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3 years ago

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yesterday Noah ran 2 1/2 miles in 3/5 hour. emily ran 3 3/4 miles in 5/6 hour. Anna ran 3 1/2 miles in 3/4 hour. how fast, in mi

les per hour, did each person run? who ran the fastest.

1 answer:

otez555 [7]3 years ago

5 0

Answer:

Anna

Step-by-step explanation:

she was the fastest

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Solve: 2 In 3 = In(x-4)

Alona [7]

$\bf \begin{array}{llll} \textit{Logarithm of rationals} \\\\ \log_a\left( \frac{x}{y}\right)\implies \log_a(x)-\log_a(y) \end{array}~\hfill \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}$

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3 years ago

Ethanhunt only answer pls don't answer just don't if u not ethanhunt don't bother pls don't

gulaghasi [49]

1) Southwest part of 2nd & 4th street = 130 [ vertically opposite of Corresponding so will be equal.

2) Southeast part of 2nd & 4th street = 190 - 130 = 50 [ supplementary with previous one ]

3) Northwest part of 2nd & 4th street = 50 [ Vertically opposite of 2nd one ]

4) Northeast part of 2nd & 4th street = 130 [ Corresponding angles ]

[ Final Answers are in bold ]

Hope this helps!

6 0

3 years ago

Read 2 more answers

Determine whether each sequence is an arithmetic sequence. Write yes or no. Explain. 7, 10, 13, 16, ...

kvv77 [185]

Answer:

The given sequence is an arithmetic sequence.

Step-by-step explanation:

Given sequence is:

7, 10, 13, 16, ...

Here

$a_1 = 7\\a_2 = 10\\a_3 = 13\\a_4 = 16$

In order for a sequence to be an arithmetic sequence the common difference between the consecutive terms of the sequence should be same. Common difference is denoted by d.

So,

$d = a_2 - a_1 = 10-7 = 3\\d = a_3 - a_2 = 13-10 = 3\\d = a_4-a_3 = 16-13 = 3$

As we can see that the common difference is same for consecutive terms.

The given sequence is an arithmetic sequence.

7 0

3 years ago

Of the cartons produced by a company, % have a puncture, % have a smashed corner, and % have both a puncture and a smashed

LuckyWell [14K]

Full Question

<em>Of the cartons produced by a company, 10% have a puncture, 6% have a smashed corner, and 0.4% have both a puncture and a smashed corner. Find the probability that a randomly selected carton has a puncture or a smashed corner. The probability that a randomly selected carton has a puncture or a smashed corner nothing ____%. (Type an integer or a decimal. Do not round.)</em>

<em />

Answer:

$P(Punctured\ or\ Smashed\ Corner) = 0.156$

Step-by-step explanation:

Given

$Puncture\ Corner = 10\%$

$Smashed\ Corner = 6\%$

$Punctured\ and\ Smashed\ Corner = 0.4\%$

Required

$P(Punctured\ or\ Smashed\ Corner)$

For non-mutually exclusive event described above, P(Punctured or Smashed Corner) can be calculated as thus;

$P(Punctured\ or\ Smashed\ Corner) = P(Punctured\ Corner) + P(Smashed\ Corner) - P(Punctured\ and\ Smashed\ Corner)$

Substitute:

10% for P(Puncture Corner),

6% for P(Smashed Corner) and

0.4% for P(Punctured and Smashed Corner)

$P(Punctured\ or\ Smashed\ Corner) = 10\% + 6\% - 0.4\%$

$P(Punctured\ or\ Smashed\ Corner) = 15.6\%$

Convert % to fraction

$P(Punctured\ or\ Smashed\ Corner) = \frac{15.6}{100}$

Convert to decimal

$P(Punctured\ or\ Smashed\ Corner) = 0.156$

3 0

3 years ago

Please help! Brainliest answer! thank you!1

kobusy [5.1K]

Answer:

I think it's A

Step-by-step explanation:

because none of the others appear to be correct

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3 years ago

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