mu = p$ and the standard deviation is $s = \sqrt{\frac{p(1-p)}{n}}$

In this problem, we have that:

$p = 0.45, n = 100$

a)By the Central Limit Theorem, the distribution would be approximately normal, with mean $\mu = 0.45$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.45*0.55}{100}} = 0.0497$

b)This is Z when X = 0.25.

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$