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3 years ago

A whole number rounded to the nearest ten thousand is 620,000. What is the least possible value for that number?

Mathematics

1 answer:

Ymorist [56]3 years ago

4 0

Answer:

615,000

Step-by-step explanation:

Given that:

Whole number rounded to the nearest ten thousand is 620,000

Let x = the whole number

HTh__ TTh ____ Th ____ H ____ T ____ U

6_____ 2 ______ 0 ____ 0 ____ 0 ____ 0

The ten thousand digit is 0 ;

To round to the nearest 10 thousand, the thousand digit is rounded up to 1 and added to the ten thousand digit if it up to 5 or above and rounded down to 0 if below 5 with all subsequent digits(hundred, tens and unit) rounded to Zero..

Hence,

The whole number will could be range between :

615,000 - 619,999

Hence, the least possible value for the number is 615,000.

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Simplify the radical

ElenaW [278]

Answer:

Step-by-step explanation:

4 to the 4th power is 256

7 0

3 years ago

Help please need to turn In tomorrow

qwelly [4]

21/72
then simplified
= 7/24

7 0

3 years ago

Un futbolista ha metido 2/9 del número de goles marcados por su equipo y otro la

storchak [24]

Answer:

77 goles

Step-by-step explanation:

Un futbolista ha marcado 2/9 del número de goles marcados por su equipo.

Otro anotó una cuarta parte del resto.

El resto = 1 - 2/9

= 7/9

De ahí que otro futbolista anotó

= 1/4 de 7/9

= 7/36

Si los otros jugadores han marcado 45 goles

Tenemos que averiguar la fracción de goles que marcó el otro jugador

Deje que el número total de goles marcados por el equipo durante la temporada = 1

Por lo tanto:

1 - (2/9 + 7/36)

1 - (8 + 7/36)

1 - 15/36

1 - 5/12

= 7/12

¿Cuántos goles marcó el equipo a lo largo de la temporada?

El número total de goles que marcó ese equipo se calcula como:

7/12 × x = 45

Donde x = número total de goles

7x / 12 = 45

Cruz multiplicar

7x = 45 × 12

x = 45 × 12/7

x = 77,142857143

Aproximadamente = 77 goles

7 0

3 years ago

What is the result of substituting for y in the bottom equation y=x+3 y=x^2+2x-4

8_murik_8 [283]

The solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

Solution:

Given that,

$y = x + 3 ------- eqn 1\\\\y = x^2 + 2x - 4 ----- eqn 2$

We have to substitute eqn 1 in eqn 2

$x + 3 = x^2 + 2x - 4$

$\mathrm{Switch\:sides}\\\\x^2+2x-4=x+3\\\\\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}\\\\x^2+2x-4-3=x+3-3\\\\\mathrm{Simplify}\\\\x^2+2x-7=x\\\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\\\x^2+2x-7-x=x-x\\\\\mathrm{Simplify}\\\\x^2+x-7=0$

$\mathrm{Solve\:with\:the\:quadratic\:formula}\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=1,\:c=-7\\\\x =\frac{-1\pm \sqrt{1^2-4\cdot \:1\left(-7\right)}}{2\cdot \:1}$

$x = \frac{-1 \pm \sqrt{ 1 + 28}}{2}\\\\x = \frac{ -1 \pm \sqrt{29}}{2}$

$x = \frac{ -1 \pm 5.385 }{2}\\\\We\ have\ two\ solutions\\\\x = \frac{ -1 + 5.385 }{2}\\\\x = 2.1925$

$Also\\\\x = \frac{ -1 - 5.385 }{2}\\\\x = -3.1925$

Substitute x = 2.1925 in eqn 1

y = 2.1925 + 3

y = 5.1925

Substitute x = -3.1925 in eqn 1

y = -3.1925 + 3

y = -0.1925

Thus the solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

6 0

4 years ago

In the June 2007 issue, Consumer Reports also examined the relative merits of top-loading and front-loading washing machines, te

saveliy_v [14]

Answer:

Yes

Step-by-step explanation:

Given that in the June 2007 issue, Consumer Reports also examined the relative merits of top-loading and front-loading washing machines, testing samples of several different brands of each type.

The difference in mean values test gave a p value of 0.32

Confidence level = 95%

Alpha = 1-0.95 = 0.05

Compare p with alpha, here p >alpha

Hence we accept null hypothesis that there is no difference in the means.

Confidence interval method also will yield the same result. i.e. confidence interval for difference of means would definitely contain 0 at 95% conf level.

So answer is yes

3 0

3 years ago