How old am i if 300 reduced by 2 times my age is 288?

Mathematics

1 answer:

aleksandr82 [10.1K]2 years ago

5 0

Answer:

Step-by-step explanation:

288 = 300-2x

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Please help ASAP for question b

zlopas [31]

Answer: The height of the new player is 210 cm

Step-by-step explanation: The previous mean of the entire team has been calculated as 200.3

What this means is that, we have a summation of the observed data and a summation of the frequency of data.

The mean was calculated as

Sum FX/Sum F = 200.3

Where Sum FX is 2604 and Sum F is 13

However, our calculation should now read thus,

Sum FX/Sum F = 201 {where 201 is the new mean}

By cross multiplication we now have

Sum FX = Sum F x 201

Remember that a new member has joined the team so our Sum F is now 14 and we can now express it as thus

Sum FX = 14 x 201

Sum FX = 2814

If the summation of the observed data after adding a new team member is now 2814, then the addition to the previous observed data would be

2814 - 2604 = 210

So the height of the new member added to the team is 210 cm.

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4 years ago

HELP PLS<br><br> if i get this correct i will give the one who gave me it brainliest

const2013 [10]

The last answer is the correct one.

If it does not make sense ask for clarification.

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2 years ago

Solve The Equation <br> 4x×9y=7<br> 4x-9y=9

hoa [83]

Answer:

$\large\boxed{x=\dfrac{9}{8}-\dfrac{\sqrt{109}}{8},\ y=-\dfrac{1}{2}-\dfrac{\sqrt{109}}{18}}\\or\\\boxed{x=\dfrac{9}{8}+\dfrac{\sqrt{109}}{2},\ y=-\dfrac{1}{2}+\dfrac{\sqrt{109}}{18}}$

Step-by-step explanation:

$\left\{\begin{array}{ccc}4x\times9y=7&(1)\\4x-9y=9&(2)\end{array}\right\\\\(2)\\4x-9y=9\qquad\text{subtract}\ 4x\ \text{from both sides}\\-9y=-4x+9\qquad\text{change the signs}\\9y=4x-9\qquad\text{substitute it to (1)}\\\\4x(4x-9)=7\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\(4x)(4x)+(4x)(-9)=7\\(4x)^2-36x=7\\(4x)^2-2(4x)(4.5)=7\qquad\text{add}\ 4.5^2\ \text{to both sides}\\(4x)^2-2(4x)(4.5)+4.5^2=7+4.5^2\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2$

$(4x-4.5)^2=7+20.25\\(4x-4.5)=27.25\to 4x-4.5=\pm\sqrt{27.25}\\\\4x-\dfrac{45}{10}=\pm\sqrt{\dfrac{2725}{100}}\\\\4x-\dfrac{45}{10}=\pm\dfrac{\sqrt{2725}}{\sqrt{100}}\\\\4x-\dfrac{45}{10}=\pm\dfrac{\sqrt{25\cdot109}}{10}\\\\4x-\dfrac{45}{10}=\pm\dfrac{\sqrt{25}\cdot\sqrt{109}}{10}\\\\4x-\dfrac{45}{10}=\pm\dfrac{5\sqrt{109}}{10}\qquad\text{add}\ \dfrac{45}{10}\ \text{to both sides}\\\\4x=\dfrac{45}{10}\pm\dfrac{5\sqrt{109}}{10}$

$4x=\dfrac{9}{2}\pm\dfrac{\sqrt{109}}{2}\qquad\text{divide both sides by 4}\\\\x=\dfrac{9}{8}\pm\dfrac{\sqrt{109}}{8}\\\\\text{Put the values of}\ x\ \text{to (2):}\\\\9y=4\left(\dfrac{9}{8}\pm\dfrac{\sqrt{109}}{8}\right)-9\\\\9y=\dfrac{9}{2}\pm\dfrac{\sqrt{109}}{2}-\dfrac{18}{2}\\\\9y=-\dfrac{9}{2}\pm\dfrac{\sqrt{109}}{2}\qquad\text{divide both sides by 9}\\\\y=-\dfrac{1}{2}\pm\dfrac{\sqrt{109}}{18}$