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1 year ago

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Playing tennis burns energy at a rate of about 25 kJ/min. Cycling burns energy at about 35kJ/min. Hans exercised by playing tenn

is and then cycling. He exercised for 50 min altogether and used a total of 1450kJ of energy. For how long did he play tennis?

1 answer:

grigory [225]1 year ago

8 0

To answer this, you need to make a systems of equations.

25x + 35y = 1450
x + y = 50 ----> y = 50 - x

25x + 35(50-x) = 1450
25x + 1750 - 35x = 1450
25x - 35x + 1750 = 1450
-10x + 1750 = 1450
- 1750 -1750

-10x = -300
x = 30

He played tennis for 30min

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Lucy has to run two errands. She starts from home and travels 3 miles south to the post office. From the post office, she travel

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Answer:

the correct choice is A. At the gas station

Step-by-step explanation:

Lucy starts at home and travels 3 miles south to the post office. From the post office, she travels 4 miles east to the gas station. As it is known south and east directions form right angle. Since the entire trip forms a triangle, this triangle is right with right angle at the post office.

Call the vertices of this triangle P - post office, G - gas station, H - home. Then HP and PG are legs of this triangle and GH is hypotenuse.

From the given data:

HP=3;

PG=4;

GH=5;

∠P=90°.

The smallest angle is opposite to the smallest side. The smallest side is leg HP, so the smallest angle is G that is the angle at gas station.

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1 year ago

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Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>

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<u>So, 90% confidence interval for the population proportion, p is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

<u>90% confidence interval for p</u> = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

$0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }$

$\sqrt{n} = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}$

$\sqrt{n}$ = 54.79

n = $54.79^{2}$

n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

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1 year ago

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Dvinal [7]

Answer:

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Step-by-step explanation:

Let $\theta$ be the angle between the two sides and x be the length of the third side. By cosine rule,

$x^2 = 12^2+14^2-2\times12\times14\cos\theta = 340 - 336\cos\theta$

$x= \sqrt{340 - 336\cos\theta}$

We differentiate x with respect to $\theta$ by applying chain rule.

$\dfrac{dx}{d\theta} = \dfrac{336\sin\theta}{2\sqrt{340 - 336\cos\theta}} = \dfrac{168\sin\theta}{\sqrt{340 - 336\cos\theta}}$

Rate of change of $\theta$ is 2

$\dfrac{\theta}{dt} = 2$

Rate of change of x is

$\dfrac{dx}{dt} = \dfrac{dx}{d\theta}\times\dfrac{d\theta}{dt}$

$\dfrac{dx}{dt} = \dfrac{168\sin\theta}{\sqrt{340 - 336\cos\theta}} \times2=\dfrac{336\sin\theta}{\sqrt{340 - 336\cos\theta}}$

At 60°,

$\dfrac{dx}{dt} = \dfrac{336\sin60}{\sqrt{340 - 336\cos60}} = 1.692 \text{ m/min}$

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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1 year ago

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