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kotegsom [21]
3 years ago
13

Suppose you are to select a password of size 4. The first character should be any lowercase letter or a digit, while the second

should be a non-zero digit. The third should be any lower case letter, and the fourth should be any lowercase letter, which is different from the third. b. Consider a group of ten students, and six of them are boys. A committee of 3 students is expected to select from the above group. How many committees can you select with at least two girls
Mathematics
1 answer:
Goshia [24]3 years ago
3 0

First question seems incomplete :

Answer:

40 ways

Step-by-step explanation:

Question B:

Number of boys = 6

Number of girls = 4

Number of people in committee = 3

Number of ways of selecting committee with atleast 2 girls :

We either have :

(2 girls 1 boy) or (3girls 0 boy)

(4C2 * 6C1) + (4C3 * 6C0)

nCr = n! ÷ (n-r)!r!

4C2 = 4! ÷ 2!2! = 6

6C1 = 6! ÷ 5!1! = 6

4C3 = 4! ÷ 1!3! = 4

6C0 = 6! ÷ 6!0! = 1

(6 * 6) + (4 * 1)

36 + 4

= 40 ways

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what is the perimeter of the triangle. i do not know how to start this problem. any help will be greatly appreciated :))
xz_007 [3.2K]

Given:

Point F,G,H are midpoints of the sides of the triangle CDE.

FG=9,GH=7,CD=24

To find:

The perimeter of the triangle CDE.

Solution:

According to the triangle mid-segment theorem, the length of the mid-segment of a triangle is always half of the base of the triangle.

FG is mid-segment and DE is base. So, by using triangle mid-segment theorem, we get

\dfrac{1}{2}DE=FG

DE=2(FG)

DE=2(9)

DE=18

GH is mid-segment and CE is base. So, by using triangle mid-segment theorem, we get

\dfrac{1}{2}CE=GH

CE=2(GH)

CE=2(7)

CE=14

Now, the perimeter of the triangle CDE is:

Perimeter=CD+DE+CE

Perimeter=24+18+14

Perimeter=56

Therefore, the perimeter of the triangle CDE is 56 units.

7 0
3 years ago
Pls help im not smart enough ;-;
evablogger [386]
PART1:

First Combination:
Pizza ($7) + Chicken Strips ($6) + Biscuits ($3) + Grapes ($4) = $20

Second Combination:
Dog Food ($13) + Bread ($3) + Crackers ($2) + Broccoli ($2) = $20

Third Combination:
Shampoo ($4) + Tissues ($3) + Pizza ($7) + Eggs ($3) + Biscuits ($3) = $20


PART 2:

First Combination:
$7.20 + $5.70 + $2.90 + $3.70 = $19.60
No, I wouldn’t have gone over the limit

Second Combination:
$13.40 + $3.50 + $2.00 + $1.90 = $20.80
Yes, I would have gone over the limit

Third Combination:
$3.50 + $2.60 + $7.20 + $2.50 + $2.90 = $18.70
No, I wouldn’t have gone over the limit

Hope this helps!!



7 0
3 years ago
Can I help someone with math
Rina8888 [55]

Answer:

please mark my answer brainliest

Step-by-step explanation:

can you tell me a subscript n =sin to the power theta +cosec to the power n theta...and a subscript 1 =2...then prove that a subscript n =2...

8 0
3 years ago
Find the 67th term of the following arithmetic sequence.<br> 14, 20, 26, 32, ...
Jlenok [28]

Answer:

410

Step-by-step explanation:

You can use the equation f(n)= 14+(n-1)6 and you substitute in 67 for n.

7 0
3 years ago
Ken has $950 in a savings account at the beginning of the summer. He wants to have at least $600 in the account by the end of th
Neko [114]
Every week he withdraws $35 from the $950 in his savings, and he wants at least $600 by the end of the summer. So we could write:

\sf 950-35x\ge 600

Where 'x' is the number of weeks. Now let's solve it for 'x'.

Subtract 950 to both sides:

\sf -35x\ge -350

Divide -35 to both sides, don't forget to switch the sign when dividing/multiplying by a negative number:

\sf x\le 10

So he can only withdraw at money for at most 10 weeks if he wants at least $600 left in his account.
6 0
3 years ago
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