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tatiyna
3 years ago
9

For which number does the 9 have the least value? 0.29 7.079 0.9 9.001

Mathematics
2 answers:
OleMash [197]3 years ago
8 0
7.079 is in the hundreths place
julia-pushkina [17]3 years ago
3 0

Answer:

9 has the least value in the number 7.079.

Step-by-step explanation:

Let's see the value of 9 in each case.

0.29

In this case, 9 is two places after the decimal point. So it has the value of 9 hundreths, so 0.09.

7.079

In this case, 9 is three places after the decimal point. So it has the value of 9 thousandths, so 0.009.

0.9

In this case, 9 is one place after the decimal point. So it has the value of 9 tenths, so 0.9.

9.001

In this case, 9 is before the decimal point. So it has the value of 9.

So 9 has the least value in the number 7.079.

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3 years ago
Read 2 more answers
Can someone please help me with this math question
inn [45]

Answer:

1. reflection across x-axis

2. translation 6 units to the right and 3 units up (x+6,y+3)

Step-by-step explanation:

The trapezoid ABCD has it vertices at points A(-5,2), B(-3,4), C(-2,4) and D(-1,2).

First transformation is the reflection across the x-axis with the rule

(x,y)→(x,-y)

so,

  • A(-5,2)→A'(-5,-2)
  • B(-3,4)→B'(-3,-4)
  • C(-2,4)→C'(-2,-4)
  • D(-1,2)→D'(-1,-2)

Second transformation is translation 6 units to the right and 3 units up with the rule

(x,y)→(x+6,y+3)

so,

  • A'(-5,-2)→E(1,1)
  • B'(-3,-4)→H(3,-1)
  • C'(-2,-4)→G(4,-1)
  • D'(-1,-2)→F(5,1)

7 0
3 years ago
The parrots at the top of the tree is about 10 inches long and has a wing span of 15 inches what fraction represents the birds l
QveST [7]
The correct answer is 10/15, simplified to 2/3.
6 0
3 years ago
A particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams. The heavies
RoseWind [281]

Answer:

The heaviest 5% of fruits weigh more than 747.81 grams.

Step-by-step explanation:

We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.

Let X = <u><em>weights of the fruits</em></u>

The z-score probability distribution for the normal distribution is given by;

                              Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean weight = 733 grams

            \sigma = standard deviation = 9 grams

Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;

                    P(X > x) = 0.05       {where x is the required weight}

                    P( \frac{X-\mu}{\sigma} > \frac{x-733}{9} ) = 0.05

                    P(Z > \frac{x-733}{9} ) = 0.05

In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;

                               \frac{x-733}{9}=1.645

                              {x-733}}=1.645\times 9

                              x = 733 + 14.81

                              x = 747.81 grams

Hence, the heaviest 5% of fruits weigh more than 747.81 grams.

8 0
2 years ago
Calcula y comprueba las ecuaciones: plisss lo necesito alguien me puede ayudar a) 2X = 6 b) 10 + Z = 20 c) P + 9 = 11 d) 3X + 8
vesna_86 [32]

Answer:

a) x=3

b) z=10

c) P= 2

d) X=7

e) U=1

Step-by-step explanation:

Resolver una ecuación consiste en hallar los valores de la variable que hacen cierta la igualdad.

a) 2x= 6

El coeficiente es el número junto a la variable. En este caso, el coeficiente es 2. Para eliminar este número en la expresión 2x, debido a que la variable x esta multiplicada por 2, deberás dividir ambos lados de la ecuación entre 2, debido a que la operación opuesta de la multiplicación es la división.

(2x)÷2=6÷2

x= 3

Comprobar la solución de una ecuación se hace al remplazar la variable en una ecuación con el valor de la solución. La solución debería satisfacer la ecuación cuando se ingresa en esta.

En este caso:

2*3= 6

6=6

b) 10 + z= 20

En este caso se debe sumar o restar la constante que se encuentra acompañando a la variable en ambos lados de la ecuación de manera de aislar el término de la variable. En este caso:

10 - 10 + z= 20 -10

z= 10

Comprobación:

10 + z=20

10 + 10=20

20=20

c) P + 9= 11

P +9 - 9= 11 -9

P=2

Comprobación:

2 + 9= 11

11=11

d) 3X + 8 = 29

En este caso, se suma o resta la constante en ambos lados de la ecuación y luego se elimina el coeficiente de la variable mediante la división o multiplicación. Esto es:

3X + 8 - 8= 29 - 8

3X= 21

3X ÷3= 21÷3

X=7

Comprobación:

3*7 + 8=29

21+8=29

29=29

e) 2U + 8= 10

2U + 8 - 8= 10 -8

2U= 2

2U ÷2= 2÷2

U=1

Comprobación:

2*1 + 8= 10

2 + 8= 10

10=10

8 0
2 years ago
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