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pantera1 [17]
3 years ago
12

Solve the equation 1/3 + 1/4

Mathematics
2 answers:
Shkiper50 [21]3 years ago
8 0
1/3 + 1/4
4/12 + 3/12
4+3 divided by 12
The answer is 7/12
Ronch [10]3 years ago
6 0
1/3+1/4 = 4/12 + 3/12 = 7/12
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James works in a sporting goods store and earns $324 a week and 5% of his sales. One week James earned $432. What were his sales
Bogdan [553]

Answer:

His sales that week were $2,160.

Step-by-step explanation:

First, you have to subtract $324 from the amount he earned that week, to find the 5% he got from sales:

$432-$324=$108

Now, you know that he received $108 that represent 5% of his sales and you can use a rule of three to find the amount that represents 100% which would be his sales that week:

 5%   →    108

100% →      x

x=(100*108)/5=2160

According to this, the answer is that his sales that week were $2,160.

7 0
3 years ago
HOW DO I SOLVE THIS PLEASE HELP
malfutka [58]

Answer:

  x = 15

Step-by-step explanation:

We assume you want to find the value of x.

Know (or prove) that in this geometry, all of the right triangles are similar. That means the ratios of corresponding sides are proportional.

  short side / hypotenuse = 9/x = x/25

  x^2 = (9)(25) . . . . . . . . . . multiply by 25x ("cross multiply")

  x = √((9)(25)) = (3)(5) . . . take the square root

  x = 15

3 0
2 years ago
Help no time plzzzz plz plz​
Sav [38]

Answer:

FALSE

Step-by-step explanation:

<E in ∆AED ≅ <E in ∆CEB.

Both are 90°.

Side ED ≅ Side EB

Side AD ≅ Side CB.

Thus, two sides (ED and AD) and a non-included angle (<E) of ∆AED are congruent to corresponding two sides (EB and CB) and a non-included angle (<E) of ∆CEB. Therefore, by A-S-S Congruence Theorem, both triangles are congruent to each other not by SSS.

8 0
3 years ago
A pole that is 3.1m tall casts a shadow that is 1.29m long. At the same time, a nearby building casts a shadow that is 44.75m lo
Mkey [24]
Let b=height of building.   By similar triangles we can say:

b/44.75=3.1/1.29

b=44.75*3.1/1.29

b≈107.54

b≈108 meters  (to the nearest whole meter)
8 0
3 years ago
Read 2 more answers
Radioactive isotopes are commonly used to estimate artifact ages. Let X be the time in years for a Carbon-14 atom to decay to Ni
Alexeev081 [22]

Answer:

a) 1 / years

b) P ( 5000 < X < 6000) =  0.06222

c) 8267 years

d) T_1/2 = 5730.25 years          

Step-by-step explanation:

Given:

- The decay constant λ = 1 / 8267

- X is an exponential random Variable

Find:

a. What are the units of What are the units of λ?

b. What is the probability the decay time is between 5000 and 6000 years?

c. Compute SD(X).

d. Compute the median decay time. The result is called the half-life of Carbon-14?

Solution:

- The decay constant λ means the rate at which the C-14 atoms decays into N-14 atom. The rate is expressed in units of 1 / year.

- The random variable X follows an exponential distribution which has a probability mass function and cumulative density functions as follows:

                    P ( X = t ) = λ*e^(-λ*t)

                    P ( X = t ) = e^(-t/8267) / 8267

                    P ( X < t ) = 1 - e^(-λ*t) = 1 - e^(-t/8267)

- The probability of the decay time between years 5000 and 6000 years is:

                    P ( 5000 < X < 6000) = 1 - e^(-6000/8267) - 1 + e^(-5000/8267)

                    P ( 5000 < X < 6000) =  0.06222

- The standard deviation of the exponential distribution is given by:

                    SD(X) = 1 / λ = 8267 years

- The median decay time for an exponential distribution is given by:

                    T_1/2 = Ln(2) / λ = Ln(2)*8267

                    T_1/2 = 5730.25 years                        

3 0
3 years ago
Read 2 more answers
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