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Aloiza [94]
3 years ago
6

Amir can run 151515 kilometers in one hour. How many kilometers can Amir run per minute?

Mathematics
1 answer:
DIA [1.3K]3 years ago
8 0

There is no way that he can run 151515km/h, but if he really can, then :

1 hour = 60 minutes

(151515km/h)/60minutes = 2525,25km/minute

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The answer is 1 unit
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Water safety standards in southern California suggest that if a concentration of adenoviruses in the water is more than 1,000 pe
Alborosie

Answer:

I should use a t-distribution test.

Step-by-step explanation:

A t-distribution test is used in lieu of a standard normal distribution test if the population standard deviation is unknown.

In this case, the population standard deviation is unknown but the sample standard deviation is known, therefore, a t-distribution hypothesis test is appropriate to use.

7 0
3 years ago
Pls help me
Vladimir [108]
<h3>Learning Task 1</h3>

Correct responses;

\displaystyle 1. \hspace{0.3 cm} 7^{-1} = \frac{1}{7}

2. (14·a·b·c)⁰ = 1

\displaystyle 3. \hspace{0.3 cm} 10^{-9} = \frac{1}{10^9}

4. 5·(x·y)⁰ = 5

5. 0¹⁵ = 0

\displaystyle 6. \hspace{0.3 cm} \frac{24 \cdot x^8 \cdot y^4}{6 \cdot x^5 \cdot y^4} = 4 \cdot x^3

\displaystyle 7. \hspace{0.3 cm} \left(\frac{5 \cdot x^0}{y} \right)^{-1} = \frac{y}{5}

8. \hspace{0.3 cm}\displaystyle \frac{120 \cdot a^5 \cdot b^6  \cdot c^{-5} }{12 \cdot a^{-2} \cdot b^0  \cdot c^{2}}  = \frac{10 \cdot a^7  \cdot b^6}{c^7}

\displaystyle 9. \hspace{0.3 cm} \frac{(4 \cdot x)^0  \cdot y^{-5} \cdot z^{-2} }{(234  \cdot x  \cdot y \cdot z)^0}  =  \frac{1}{y^{5} \cdot z^2}

\displaystyle 10. \hspace{0.3 cm} \left(\frac{(5 \cdot x \cdot y)^0}{10} \right)^{-2} = 100

11. \displaystyle \hspace{0.3 cm} \left(\frac{(a \cdot b)}{9} \right)^{-1} = \frac{9}{a \cdot b}

\displaystyle 12. \hspace{0.3 cm} \frac{9}{x^{-2}}  = 9 \cdot x^2

13. \displaystyle \hspace{0.3 cm} \frac{12 \cdot b^{-3}}{a^{-4}}  = \frac{12 \cdot a^4}{b^3}

14. \displaystyle \hspace{0.3 cm} \frac{1}{a^{-5 \cdot n}}  =a^{5 \cdot n}

15. \displaystyle \hspace{0.3 cm} \left(\frac{1}{2} \right)^{-5} = 32

<h3>Methods by which the above expressions are simplified</h3>

The given expressions expressed into non zero and non negative exponents are;

\displaystyle 1. \hspace{0.3 cm} \mathbf{7^{-1}} = \underline{\frac{1}{7}}

2. (14·a·b·c)⁰ = 14⁰ × a⁰ × b⁰ × c⁰ = 1 × 1 × 1 × 1 =<u> 1</u>

\displaystyle 3. \hspace{0.3 cm}  \mathbf{ 10^{-9}} =  \underline{ \frac{1}{10^9}}

4. 5·(x·y)⁰ = 5 × x⁰ × y⁰ = 5 × 1 × 1 = <u>5</u>

5. 0¹⁵ = <u>0</u>

\displaystyle 6. \hspace{0.3 cm}  \mathbf{\frac{24 \cdot x^8 \cdot y^4}{6 \cdot x^5 \cdot y^4}} = \frac{ 4 \times 6 \times x^5 \times x^3 \times y^4}{6 \times x^5 \times y^4} = \underline{4 \cdot x^3}

\displaystyle 7. \hspace{0.3 cm}  \mathbf{\left(\frac{5 \cdot x^0}{y} \right)^{-1} }= \frac{1}{\frac{5 \times 1}{y} } = \underline{\frac{y}{5}}

8. \hspace{0.3 cm}\displaystyle  \mathbf{\frac{120 \cdot a^5 \cdot b^6  \cdot c^{-5} }{12 \cdot a^{-2} \cdot b^0  \cdot c^{2}}}  = 10 \cdot a^{5 - (-2)} \cdot  b^{6 - 0}  \cdot c^{-5 -2} = \underline{\frac{10 \cdot a^7  \cdot b^6}{c^7}}

\displaystyle 9. \hspace{0.3 cm}  \mathbf{\frac{(4 \cdot x)^0  \cdot y^{-5} \cdot z^{-2} }{(234  \cdot x  \cdot y \cdot z)^0}}  = y^{-5} \cdot z^{-2} = \underline{\frac{1}{y^{5} \cdot z^2}}

\displaystyle 10. \hspace{0.3 cm} \mathbf{\left(\frac{(5 \cdot x \cdot y)^0}{10} \right)^{-2}} = \left(\frac{1}{10} \right)^{-2} = \frac{1}{\left(\frac{1}{10} \right)^2 } = \frac{1}{\frac{1}{100} } = \underline{100}

11. \displaystyle \hspace{0.3 cm}  \mathbf{\left(\frac{(a \cdot b)}{9} \right)^{-1}} = \frac{1}{\left(\frac{(a \cdot b)}{9} \right) }  = \underline{\frac{9}{a \cdot b}}

\displaystyle 12. \hspace{0.3 cm} \mathbf{\frac{9}{x^{-2}}}  = \frac{9}{\frac{1}{x^2} }  = \underline{9 \cdot x^2}

13. \displaystyle \hspace{0.3 cm}  \mathbf{\frac{12 \cdot b^{-3}}{a^{-4}}}  = 12 \cdot \frac{b^{-3}}{a^{-4}} =12 \cdot \frac{1}{\frac{b^3}{a^4} }  = 12 \cdot \frac{a^4}{b^3}  = \underline{\frac{12 \cdot a^4}{b^3}}

14. \displaystyle \hspace{0.3 cm}  \mathbf{\frac{1}{a^{-5 \cdot n}}}  = \frac{1}{\frac{1}{a^{5 \cdot n}} } = \underline{a^{5 \cdot n}}

15. \displaystyle \hspace{0.3 cm}  \mathbf{\left(\frac{1}{2} \right)^{-5}} = \frac{1}{\left(\frac{1}{2}  \right)^5}  = 2^5 = \underline{32}

Learn more about the laws of indices here:

brainly.com/question/8959311

6 0
2 years ago
Solve the equation. −49f=−3 f=
attashe74 [19]

Answer:

f = 3/49

Step-by-step explanation:

-49f = -3

divide each side by -49

-49f/-49 = -3/-49

f = 3/49


3 0
3 years ago
r E c B . . a 5 3 H a 3 5 C g G m m 8 a 5 m m . 9.. B gt 8 a r n 8 . Jim r . g W 7 i 9 m 9.8. g- a w o . 1. l I i
xeze [42]
Im sorry but i dont understand your question. can you be more pacific.
6 0
3 years ago
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