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gizmo_the_mogwai [7]
2 years ago
12

2. Show that every even number greater than 6 and less than 30 is the sum of two

Mathematics
1 answer:
Archy [21]2 years ago
7 0

Answer:

I don't know

Step-by-step explanation:

khhikhhjjhhhjjhhhgggggghhhhh

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What is the product?
hammer [34]

Answer:

3/88 is your answer

Step-by-step explanation:

4 0
3 years ago
Sociologists say that 85% of married women claim that their husband's mother is the biggest bone of contention in their marriage
hoa [83]

Answer:

(A) 0.377,

(B) 0.000,

(C) 0.953,

(D) 0.047

Step-by-step explanation:

We assume that having a bone of intention means not liking one's Mother-in-Law

(A) P(all six dislike their Mother-in-Law) = (85%)^6 = (.85)^6 = 0.377

(B) P(none of the six dislike their Mother-in-Law) =

(100% - 85%)^6 =

0.15^6 =

0.000

(C) P(at least 4 dislike their Mother-in-Law) =

P(exactly 4 dislike their Mother-in-Law) + P(exactly 5 dislike their Mother-in-Law) + P(exactly 6 dislike their Mother-in-Law) =

C(6,4) * (.85)^4 * (1-.85)^2 + C(6,5) * (.85)^5 * (.15)^1 + C(6,6) * (.85)^6 = (15) * (.85)^4 * (.15)^2 + (6) * (.85)^5 * .15 + (1) * (.85)^6 =

0.953

(D) P(no more than 3 dislike their Mother-in-Law) =

P(exactly 0 dislikes their Mother-in-Law) + P(exactly 1 dislikes her Mother) + P(exactly 2 dislike their Mother-in-Law) + P(exactly 3 dislike their Mother-in-Law) =

C(6,0) * (.85)^0 * (.15)^6 + C(6,1) * (.85)^1 * (.15)^5 + C(6,2) * (.85)^2 * (.15)^4 + C(6,3) * (.85)^3 * (.15)^3 =

(1)(1)(.15)^6 + (6)(.85)(.15)^5 + (15)(.85)^2 *(.15)^4 + (20)(.85)^3 * (.15)^3 =

0.047

3 0
2 years ago
Boris choos three diffrent numbers.the sum of the three numbers is 36.One of trh numbers is a cube number.the other two number a
inna [77]

Answer:

4,5,27

Problem:

Boris chose three different numbers.

The sum of the three numbers is 36.

One of the numbers is a perfect cube.

The other two numbers are factors of 20.

Step-by-step explanation:

Let's pretend those numbers are:

a,b, \text{ and } c.

We are given the sum is 36: a+b+c=36.

One of our numbers is a perfect cube. a=n^3 where n is an integer.

The other two numbers are factors of 20. bk=20 and ci=20 where a,c,i, \text{ and } k \text{ are integers}.

n^3+\frac{20}{k}+\frac{20}{i}=36

From here I would just try to find numbers that satisfy the conditions using trial and error.

3^3+\frac{20}{2}+\frac{20}{2}

27+10+10

47

3^3+\frac{20}{4}+\frac{20}{5}

27+5+4

36

So I have found a triple that works:

27,5,4

The numbers in ascending order is:

4,5,27

4 0
3 years ago
Sky is a member of her school’s golf team. The table shows her scores from each of her matches. What was her average golf score?
barxatty [35]
70 70 70 70 70 70 70
6 0
1 year ago
Read 2 more answers
Please help! Math problem is attached!!
sesenic [268]

Answer: i gotchu, it would be 22 ounces

Step-by-step explanation:

3 0
3 years ago
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