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Murrr4er [49]
1 year ago
13

The volume of water in a tank is modeled by the function y = 1.15x3 – 0.1x2 + 2 where x is the number of minutes after the fauce

t filling the tank is turned on. What is the y-intercept of the graph, and what does it represent in this context? Choose from the drop-down menus.
The y-intercept is
Group of answer choices

0

0.5

1.5

2
Mathematics
2 answers:
baherus [9]1 year ago
8 0

Answer:

2

Step-by-step explanation:

in this context the y intercept represents the volume of water already in the tank

Arlecino [84]1 year ago
8 0

Answer:

2

Step-by-step explanation:

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A = a^2.....where " a " is the length of one side
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D is the answer

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Solve the system using elimination. x + 2y = –6 3x + 8y = –20 (–4, –1) (–4, 4) (–1, –4) (3, 1)
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x + 2y =  - 6 >  >  > 3x =  - 18 - 6y
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A pharmaceutical company proposes a new drug treatment for alleviating symptoms of PMS (premenstrual syndrome). In the first sta
Akimi4 [234]

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;

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

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

          n = sample of women = 14

<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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