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postnew [5]
3 years ago
12

What is the smallest power of 10 that would exceed 999,999,999,991

Mathematics
2 answers:
Tju [1.3M]3 years ago
6 0
10 to the power of 11 = 100000000000
Anit [1.1K]3 years ago
3 0
69..............should of listened in class
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Based on the limited amount of available student parking spaces on the GSU campus, students are being encouraged to ride their b
vfiekz [6]

Answer:

1

Step-by-step explanation:

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3 years ago
The sum of three integers is 74. The first integer is twice the second and the third is six more than the second. What are the t
Levart [38]
Let thw second integer be x
then the first = 2x
and the third integer is = 6+x
their sum = 74
x+2x+6+x = 74
4x +6 = 74
4x = 74-6 =68
x =68÷4
x = 17
then the first integer is 2×17 = 34
the third integer is 17+6 = 23

the integers are 17 , 23 and 34
5 0
3 years ago
What is the distance between (3,-9) and (6,-2)?
exis [7]

Answer:

7/3

Step-by-step explanation:

y2-y1/x2-x1

-2-(-9)/6-3 = -2+9/3 = 7/3

8 0
3 years ago
Is x = -3 a solution to 5x + 7 = 4
SVEN [57.7K]
So, 5 x (-3) + 7 = -15 + 7 = - 8 , but not 4;
Then, -3 is not a solution to 5x + 7 = 4;
6 0
4 years ago
2. (For this problem, leave your answers in Combinatorics format – do NOT simplify). A conference is being held that faculty mem
SSSSS [86.1K]

Answer:

Follows are the solution to this question:

Step-by-step explanation:

please find the complete question in the attached file.

female = 23  and male = 19 faculty members

Randomly selected member are = 6

In point 1:

The Probability to equally selecting men and women:

= \frac{(23C3+19C3)}{42C6}

In point 2:

The Probability of selecting at least 4 men:

= \frac{((19C4 \times 23C2)+(19C5 \times 23C1)+(19C6))}{42C6}

In point 3:

The probability in which no women are chosen implies that 6 selected members are from 23 women:

= \frac{23C6 }{42C6}

In point 4:

The Probability in which no more than 4 women are selected:

\frac{((19C6)+(23C1 \times 19C5)+(23C2 \times 19C4)+(23C3 \times 19C3)+(23C4 \times 19C2))}{42C6}

8 0
3 years ago
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