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uranmaximum [27]
3 years ago
15

30 students went on the school trip. If the total cost of the trip was $805.50.How much money does each student need to contribu

te
Mathematics
1 answer:
enot [183]3 years ago
7 0

Answer:

$26.85 per student

Step-by-step explanation:

805.50 / 30 = 26.85 per student

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-8(3m-3)=-5m+10(-3m-5) ? :(
yanalaym [24]

Answer:

m=-74/11

Step-by-step explanation:

Distribute the 8 through the Parenthesis.

-24=24=-5m+10(3m-5)

Collect Like terms

-24+24=-35m-50

Move terms

-24m+35m=-50-24

11m=-50-24

11=-74

Divide by 11

Answer- -74/11

8 0
3 years ago
I need help ASAP I’ll give brainliest to that person :)
riadik2000 [5.3K]

Answer:

3

Step-by-step explanation:

first work out wax used for cone using formula

1/3 x 3.14 x r squared x height

using this the total for the candle must be 113

formula for a sphere is 4/3 x 3.14 x r cubed

so we know 113 /3.14 / 4/3 = r cubed

cube root 27 = 3

6 0
3 years ago
How to add a + b using 45a +90b
yulyashka [42]

Answer:

45(a + 2b)

Step-by-step explanation:

45a

step1 in order to factor an integer, we need to repeatedly divide it by the ascending sequence of primes (2, 3, 5...) The number of times that each prime divides the original integer becomes its exponent in the final result

Prime number 3 to the power of 2 equals 9 .

Prime number 5 to the power of 1 equals 5 .

3x^{2} *5

90b

do the same as above

 Prime number 2 to the power of 1 equals 2 .

    Prime number 3 to the power of 2 equals 9 .

    Prime number 5 to the power of 1 equals 5 .

3x^{2} *5*2

3 0
3 years ago
The table below shows all outcomes for rolling 2 dice. The bold numbers 1 through 6 show the possible results for each die. All
Leokris [45]

Part I:

1. If you have to circle each outcome in the table that at least 1 die is a 4, then you have to circle row corresponding to the number 4 and column corresponding to the number 4.

2. If you have to cross out each outcome where the sum of the dice is greater than 5, then you have to cross out all not bold numbers 6, 7, 8, 9, 10, 11 and 12 in the table.

Part II:

Here you can see 36 possible outcomes.

1. The probability that at least 1 die is a 4 is:

Pr(A)=\dfrac{11}{36} - (favorable outcomes are 4+1=5, 4+2=6, 4+3=7, 4+4=8, 4+5=9, 4+6=10, 1+4=5, 2+4=6, 3+4=7, 5+4=9, 6+4=10).

2. The probability that the sum of the dice is greater than 5 is:

Pr(B)=\dfrac{26}{36}=\dfrac{13}{18}.

3. The probability that at least 1 die is a 4 and that the sum of the dice is greater than 5 is

Pr(A\cap B)=\dfrac{9}{36}=\dfrac{1}{4}.

4. The probability that at least 1 die is a 4 or that the sum of the dice is greater than 5 is:

Pr(A\cup B)=\dfrac{28}{36}=\dfrac{7}{9}.

4 0
3 years ago
PLEASE HELP ASAP I WILL MARK BRAINLIEST!!!
Setler79 [48]

the answer on number 14. is ×=3 your welcome now please give me brainlest

3 0
3 years ago
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