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3 years ago

Plz help for 60 points

Mathematics

2 answers:

ycow [4]3 years ago

8 0

4. 22
5. 37
6. 108
7. 494
8. 168

I am not sure what seven is since the picture has cropped it out, sorry. Hope this helps!

Happy learning! - Ann

Brilliant_brown [7]3 years ago

7 0

Answer:

1. 22

2. 37.

and....108 are the answers for this question.

Step-by-step explanation:

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tatyana61 [14]

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2 years ago

Please solve with explanation (high points)

IgorLugansk [536]

So, we know that:

$A=\dfrac{1}{2}ab\\P=a+b+c\\a^2+b^2=c^2$

We also know that $(a+b)^2=a^2+2ab+b^2$ . We will use that fact to find $c$ in terms of $A$ and $P$ .

$A=\dfrac{1}{2}ab\Rightarrow 2ab=4A\\P=a+b+c\Rightarrow a+b=P-c$

Therefore:

$(P-c)^2=c^2+4A\\P^2-2cP+c^2=c^2+4a\\2cP=P^2-4A\\c=\dfrac{P^2-4A}{2P}$

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2 years ago

Give the amplitude, vertical and horizontal shift, and period of the function.

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Vertical shift up then down

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3 years ago

Read 2 more answers

A chemist is using 362 milliliters of a solution of acid and water. If 12.1% of the solution is acid, how many milliliters of ac

PSYCHO15rus [73]

Answer:

The answer to your question is 43.8 mL of acid.

Step-by-step explanation:

Data

Total volume = 362 ml

12.1 % is acid

The Volume of acid = ?

Process

1.- Use the rule of three and cross multiplication to solve this problem.

362 ml ------------------- 100%

x ------------------- 12.1 %

x = (12.1 x 362) / 100

x = 4380.2 / 100

x = 43.802 ml

2.- Round your answer to the nearest tenth.

43.802 only consider the first decimal

43.8 mL

3.- Conclusion

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3 years ago

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The class sizes of elementary school classes in a public school district are normally distributed with an unknown population mea

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Answer: $(17.14,\ 22.86)$

Step-by-step explanation:

The confidence interval estimate for the population mean is given by :-

$\overline{x}\pm ME$ , where $\overline{x}$ is the sample mean and ME is the margin of error.

Given : Sample mean: $\overline{x}=20$

The margin of error for a 98% confidence interval estimate for the population mean using the Student's t-distribution : $ME=2.86$

Now, the confidence interval estimate for the population mean will be :-

$20\pm2.86=(20-2.86,\ 20+2.86)=(17.14,\ 22.86)$

Hence, the 98% confidence interval estimate for the population mean using the Student's t-distribution = $(17.14,\ 22.86)$

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3 years ago