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Anarel [89]
4 years ago
15

Simplify completely 4 m cubed over 3 m squared n times 9 n squared over 30 m squared n squared.

Mathematics
2 answers:
V125BC [204]4 years ago
5 0

Answer: Our simplified form will be

\frac{2}{5mn}

Step-by-step explanation:

Since we have given that

4 m cubed over 3 m squared n times 9 n squared over 30 m squared n squared.

\frac{4m^3}{3m^2n}\times \frac{9n^2}{30m^2n^2}\\\\\text{we will use }a^m\times b^n=a^{m+n}\\\\=\frac{4m}{3n}\times \frac{3}{10m^2}\\\\=\frac{2}{5mn}

Hence, our simplified form will be

\frac{2}{5mn}

astra-53 [7]4 years ago
3 0
\frac{4 m^{3} }{3 m^{2}n }  \times  \frac{9 n^{2} }{30 m^{2}  n^{2} }  \\ = \frac{36}{90}  \times  \frac{m^{3 }}{m^{4}} \times  \frac{n^{2}}{n^{3}}  \\ = \frac{2}{5mn}
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I hope this helps you

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3 years ago
An agency is studying the income of store managers in the retail industry. A random sample of 25 managers reveals a sample mean
aalyn [17]

Answer:

The Confidence Interval = ($44,745.55 , $46,094.45)

Step-by-step explanation:

The formula for Confidence Interval =

Confidence Interval = Mean ± z × Standard deviation/√n

Where n = number of samples = 25 managers

Standard deviation = $2,050

Mean = $45,420

z = z score of the given confidence interval

= z score of 90% confidence interval

= 1.645

Confidence Interval = $45,420 ± 1.645 × $2,050/√25

= $45,420 ± 1.645 × $2,050/5

= $45,420 ± 674.45

Confidence Interval =

$45,420 - 674.45 = $44,745.55

$45,420 + 674.45 = $46,094.45

Therefore, the Confidence Interval = ($44,745.55 , $46,094.45)

3 0
3 years ago
Lisa purchased almonds for $3.50 per pound. She spent a total of $14.70. How many
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Answer:

4.2

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Find the exact values of a) sec of theta b)tan of theta if cos of theta= -4/5 and sin<0
Gre4nikov [31]

Answer:

Using trigonometric ratio:

\sec \theta = \frac{1}{\cos \theta}

\tan \theta = \frac{\sin \theta}{\cos \theta}

From the given statement:

\cos \theta = -\frac{4}{5} and sin < 0

⇒\theta lies in the 3rd quadrant.

then;

\sec \theta = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}

Using trigonometry identities:

\sin \theta = \pm \sqrt{1-\cos^2 \theta}

Substitute the given values we have;

\sin \theta = \pm\sqrt{1-(\frac{-4}{5})^2 } =\pm\sqrt{1-\frac{16}{25}} =\pm\sqrt{\frac{25-16}{25}} =\pm \sqrt{\frac{9}{25} } = \pm\frac{3}{5}

Since, sin < 0

⇒\sin \theta = -\frac{3}{5}

now, find \tan \theta:

\tan \theta = \frac{\sin \theta}{\cos \theta}

Substitute the given values we have;

\tan \theta = \frac{-\frac{3}{5} }{-\frac{4}{5} } = \frac{3}{5}\times \frac{5}{4} = \frac{3}{4}

Therefore, the exact value of:

(a)

\sec \theta =-\frac{5}{4}

(b)

\tan \theta= \frac{3}{4}

7 0
4 years ago
Y= 1/10 tan (60– 60)
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Answer:

0

Step-by-step explanation:

tan(60 - 60)

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