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siniylev [52]
3 years ago
9

1.all real numbers that are less than -3 or greater than or equal to 5

Mathematics
2 answers:
Paha777 [63]3 years ago
7 0

1   is   a

2   is   b

3   is  b

hope   it   helped


SpyIntel [72]3 years ago
3 0
1) Is A- x < -3, or x> 5 (d=cant put the line under the second one) 
2) C- 25<*with line under* x >with line under*30 Because its between 25, so must be more, and 30, so has to be less than 30
3)?
4)?
                              
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Un padre para estimular a su hijo para estudiar matemáticas le dice: por cada ejercicio que resuelvas bien te daré 7 € y por cad
Naya [18.7K]

Answer:

8 ejercicios

Step-by-step explanation:

De acuerdo a la información dada, puedes plantear las siguientes ecuaciones:

x+y= 45 (1)

7x-1y= 19 (2)

x= ejercicios que resolvió bien

y= ejercicios que resolvió mal

Primero, puedes despejar x en (1):

x= 45-y (3)

Segundo, debes reemplazar (3) en (2):

7(45-y)-1y=19

315-7y-1y= 19

315-8y= 19

315-19= 8y

296= 8y

y= 296/8

y= 37

Tercero, debes reemplazar el valor de y en (3) para encontrar x:

x= 45-37

x= 8

De acuerdo a esto, la respuesta es que el hijo ha resuelto bien 8 ejercicios.

8 0
3 years ago
How do yu solve log7-2log 12 ?
Alekssandra [29.7K]
Log 7 - 2log 12 = log \frac{7}{ 12^{2} } ≈ -1.31
4 0
3 years ago
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kati45 [8]

Answer:

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Step-by-step explanation:

4 0
3 years ago
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2765 divided by 9 partial quoents
storchak [24]

Answer:307.2222 or just 307

Step-by-step explanation:

3 0
3 years ago
xP(x)00.2510.0520.1530.55Find the standard deviation of this probability distribution. Give your answer to at least 2 decimal pl
Nookie1986 [14]

The Solution:

Given:

Required:

Find the standard deviation of the probability distribution.

Step 1:

Find the expected value of the probability distribution.

E(x)=\mu=\sum_{i\mathop{=}0}^3x_iP_(x_i)\begin{gathered} \mu=(0\times0.25)+(1\times0.05)+(2\times0.15)+(3\times0.55) \\  \\ \mu=0+0.05+0.30+1.65=2.0 \end{gathered}

Step 2:

Find the standard deviation.

Standard\text{ Deviation}=\sqrt{\sum_{i\mathop{=}0}^3(x_i-\mu)^2P_(x_i)}=(0-2)^2(0.25)+(1-2)^2(0.05)+(2-2)^2(0.15)+(3-2)^2(0.55)=4(0.25)+1(0.05)+0(0.15)+1(0.55)=1+0.05+0+0.55=1.60

Thus, the standard deviation is 1.60

Answer:

1.60

8 0
1 year ago
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