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attashe74 [19]
3 years ago
8

The following represents age distribution of students in an elementary class. Find the mode of the values: 7, 9, 10, 13,

Mathematics
1 answer:
Vitek1552 [10]3 years ago
3 0
The mode of the values listed is 9.
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Peyton received a graduation gift from his grandparents of $12,700. He
arlik [135]

Answer:

Peyton's account will have $13,842.18 after a year.

Step-by-step explanation:

Given that Peyton received $ 12,700 and decided to invest it for a year in an account that grants an interest of 8.8% per year, compounded semiannually, to determine the amount of money that will be in said account after the passage of one year, it is necessary to perform the following calculation:

X = 12,700 (1 + 0.088 / 2) ^ 1x2

X = 13,842.18

Therefore, after a year has passed, Peyton's account will be $ 13,842.18.

7 0
3 years ago
4(2x+8)=5(x+4)<br>solve for X​
DerKrebs [107]

x = -4

Have a Merry Christmas!

6 0
3 years ago
Neal opened a savings account 11 years ago with a deposit of $2,335.69. The account has an interest rate of 4.3% compounded dail
Vinvika [58]

Answer: $1,412.52

Step-by-step explanation:

Formula to calculate the accumulated amount if <em>P</em> principal invested for <em>t </em>years at a rate of interest <em>r</em> that compounded daily is given by:-

A=P(1+\dfrac{r}{365})^{365t}

Given: P= $2,335.69

r= 4.3%= 0.043

t= 11 years

Then,

A=2335.69(1+\dfrac{0.043}{365})^{365\times11}\\\\=2335.69(1+0.000117808219178)^{4015}\\\\=2335.69(1.000117808219178)^{4015}\\\\= 2335.69(1.6047566747)\\\\=3748.09044309\approx3748.21\\\\\Rightarrow\ A=\$3748.21

Interest earned = A-P

= $3748.21- 2335.69.

= $1412.52

Hence, Neal earned $1,412.52 as interest.

8 0
3 years ago
Need help explaining why this happens
Charra [1.4K]
Just like 2 squared = 4, the square root of 2 multiplied by itself will result in a product of 2. Squaring a number always means multiplying that number by the same number.
8 0
3 years ago
Determine the zeroes of the polynomial
Anna71 [15]

Answer:

3,7/6

Step-by-step explanation:

(\sqrt{x^2-4x+3} )+(\sqrt{x^{2} -9} )-(\sqrt{4x^2-14x+6} )\\=(\sqrt{x^2-x-3x+3} )+(\sqrt{(x^2-3^2})-(\sqrt{4x^2-2x-12x+6})\\ =(\sqrt{x(x-1)-3(x-1)} )+\sqrt{(x+3)(x-3)}-\sqrt{2x(2x-1)-6(2x-1)}  \\=\sqrt{(x-1)(x-3)}+\sqrt{(x+3)(x-3)}  -\sqrt{2(2x-1)(x-3)} \\=\sqrt{x-3} (\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} )\\

\sqrt{x-3} =0~gives~x=3\\or~\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} =0\\or~ \sqrt{x-1} +\sqrt{x+3} =\sqrt{2(2x-1)} \\squaring\\x-1+x+3+2\sqrt{x-1} \sqrt{x+3} =2(2x-1)\\2x+2+2\sqrt{(x-1)(x+3)} =4x-2\\2\sqrt{x^2-x+3x-3} =2x-4\\\sqrt{x^2+2x-3} =x-2\\again ~squaring\\x^2+2x-3=x^2-4x+4\\\\2x+4x=4+3\\6x=7\\x=\frac{7}{6}

8 0
3 years ago
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