All categories

2 years ago

What two mixed numbers have a sum of 3

1 answer:

4 0

The answer is one and 1/3 + one and 2/3.
To get this answer, we must first need to understand improper fractions are (i guess) an 'uglier' version of mixed numbers because improper fractions can be converted to mixed numbers and vice versa making them pretty much the same thing.
So an improper fraction is when the numerator is bigger than denominator.
(we also need to understand that for example: 12/3 is the same as 12 divided by 3)
So, what divided by what gives 3? Well 3*3=9 so if 9 is divided by 3 it will equal 3. Therefore we have an improper fraction of 9/3.

The basics of addition in fraction is when the numerator is the same, just add the top. So for simplicity's sake we will make out sum where both improper fraction's denominator is 3. So... what plus what equals 9? 4+5. 4 and 5 are chosen because both numbers are bigger than 3.
so now we have:
4/3 + 5/3
Now just turn the improper fraction into a mixed number and we get the answer of: one and 1/3 + <span>one and 2/3, which equals 3</span>

You might be interested in

Helllppppppppppppppppppppppppppppp

Georgia [21]

The equation: n - 4 = -15

Now let's solve our equation for n.

n - 4 = -15

Add 4 to both sides.

n = -11

4 0

3 years ago

._. Help me please I’m putting my faith in this app.

IRINA_888 [86]

Vas happenin!
Hope your day is going well
I think it’s A
Sorry if it’s wrong
Hope this helps *smiles*

8 0

2 years ago

What is -7(5x-4) simplifyed

HACTEHA [7]

Answer:

-35x+28

Step-by-step explanation:

have a great day :'D

7 0

2 years ago

Read 2 more answers

An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2000 individ

Tema [17]

Answer:

a) P=0.558

b) P=0.021

Step-by-step explanation:

We can model this random variable as a Poisson distribution with parameter λ=1/500*2000=4.

The approximate distribution of the number who carry this gene in a sample of 2000 individuals is:

$P(x=k)=\frac{\lambda^ke^{-\lambda}}{k!} =\frac{4^ke^{-4}}{k!}$

a) We can calculate that the approximate probability that between 4 and 9 (inclusive) as:

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)\\\\\\ P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\P(9)=4^{9} \cdot e^{-4}/9!=262144*0.0183/362880=0.013\\\\\\$

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)=0.195+0.156+0.104+0.060+0.030+0.013=0.558$

b) The approximate probability that at least 9 carry the gene is:

$P(x\geq9)=1-P(x\leq 8)\\\\\\$

$P(0)=4^{0} \cdot e^{-4}/0!=1*0.0183/1=0.018\\\\P(1)=4^{1} \cdot e^{-4}/1!=4*0.0183/1=0.073\\\\P(2)=4^{2} \cdot e^{-4}/2!=16*0.0183/2=0.147\\\\P(3)=4^{3} \cdot e^{-4}/3!=64*0.0183/6=0.195\\\\P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\$

$P(x\geq9)=1-P(x\leq 8)\\\\P(x\geq9)=1-(0.018+0.073+0.147+0.195+0.195+0.156+0.104+0.060+0.030)\\\\P(x\geq9)=1-0.979=0.021$

8 0

2 years ago

Someone please I’m begging this is due by midnight and I can’t have it be late

Thepotemich [5.8K]

Answer:

(1) Coeficcient (2) Constant (3) 4 terms (4) In descending order, 2, 1, 3, 4 (5) -2 (6) 1 (7) 5 (8) 2 (9) -15, 5x, -7x^2, 2x^3 (10) -4x^2y, 2xy, 5x, 7

Step-by-step explanation:

Sorry if I got one wrong

8 0

2 years ago

Read 2 more answers