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nikitadnepr [17]
3 years ago
9

A mountain climber ascended 3 4 of a kilometer in 1 12 of an hour. What was the mountain climber’s ascent rate in kilometers per

hour?
Mathematics
2 answers:
Oksana_A [137]3 years ago
8 0

A mountain climber ascended 3/4 of a kilometer in 1/12 of an hour. The mountain climber's ascent rate is ; C. 9 kilometers per hour.

Debora [2.8K]3 years ago
4 0
The mountain climber's ascent rate is calculated by dividing 3/4 kilometer by 1/12 of an hour. That is mathematically expressed as,
                                   ascent rate = (3/4 km) / (1/12 h) 
                                                     = 9 km/h
Thus, the climber ascended at a rate of 9 km/h. 
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Flauer [41]

Answer:

Yes

Step-by-step explanation:

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7 0
3 years ago
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Brilliant_brown [7]

Answer:

y = 2x + 54

Step-by-step explanation:

slope of equation:

(y2-y1)/(x2-x1)

= (72-58)/(9-2)

= 14/7

= 2

y = 2x + b

58 = 2(2) + b

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6 0
3 years ago
Let f(x) = tan(x) - 2/x. Let g(x) = x^2 + 8. What is f(x)*g(y)?
Tema [17]

Answer:

f(x)\times g(y)=y^2tan(x)+8tan(x)-\frac{2y^2}{x}-\frac{16}{x}

Step-by-step explanation:

We are given that

f(x)=tan(x)-\frac{2}{x}

g(x)=x^2+8

We have to find f(x)\times g(y)

To find the value of f(x)\times g(y) we will multiply f(x) by g(y)

g(y)=y^2+8

Now,

f(x)\times g(y)=(tanx-\frac{2}{x})(y^2+8)

f(x)\times g(y)=tan(x)(y^2+8)-\frac{2}{x}(y^2+8)

f(x)\times g(y)=y^2tan(x)+8tan(x)-\frac{2y^2}{x}-\frac{16}{x}

Hence,

f(x)\times g(y)=y^2tan(x)+8tan(x)-\frac{2y^2}{x}-\frac{16}{x}

5 0
3 years ago
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kow [346]

Answer:

the 2nd option f(-4) = g(-4) and f(0) = g(0)

Step-by-step explanation:

the right answer was the same x and sampe y. it means in their intersection for both f(x) anad g(x)

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4 0
3 years ago
Use the given degree of confidence and sample data to construct interval for the population proportion. Of 369 randomly selected
Alexxx [7]

Answer:

e)  (3.77%, 8.70%)

95% confidence interval for the percentage of all medical students who plan to work in a rural community

(3.83% , 8.69%)

95% confidence level for the population proportion who blame oil companies for the recent increase in gasoline prices

(0.406 , 0.454)

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given random sample size 'n' = 369

Sample proportion

                              p = \frac{x}{n} = \frac{23}{369} = 0.0623

95% confidence intervals are determined by

(p^{-} - Z_{0.05} \sqrt{\frac{p(1-p)}{n} } , p^{-} + Z_{0.05} \sqrt{\frac{p(1-p)}{n} })

(0.0623 - 1.96 \sqrt{\frac{0.0623(1-0.0623)}{369} } , 0.0623 + 1.96\sqrt{\frac{0.0623(1-0.0623)}{369} })

(0.0623 - 0.0246 , 0.0623 + 0.0246)

(0.0383 , 0.0869)

(3.83% , 8.69%)

95% confidence interval for the percentage of all medical students who plan to work in a rural community

(3.83% , 8.69%)

<u>Step(ii):-</u>

Given 43% of those polled blamed of companies the most for the recent increase in gasoline prices

sample proportion 'p' = 0.43

Given Margin of error (M.E) = 0.024

95% confidence intervals are determined by

(p^{-} - Z_{0.05} \sqrt{\frac{p(1-p)}{n} } , p^{-} + Z_{0.05} \sqrt{\frac{p(1-p)}{n} })

(0.43 - 0.024 } , 0.43 +0.024 )

(0.406 , 0.454)

<u>Final answer</u>:-

95% confidence level for the population proportion who blame oil companies for the recent increase in gasoline prices

(0.406 , 0.454)

6 0
3 years ago
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