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satela [25.4K]
3 years ago
9

Mandi already has a practice ball. She is buying the light jacket, but not the lined one. She needs to buy all of the other item

s on the soccer fee list. What are her total fees for soccer. Soccer fees: game uniform- $37.50 Light jacket- $23.25 Lined jacket- $42.00 Practice ball- $22.75 Pictures- $40.50 Hair band- $2.50
Mathematics
1 answer:
seraphim [82]3 years ago
4 0
Hayskdbdbcjdnsnahshkdnf
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What is the simplest form of 4 square root 81x^8y^5?
olasank [31]
Firstly,
thanks for posting this.

81 = 9x9 or 9 squared so at this stage I'm guessing the last two. 
x^ 8 = (x^ 4) x (x^ 2) (its called the multiplication rule of indices). 
Okay I have approx 14 mins to answer.
No panic
√ 81.x^8 x <span>√ y^5
</span>√ (9x^2 x 9x^6) x <span>√ y^5
</span>= 9x x <span>√ 9x^6 x  </span><span>√ y^5
</span>did I just gain time?
√ 9x^6 = <span>√ </span>3x ^3 x <span>√ 3x ^ 3
</span>or 2 . √ 3x^3. <span>√ y^5
</span>more time
okay, so that wasnt successful. 
√ 9x^6 =<span>√ </span> (3x^2) x <span>√ </span>(3x^4)




5 0
3 years ago
"blank" are the products of the smaller problems in the place value rows methed
Talja [164]

Answer:

hindi sulotion kung hindi man sorry nalang

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Step-by-step explanation:

im sosory

8 0
3 years ago
What is the solution of the following system?
alexandr402 [8]
<span>4x-y = 15 3x+2y = -8 Multiplying the first equation by two makes for an easy elimination of the y variable. 8x - 2y = 30 3x + 2y = -8 Add vertically. 11x = 22. Divide by 11 on both sides and get x = 2. Plug into an equation. 4(2) - y = 15 8 - y = 15 8 = 15 + y 8 - 15 = y -7 = y. Thus the solution is indeed (2, -7).</span>
5 0
3 years ago
What is the LCM of 539, and 15?
aev [14]
<u>LCM on 539 and 15 :</u>

539 x 15
= 8 085

Answer : 8 085

8 0
3 years ago
Read 2 more answers
Construct a 99% confidthence interval for the population mean .Assume the population has a normal distribution. A group of 19 ra
Kobotan [32]
<h2>Answer with explanation:</h2>

Confidence interval for mean, when population standard deviation is unknown:

\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}

, where \overline{x} = sample mean

n= sample size

s= sample standard deviation

t_{\alpha/2} = Critical t-value for n-1 degrees of freedom

We assume the population has a normal distribution.

Given, n= 19 , s= 3.8 , \overline{x}=22.4

\alpha=1-0.99=0.01

A) Critical t value for \alpha/2=0.005 and degree of 18 freedom

t_{\alpha/2} = 2.8784

B) Required confidence interval:

22.4\pm ( 2.8744)\dfrac{3.8}{\sqrt{19}}\\\\=22.4\pm2.5058\\\\=(22.4-2.5058,\ 22.4+2.5058)=(19.8942,\ 24.9058)\approx(19.9,\ 24.9)

Lower bound = 19.9 years

Uppen bound = 24.9 years

C) Interpretation: We are 99% confident that the true population mean of lies in (19.9, 24.9) .

5 0
3 years ago
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