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Charra [1.4K]
2 years ago
14

Aiden is a spice trader. He charges $5 for 1 kilogram of cumin seeds, which is the smallest amount he sells. The largest amount

of cumin seeds Aiden can sell is 1000 kilograms, which cost $2000 . P(w) models the price (in dollars) of  'w' kilograms of cumin seeds in Aiden's shop. Which number type is more appropriate for the domain of p ?
Integers/Real Numbers

A. 1≤w≤1000

B. 5≤w≤1000

C. 1≤w≤2000
Mathematics
1 answer:
Contact [7]2 years ago
7 0

Answer:

1 \leq\ w\leq\ 1000

Step-by-step explanation:

Given

Charges: $5 for 1 kg

Charges: $2000 for 1000 kg

w represents kilograms

P(w) represents Price

Required

What's the domain of P

Start by representing P as a function of w

When

w = 1\ kg

P(1) = \$5

w = 1000\ kg

P(1000) = \$2000

The domain of P are the values of w and is represented as follows;

1 \leq\ w\leq\ 1000

Hence;

<em>Option A answers the question</em>

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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>

<em />

<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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Answer:

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The probability of getting two 3 as outcome is \frac{1}{4} \times\frac{1}{4} = \frac{1}{16}.

In the rest of the 4 sequences, will not be any 3 as outcome.

Probability of not getting a outcome rather than 3 is 1 - \frac{1}{4} = \frac{3}{4}.

Hence, the required probability is 15\times\frac{1}{16}(\frac{3}{4})^4 = \frac{1215}{4096}≅0.2966 or, 0.23.

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Well, let us solve this step by step.

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