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kvv77 [185]
3 years ago
9

Rachel spent $628 more than Samuel if Samuel spent $329 then how much did Rachel spent

Mathematics
2 answers:
RSB [31]3 years ago
7 0

Answer:

$957

Step-by-step explanation:

As the problem says Rachael spent more than Samuel then that means R=S+628. 329+628=957

sukhopar [10]3 years ago
7 0

Step-by-step explanation:

S:R+628

329:329+628=957

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Joe buys groceries for his grandfather the groceries cost $42 plus 5% tax joe wants to know the total cost of the groceries incl
vekshin1
<h3>Answer: $44.10</h3>

====================================================

Work Shown:

5% = 5/100 = 0.05

5% of 42 = 0.05*42 = 2.10

tax = $2.10

total cost including tax = $42 + $2.10 = $44.10

----------

An alternative is to multiply 42 by 1.05 to get

1.05*42 = 44.10

We get the same answer. The 1.05 refers to 1 + 0.05 where the "1" is 100% and the 0.05 is 5%, so the two combine to 105%, meaning the total cost is 105% of the cost before tax is considered.

7 0
3 years ago
Help me please ASAP!!
Tcecarenko [31]
Hello there k12 user. - i am in k12 too :)

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6 0
3 years ago
Can anyone help with math models?? Please…?
Solnce55 [7]

Answer:

12*1/4= 3

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31 is B

8 0
2 years ago
Read 2 more answers
Of all college degrees awarded in the United States, 50% are bachelor's degrees, 59% are earned by women, and 29% are bachelor's
Natasha2012 [34]

Answer:

  P(W|B) = 0.58

Step-by-step explanation:

The conditional probability formula tells you ...

  P(W | B) = P(W&B) / P(B)

  = 0.29/.050

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7 0
3 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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