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Colt1911 [192]
1 year ago
14

Jorden drives to the store at 30 miles per hour. On her way home she averages only 20 miles per hour. If the total driving time

takes half an hour, how far does she live from the store?
Mathematics
2 answers:
kozerog [31]1 year ago
7 0

Answer:

Rate column: 30, 20.

Distance column: x miles for both.

Step-by-step explanation:

Trava [24]1 year ago
4 0
For this case we have the following definition:
 d = v * t

 Where,
 d: distance
 v: speed
 t: time
 Clearing the time we have:
 t = d / v

 The total driving time takes half an hour:
 t = d / v1 + d / v2

1.5 = d / 30 + d / 20
 Clearing d we have:
 1.5 * 30 * 20 = 20d + 30d

1.5 * 30 * 20 = 50d

d = (1.5 * 30 * 20) / (50)

d = 18 miles
 Answer:
 
she lives 18 miles far from the store
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The orbit of the planet Venus is nearly circular. An astronomer develops a model for the orbit in which the sun has coordinates
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Since the Venus orbits round the sun, the sun is the center of the circular path of the revolution of the planet, Venus.

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A pharmaceutical company proposes a new drug treatment for alleviating symptoms of PMS (premenstrual syndrome). In the first sta
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Answer:

95% confidence interval for p, the true proportion of all women who will find success with this new treatment is [0.238 , 0.762].

Step-by-step explanation:

We are given that a pharmaceutical company proposes a new drug treatment for alleviating symptoms of PMS (premenstrual syndrome).

In the first stages of a clinical trial, it was successful for 7 out of the 14 women.

Firstly, the pivotal quantity for 95% confidence interval for the true proportion is given by;

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where, \hat p = sample proportion of women who find success with this new treatment = \frac{7}{14} = 0.50

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<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the true proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5%

                                            level of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

<u />

<u>95% confidence interval for p</u> = [\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

= [ 0.50-1.96 \times {\sqrt{\frac{0.50(1-0.50)}{14} } } , 0.50+1.96 \times {\sqrt{\frac{0.50(1-0.50)}{14} } } ]

 = [0.238 , 0.762]

Therefore, 95% confidence interval for p, the true proportion of all women who will find success with this new treatment is [0.238 , 0.762].

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