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2 years ago

What is the measure of RVE? А. 45° В. 60° ) С. 750 OD. 309

Mathematics

1 answer:

icang [17]2 years ago

7 0

Answer: A. 45

Explanation:
First find AVL because AVL = RVE.
AVL = 180 - 60 - 75
AVL = 45

Now, AVL = RVE, so RVE = 45

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What is the value of 6i/1+i? 2 – 6i 2 + 6i 3 – 3i 3 + 3i

mihalych1998 [28]

6i/ (1+i)

multiply by the complex conjugate (1-i)/(1-i)

6i/(1+i) * (1-i)/(1-i)

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(6+6i)/2

3+3i

Answer: 3+3i

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1.which expression is equivalent to (4-3i) 2

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What are the expressions that are possible you can solve by simplyfiying

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For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding.

ollegr [7]

Answer: a. A point estimate for p is 0.597 .

b. The 95% confidence interval for p.

Lower limit = 0.48

Upper limit = 0.71

Step-by-step explanation:

Given : Sample size of professional actors : n= 67

Number of extroverts : x= 40

Let p represent the proportion of all actors who are extroverts.

a. The point estimate for p = sample proportion = $\hat{p}=\dfrac{x}{n}$

$=\dfrac{40}{67}=0.597$

b. Confidence interval for population proportion :

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

Since the critical value for 95% confidence interval is 1.96 , so the 95% confidence interval for p would be

$0.597\pm (1.96)\sqrt{\dfrac{0.597(1-0.597)}{67}}$

$0.597\pm (1.96)\sqrt{0.0036}$

$0.597\pm (1.96)(0.06)$

$0.597\pm 0.1176$

$(0.597-0.1176,\ 0.597+0.1176)\\\\=(0.4794,\ 0.7146)\\\\\approx(0.48,\ 0.71)$

In the 95% confidence interval for p.

Lower limit = 0.48

Upper limit = 0.71

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2 years ago

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180 because you multiply the 12 packs by 10 to get 120. Then you subtract 120 from 300

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2 years ago

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An automated egg carton loader has a 1% probability of cracking an egg, and a customer will complain if more than one egg per do

vaieri [72.5K]

Answer:

a) Binomial distribution B(n=12,p=0.01)

b) P=0.007

c) P=0.999924

d) P=0.366

Step-by-step explanation:

a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.

b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.

$P(k=0)=\binom{12}{0}p^0(1-p)^{12}=1*1*0.99^{12}=1*0.886=0.886\\\\P(k=1)=\binom{12}{1}p^1(1-p)^{11}=12*0.01*0.99^{11}=12*0.01*0.895=0.107$

Then,

$P(k>1)=1-(P(k=0)+P(k=1))=1-(0.886+0.107)=1-0.993=0.007$

c) In this case, the distribution is B(1200,0.01)

$P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924$

d) In this case, the distribution is B(100,0.01)

We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.

$P(k=0)=\binom{100}{0}p^0(1-p)^{100}=0.99^{100}=0.366$

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2 years ago