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Delicious77 [7]
3 years ago
14

If there are 400 people that attend the rally that you have to provide paraphernalia to and you spend $8,000, how many shirts di

d you order? How many hats did you order?
Mathematics
2 answers:
den301095 [7]3 years ago
6 0

Answer:

OK

Step-by-step explanation:

Rufina [12.5K]3 years ago
3 0
Uh i just did 8000 divided by 400 and it got me 20 but idk if it’s right
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Multiply————-(b+12)^2
Talja [164]

Use Square of Sum: (a + b)^2 = a^2 + 2ab + b^2

b^2 + 2b * 12 + 12^2

Simplify 12^2 to 144

b^2 + 2b * 12 + 144

Simplify 2b * 12 to 24b

b^2 + 24b + 144

5 0
3 years ago
Help please <br><br>numbers 12, 13, 14, 15<br><br>please show work
kari74 [83]
Those are the answers for 12,13,14,15

5 0
3 years ago
А square tile has a width of foot. How many tiles will fit end-to-end along a<br> 4-foot wall?
irina1246 [14]

Answer:

4 tiles

Step-by-step explanation:

4/1 = 4

5 0
3 years ago
Read 2 more answers
The night watchman in a factory cannot guard both the safe in back and the cash register in front. The safe contains $6000, whil
Zolol [24]

Answer:

The guard should be positioned at the safe

Step-by-step explanation:

The night watchmen should be positioned to guard whichever place gives the highest expected value to the thief.

The expected value of robbing the safe is:

EV_{s} = \$ 6000*0.2\\EV_{s} = \$ 1200

The expected value of robbing the cash register is:

EV_{r} = \$ 1000*0.8\\EV_{r} = \$ 800

Therefore, the guard should be positioned at the safe since it yields a higher expected value to the thief in case he tries to rob it.

7 0
3 years ago
A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee
katrin [286]

Answer: A) 1260

Step-by-step explanation:

We know that the number of combinations of n things taking r at a time is given by :-

^nC_r=\dfrac{n!}{(n-r)!r!}

Given : Total multiple-choice questions  = 9

Total open-ended problems=6

If an examine must answer 6 of the multiple-choice questions and 4 of the open-ended problems ,

No. of ways to answer 6 multiple-choice questions

= ^9C_6=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!3!}=84

No. of ways to answer 4 open-ended problems

= ^6C_4=\dfrac{6!}{4!(6-4)!}=\dfrac{6\times5\times4!}{4!2!}=15

Then by using the Fundamental principal of counting the number of ways can the questions and problems be chosen = No. of ways to answer 6 multiple-choice questions x No. of ways to answer 4 open-ended problems

= 84\times15=1260

Hence, the correct answer is option A) 1260

5 0
4 years ago
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