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Rama09 [41]
3 years ago
8

Solve 4(a+4)-2=34 what is a

Mathematics
2 answers:
Vikentia [17]3 years ago
7 0

4(a+4)−2=34

Simplify:

4(a+4)−2=34

(4)(a)+(4)(4)+−2=34(Distribute)

4a+16+−2=34

(4a)+(16+−2)=34(Combine Like Terms)

4a+14=34

Subtract 14 from both sides.

4a+14−14=34−14

4a=20

Divide both sides by 4.

4a/4=20/4

a=5

Marianna [84]3 years ago
4 0
4(a+4)-2=34 
4a+16-2=34
4a=34-16+2
4a=20
a=5

good luck

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Answer:

y = 11.54°

Step-by-step explanation:

Reference angle = y°

Opposite side length = 4

Hypotenuse = 20

Apply trigonometric function SOH:

Sin(y) = \frac{Opp}{Hyp}

Sin(y) = \frac{4}{20}

Sin(y) = 0.2

y = Sin^{-1}(0.2)

y = 11.536959° ≈ 11.54° (nearest hundredth)

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3 years ago
We have n = 100 many random variables Xi ’s, where the Xi ’s are independent and identically distributed Bernoulli random variab
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Answer:

(a) The distribution of X=\sum\limits^{n}_{i=1}{X_{i}} is a Binomial distribution.

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(c) The value of P(\bar X>0.50) is 0.50.

Step-by-step explanation:

It is provided that random variables X_{i} are independent and identically distributed Bernoulli random variables with <em>p</em> = 0.50.

The random sample selected is of size, <em>n</em> = 100.

(a)

Theorem:

Let X_{1},\ X_{2},\ X_{3},...\ X_{n} be independent Bernoulli random variables, each with parameter <em>p</em>, then the sum of of thee random variables, X=X_{1}+X_{2}+X_{3}...+X_{n} is a Binomial random variable with parameter <em>n</em> and <em>p</em>.

Thus, the distribution of X=\sum\limits^{n}_{i=1}{X_{i}} is a Binomial distribution.

(b)

According to the Central Limit Theorem if we have an unknown population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.  

The sample size is large, i.e. <em>n</em> = 100 > 30.

So, the sampling distribution of the sample mean will be approximately normal.

The mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu=p=0.50

And the standard deviation of the distribution of sample mean is given by,

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(c)

Compute the value of P(\bar X>0.50) as follows:

P(\bar X>0.50)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{0.50-0.50}{0.05})\\

                    =P(Z>0)\\=1-P(Z

*Use a <em>z</em>-table.

Thus, the value of P(\bar X>0.50) is 0.50.

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