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3 years ago

You jog 6 and 2 thirds miles around a track each day. If you jogged that distance 4 times last week, how many miles did you jog?

Mathematics

2 answers:

Hunter-Best [27]3 years ago

7 0

You jogged 26 2/3 miles because 6 2/3x4= 26 2/3

Alex17521 [72]3 years ago

3 0

Multiply 2/3 by 4 to get a total jogging distance of 8/3 miles. Convert it to a mixed number, and your answer will be 2 1/3 miles.

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Convert 300 seconds to minutes. A) 5 minutes B) 15 minutes C) 30 minutes D) 150 minutes

Olin [163]

1 minute = 60 seconds

300 seconds / 60 seconds = 5 minutes

7 0

3 years ago

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One day in December the temperature in Dublin was -3°C, in Salerno it was 7°C, while in Murmansk it was -10°C.

sukhopar [10]

Answer:

Step-by-step explanation:

-10° C is the lowest temperature and 7°C is the highest temperature.

6 0

3 years ago

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It is known that the population variance equals 484. With a .95 probability, the sample size that needs to be taken to estimate

Hitman42 [59]

Answer:

The minimum sample size is n = 75 so that the desired margin of error is 5 or less.

Step-by-step explanation:

We are given the following in the question:

Population variance = 484

Standard deviation =

$\sigma^2 = 484\\\sigma =\sqrt{484} = 22$

Confidence level = 0.95

Significance level = 0.05

Margin of error = 5

Formula:

Margin of error =

$E = z\times \dfrac{\sigma}{\sqrt{n}}\\\\n = (\dfrac{z\times \sigma}{E})^2$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting values, we get

$E\leq 5\\\\1.96\times \dfrac{22}{\sqrt{n}} \leq 5\\\\\sqrt{n} \geq 1.96\times \dfrac{22}{5}\\\\n \geq (1.96\times \dfrac{22}{5})^2\\\\n \geq 74.373$

Thus, the minimum sample size is n = 75 so that the desired margin of error is 5 or less.

3 0

2 years ago

Find f. A) 7.4 B) 8.2 C) 10.5 D) 11.1

tatyana61 [14]

F=72

g=6

------------

$\cos { \left( F \right) } =\frac { { e }^{ 2 }+{ g }^{ 2 }-{ f }^{ 2 } }{ 2eg }$

Therefore:

$\cos { \left( 72 \right) } =\frac { { e }^{ 2 }+{ 6 }^{ 2 }-{ f }^{ 2 } }{ 2\cdot e\cdot 6 } \\ \\ \cos { \left( 72 \right) } =\frac { { e }^{ 2 }+36-{ f }^{ 2 } }{ 12e }$

$\\ \\ 12e\cdot \cos { \left( 72 \right) } ={ e }^{ 2 }+36-{ f }^{ 2 }\\ \\ \therefore \quad { f }^{ 2 }={ e }^{ 2 }-12e\cdot \cos { \left( 72 \right) } +36\\ \\ \therefore \quad f=\sqrt { { e }^{ 2 }-12e\cdot \cos { \left( 72 \right) +36 } } \\ \\ \therefore \quad f=\sqrt { e\left( e-12\cos { \left( 72 \right) } \right) +36 }$

But what is e?

E=76

G=32

g=6

And:

$\frac { e }{ \sin { \left( E \right) } } =\frac { g }{ \sin { \left( G \right) } }$

Which means that:

$\frac { e }{ \sin { \left( 76 \right) } } =\frac { 6 }{ \sin { \left( 32 \right) } } \\ \\ \therefore \quad e=\frac { 6\cdot \sin { \left( 76 \right) } }{ \sin { \left( 32 \right) } }$

If you take this value into account, you will discover that f is...

$f=\sqrt { \frac { 6\cdot \sin { \left( 76 \right) } }{ \sin { \left( 32 \right) } } \left( \frac { 6\cdot \sin { \left( 76 \right) } }{ \sin { \left( 32 \right) } } -12\cos { \left( 72 \right) } \right) +36 } \\ \\ \therefore \quad f=10.8\quad \left( 1\quad d.p \right)$

So I would have to say that the answer is approximately (c).

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3 years ago

The graph shows f(x) = (one-half) Superscript x and its translation, g(x). Which describes the translation of f(x) to g(x)?

sveta [45]

There is no graph.
The statement that describes the translation is that it does not exist.

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3 years ago

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