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Natali [406]
2 years ago
11

Joshua and his four friends have 1/2 pakage of a crackers. How many pakages of crackers do they have in all

Mathematics
1 answer:
maks197457 [2]2 years ago
7 0

answer:

2.5 or 2 1/2

work:

4(joshuas friends)+1(joshua)=5

5×1/2=5/2

5/2= 2.5

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The solution set of a linear system whose augmented matrix is [a b c d] is the same as the solution set of Ax = d, where A = [a
Charra [1.4K]

Answer:

True

Step-by-step explanation:

First statement

[a b c | d][x]

[a b c]x=d

ax+bx+cx=d

Second statement

Ax=d

Given that A = [a b c]

[a b c]x=d

ax+bx+cx=d

ax+bx+cx=d

Then, they are going to have the same solutions

6 0
3 years ago
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
3 years ago
What is the least common multiple of 5, 10, and 12? What is the least common multiple of 6, 10, and 12? What is the least common
Mila [183]

Answer:

LCM of 5, 10, and 12 is 60

6, 10, and 12 is 60

2, 4, and 9 is 36

6 0
3 years ago
Read 2 more answers
If the range of the function f(x) = 4x – 3 is {11.4, 15, 17, 29}, what is its domain?
malfutka [58]
The range of the function is the y-component or the ordinate of the sets of points. The domain of the function meanwhile is the x-component ot the abscissa of the sets of points. In this case, we have the function <span>f(x) = 4x – 3 and given each y, we get the x. Subsituting each y to the function we get the domain equal to {3.535, 4.5, 5, 8}.</span>
5 0
3 years ago
A. 60<br> b. 180<br> c. 100<br> d. 120
Leno4ka [110]

Answer:

120 is the correct answer

thanks. please mark brainlist

6 0
3 years ago
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