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1 year ago

A 6th-grade class did a weekend fitness challenge.

Mathematics

1 answer:

labwork [276]1 year ago

4 0

Using percentages we can conclude that Amanda achieved 77.34% of her goal.

<h3>What is the percentage?</h3>

A percentage is a figure or ratio stated as a fraction of 100 in mathematics.

Although the abbreviations "pct.", "pct.", and occasionally "pc" is also used, the percent symbol, "%," is frequently used to indicate it.

A % is a number without dimensions and without a standard measurement.

For instance, you would have received a 75% grade if you answered 75 out of 100 questions correctly on a test (75/100).

So, percent of goals Amanda achieved:

She exercised for 58 minutes.

Then,

58/75 * 100

77.34%

Therefore, using percentages we can conclude that Amanda achieved 77.34% of her goal.

Know more about percentages here:

brainly.com/question/9553743

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Correct question:
A 6th-grade class did a weekend fitness challenge. Each student set a goal of `75` minutes of fitness. Amanda exercised for `58` minutes. What percent of the goal did she achieve?

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Answer:

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Step-by-step explanation:

Given

$A = 8!$

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Required

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Answer:

$\sf Since \;\sqrt{\boxed{64}}=\boxed{8}\;and\;\sqrt{\boxed{81}}=\boxed{9}\; \textsf{it is known that $\sqrt{75}$ is between}\\\\\sf \boxed{8}\;and\;\boxed{9}\;.$

Step-by-step explanation:

<u>Perfect squares</u>: 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, ...

To find $\sf \sqrt{75}$ , identify the perfect squares immediately <u>before</u> and <u>after</u> 75:

64 and 81

$\begin{aligned}\sf As\;\; 64 < 75 < 81\; & \implies \sf \sqrt{64} < \sqrt{75} < \sqrt{81}\\&\implies \sf \;\;\;\;\;8 < \sqrt{75} < 9 \end{aligned}$

$\sf Since \;\sqrt{\boxed{64}}=\boxed{8}\;and\;\sqrt{\boxed{81}}=\boxed{9}\; \textsf{it is known that $\sqrt{75}$ is between}\\\\\sf \boxed{8}\;and\;\boxed{9}\;.$

See the attachment for the correct placement of $\sf \sqrt{75}$ on the number line.

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