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Umnica [9.8K]
2 years ago
5

Translated up 3 units and left 2 units

Mathematics
1 answer:
seraphim [82]2 years ago
6 0
What are we trying to answer here lol
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3 4/6 mini-pizzas were left in the fridge; the children ate 1 2/3 of them. How much pizza is left for the adults? Give your answ
Naya [18.7K]

Answer:

It would be 2 whole pizza left for the adults. the improper fraction would be 2 1/2 or 5/2

Step-by-step explanation:

8 0
2 years ago
PLS HELP 55 POINTS PLSS NO LINKS PLS HELPP
Semenov [28]

Answer:

12. 114 Laps

13. 234 students don't ride the bus.

14. 1.5% of 26,500.60 is 397.50 so the answer is $26,500.60

15. 24% × 63.4 = 15.2 so the original weight of the bag is 63.4 pounds

Step-by-step explanation:

4 0
2 years ago
40% of 50 is what number
nikitadnepr [17]

20

Step-by-step explanation:

Step 1:

Let the number be 50 and to find 40% of  50 is given interms of expression as follows

To express the percentage the following strategy is used

Eg: 40% = 40/100

∴ To express 40% of 50 is

(40/100)*50

Step 2:

On simplification the above expression we could get

0.4*50

= 20

4 0
2 years ago
Read 2 more answers
What is the length of line segment ab
Leokris [45]
The length of the line segment here is 13.

Distance Formula: 
√((x₂ - x₁)² + (y₂ - y₁))
√((5 - 0)² + (0 - 12)²) = 13
5 0
3 years ago
Read 2 more answers
For a binomial distribution with p = 0.20 and n = 100, what is the probability of obtaining a score less than or equal to x = 12
notsponge [240]
The binomial distribution is given by, 
P(X=x) =  (^{n}C_{x})p^{x} q^{n-x}
q = probability of failure = 1-0.2 = 0.8
n = 100
They have asked to find the probability <span>of obtaining a score less than or equal to 12.
</span>∴ P(X≤12) = (^{100}C_{x})(0.2)^{x} (0.8)^{100-x}
                    where, x = 0,1,2,3,4,5,6,7,8,9,10,11,12                  
∴ P(X≤12) = (^{100}C_{0})(0.2)^{0} (0.8)^{100-0} + (^{100}C_{1})(0.2)^{1} (0.8)^{100-1} + (^{100}C_{2})(0.2)^{2} (0.8)^{100-2} + (^{100}C_{3})(0.2)^{3} (0.8)^{100-3} + (^{100}C_{4})(0.2)^{4} (0.8)^{100-4} + (^{100}C_{5})(0.2)^{5} (0.8)^{100-5} + (^{100}C_{6})(0.2)^{6} (0.8)^{100-6} + (^{100}C_{7})(0.2)^{7} (0.8)^{100-7} + (^{100}C_{8})(0.2)^{8} (0.8)^{100-8} + (^{100}C_{9})(0.2)^{9} (0.8)^{100-9} + (^{100}C_{10})(0.2)^{10} (0.8)^{100-10} + (^{100}C_{11})(0.2)^{11} (0.8)^{100-11} + (^{100}C_{12})(0.2)^{12} (0.8)^{100-12}


Evaluating each term and adding them you will get,
P(X≤12) = 0.02532833572
This is the required probability. 
7 0
3 years ago
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