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3 years ago

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On Monday Sarah had homework in 7⁄10 of her classes, Tuesday 3⁄5 , Wednesday 9⁄11 , and Thursday 1⁄2 . Which day did she have th

e mosthomework

1 answer:

expeople1 [14]3 years ago

5 0

9/11, just choose the biggest fraction

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Find 4 irrational number between 0.1 and 1.12

Lelu [443]

I'm sorry but I do not know how to do this

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3 years ago

PLEASEEEE HELPPP Angle A and angle B are two angles that are complementary. We know angle A measures (x+2)° and Angle B measures

saul85 [17]

Answer:

Step-by-step explanation:

x + 2 + 57 = 90

x + 59 = 90

x = 31°

x + 2 = 33°

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3 years ago

Read 2 more answers

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

EleoNora [17]

Answer:

a) $P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

b) $P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

c) $P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

d) $E(X) = 20*0.2= 4$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=20, p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

We want this probability:

$P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

Part b

We want this probability:

$P(X=4)$

And using the probability mass function we got:

$P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

Part c

We want this probability:

$P(X>3)$

We can use the complement rule and we got:

$P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

Part d

The expected value is given by:

$E(X) = np$

And replacing we got:

$E(X) = 20*0.2= 4$

3 0

2 years ago

4 cot theta<br>=<br>tan theta<br>how do I I find the value of quadrant?

Ray Of Light [21]

Answer:

4cotα=tanα

4(1/tanα)=tanα

(4/tanα)=tanα

cross multiply

=> 4=tan²α

√4=√tan²α

±2=tanα

α=arc( tan) |2|

α=63.4° ( in first quadrant)

and

α=180+63.4=243.4 in the third quadrant

since we also found a negative answer( i.e –2) then α also lies in quadrants where it gives a negative value(i.e second and fourth quadrants)

α=180–63.4=116.6° in the second quadrant

α=360–63.4=296.6 in the fourth quadrant

therefore theta( in my case, alpha) lies in all four quadrants and is equal to:

α=63.4°,243.4°,116.6°and 296.6°

5 0

3 years ago

Solve the following equation: <br>(n + 1) + 3(n - 1) = 2(n - 1)(n + 1)

Mnenie [13.5K]

Answer:

n = 0, n =2

Step-by-step explanation:

Given

(n + 1) + 3(n - 1) = 2(n - 1)(n + 1) ← distribute parenthesis on both sides

n + 1 + 3n - 3 = 2(n² - 1), that is

4n - 2 = 2n² - 2 ( subtract 4n - 2 from both sides )

0 = 2n² - 4n ← factor out 2n from each term )

0 = 2n(n - 2)

Equate each factor to zero and solve for n

2n = 0 ⇒ n = 0

n - 2 = 0 ⇒ n = 2

7 0

3 years ago