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aalyn [17]
3 years ago
14

Dividing mixed fractions

Mathematics
1 answer:
Sonbull [250]3 years ago
7 0
A) 2 1/3 = 7/3
   7/3 / 1/4 = 7/3 x 4/1 = 28/3 = 9 and 1/3

b) 4 2/5 = 22/5
   1 1/5 = 6/5
22/5 / 6/5 = 22/5 x 5/6 = 110/30 = 3 and 2/3
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Someone thats good at geometry please help asap!
frosja888 [35]

Answer:

x=125

Step-by-step explanation:

∠PAT=180° - ∠ATP - ∠APT

∠PAT=180 - 3x - 3 -4x -4= (173 - 7x)°

3x + 3 + 4x + 4 + 173-7x = 180  

⇔x = 125

4 0
3 years ago
Last week, Rick's Math Website had the following number of visitors: Monday = 250, Tuesday = 501, Wednesday = 260, Thursday = 49
Novosadov [1.4K]
You add them all together and you get 2221
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3 years ago
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AleksAgata [21]
Meter
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Kilowatt
8 0
3 years ago
How do I find the correct Domain.
Damm [24]

Answer:

The domain of a function is the set of all possible inputs for the function. For example, the domain of f(x)=x² is all real numbers, and the domain of g(x)=1/x is all real numbers except for x=0. We can also define special functions whose domains are more limited.

Step-by-step explanation:

Another way to identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis.

hope this helps a little

5 0
3 years ago
A sample of 4 different calculators is randomly selected from a group containing 18 that are defective and 35 that have no defec
Sphinxa [80]
<span>1.

P(</span>at least one of the calculators is defective)= 

1- P(none of the selected calculators is defective).

2.

P(none of the selected calculators is defective)

           =n(ways of selecting 4 non-defective calculators)/n(total selections of 4)

3.

selecting 4 non-defective calculators can be done in C(35, 4) many ways, 

where  C(35, 4)= \frac{35!}{4!31!}= \frac{35*34*33*32*31!}{4!*31!}=  \frac{35*34*33*32}{4!}=  \frac{35*34*33*32}{4*3*2*1}=  52,360

while, the total number selections of 4 out of 18+35=53 calculators can be done in C(53, 4) many ways,

C(53, 4)= \frac{53!}{4!*49!}= \frac{53*52*51*50}{4*3*2*1}= 292,825

4. so, P(none of the selected calculators is defective)=\frac{52,360}{292,825} =0.18


5. P(at least one of the calculators is defective)= 

1- P(none of the selected calculators is defective)=1-0.18=0.82



Answer:0.82
7 0
3 years ago
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