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icang [17]
2 years ago
7

Simplify the following expression 4/1/4-5/2

Mathematics
1 answer:
lorasvet [3.4K]2 years ago
5 0

Answer:

1\frac{3}{4}

Step-by-step explanation:

4 \frac{1}{4} - \frac{5}{2} = 1 \frac{3}{4}

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Including a 6% sales tax, an inn charges $155.82 per night. Find the inn's nightly cost before tax is added.
Feliz [49]

we know the Inn is charging 155.82 and that's 100% plus 6%, namely 106% of an amout of say that is "x", being the 100%.

since we know that 155.82 is 106%, what is "x"?

\begin{array}{ccll} amount&\%\\ \cline{1-2} 155.82&106\\ x&100 \end{array}\implies \cfrac{155.82}{x}=\cfrac{106}{100}\implies \cfrac{155.82}{x}=\cfrac{53}{50} \\\\\\ 7791=53x\implies \cfrac{7791}{53}=x\implies 147=x

3 0
2 years ago
The triangle on the grid will be translated two units left. which shows the triangle translated 2 units left
Mariana [72]

Answer:

Take the coordinates of each vertex (corner point) and subtract 2 from the x-coordinate; leave the y-coordinate alone.

Step-by-step explanation:

Example: If one vertex is the point (5, 3), then it moves left 2 units to (3, 3).

If a vertex is at (-3, 1), then it moves 2 units left to (-5, 1).

5 0
3 years ago
Frist one gets brainist the questions are all the the picture ​
Nina [5.8K]
4) 4m, 18, and g
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7 0
3 years ago
Read 2 more answers
Pls help: Write 2^100 as a power with the following base. For example: 4^50<br> 1)16 2)32 3)1024
geniusboy [140]

2^{100} = 16^{25} =32^{20} =1024^{10}

P/s: 2^{100} = 2^{4.25} =16^{25} ......

ok done. Thank to me :>

7 0
2 years ago
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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