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mrs_skeptik [129]
2 years ago
12

A study was conducted to determine the percent of children that want to grow up work in the same career as a parent. In a sample

of 200 children, it was calculated that 43% wanted to eventually work in the same career as a parent. Construct the 95% confidence interval for the population proportion. Solution: phat = 0.43/200 = 0.00215 phat – qnorm(1.95/2)*sqrt(phat*(1-phat)/200) = -0.00426926 phat + qnorm(1.95/2)*sqrt(phat*(1-phat)/200) = 0.00856926 [-0.0043, 0.0086] What is wrong with this solution?
Mathematics
1 answer:
Irina-Kira [14]2 years ago
5 0

Answer:

Step-by-step explanation:

From the information given,

Number of sample, n = 200

Probability of success, p = 43/100 = 0.43

q = 1 - p = 1 - 0.43

q = 0.57

For a confidence level of 95%, the corresponding z value is 1.96.

The formula for determining the error bound for the proportion is

z × √pq/n

= 1.96 ×√(0.43 × 0.57)/200

= 1.96 × 0.035 = 0.0686

The upper boundary of the population proportion is

0.43 + 0.0686 = 0.5

The lower boundary of the population proportion is

0.43 - 0.0686 = 0.4

The error in the solution is

phat = 0.43/200 = 0.00215

Also,

[-0.0043, 0.0086] is wrong

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Stacey has a square piece of cloth. She cuts 3 inches off of the length of the square and 3 inches off of the width. The area of
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Answer:

The side length of the original square was 6 inches

Step-by-step explanation:

we know that

The area of a square is

A=b^2

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Let

x ---> the length of the original square

The area of the original square is

A=x^{2}\ in^2

The length of the smaller square is

b=(x-3)\ in

The area of the smaller square is

A=(x-3)^2\ in^2

The area of the smaller square is 1/4 the area of the original square

so

(x-3)^2=\frac{1}{4} x^{2}

solve for x

x^2-6x+9=\frac{1}{4} x^{2}

Multiply by 4 both sides

4x^2-24x+36=x^{2}

4x^2-x^2-24x+36=0\\3x^2-24x+36=0

Solve the quadratic equation by graphing

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x=2, x=6

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The solution is x=6 in

Remember that the solution must be greater than 3 inches (because Stacey cuts  3 inches off of the length of the square and 3 inches off of the width)

therefore

The side length of the original square was 6 inches

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2 years ago
What kind of salad do snowmen eat? Punchline 73 answer key
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Performance task: A parade route must start And and at the intersections shown on the map. The city requires that the total dist
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Answer:

Part A: The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles

Part B: For the total distance is as close to 3 miles as possible, the start point of the parade should be at the point on Broadway with coordinates (9.941, 4.970)

Part C: The coordinates of the cameras stationed half way down each road are;

For central avenue; (4, 2)

For Broadway; (7.97, 2.49)

Step-by-step explanation:

Part A: The length of the given route can be found using the equation for the distance, l, between coordinate points as follows;

l = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

Where for the Broadway potion of the parade route, we have;

(x₁, y₁) = (12, 3)

(x₂, y₂) = (6, 0)

l_1 = \sqrt{\left (0 -3\right )^{2}+\left (6-12 \right )^{2}} = 3 \cdot \sqrt{5}

For the Central Avenue potion of the parade route, we have;

(x₁, y₁) = (6, 0)

(x₂, y₂) = (2, 4)

l_2 = \sqrt{\left (4 -0\right )^{2}+\left (2-6 \right )^{2}} = 4 \cdot \sqrt{2}

Therefore, the total length of the parade route =-3·√5 + 4·√2 = 12.265 unit

The scale of the drawing is 1 unit = 0.25 miles

Therefore;

The actual length of the initial parade =0.25×12.265 unit = 3.09 miles

The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles

Part B:

For an actual length of 3 miles, the length on the scale drawing should be given as follows;

1 unit = 0.25 miles

0.25 miles = 1 unit

1 mile =  1 unit/(0.25) = 4 units

3 miles = 3 × 4 units = 12 units

With the same end point and route, we have;

l_1 = \sqrt{\left (0 -y\right )^{2}+\left (6-x \right )^{2}} = 12 - 4 \cdot \sqrt{2}

y² + (6 - x)² = 176 - 96·√2

y² = 176 - 96·√2 - (6 - x)²............(1)

Also, the gradient of l₁ = (3 - 0)/(12 - 6) = 1/2

Which gives;

y/x = 1/2

y = x/2 ..............................(2)

Equating equation (1) to (2) gives;

176 - 96·√2 - (6 - x)² = (x/2)²

176 - 96·√2 - (6 - x)² - (x/2)²= 0

176 - 96·√2 - (1.25·x²- 12·x+36) = 0

Solving using a graphing calculator, gives;

(x - 9.941)(x + 0.341) = 0

Therefore;

x ≈ 9.941 or x = -0.341

Since l₁ is required to be 12 - 4·√2, we have and positive, we have;

x ≈ 9.941 and y = x/2 ≈ 9.941/2 = 4.97

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Part C: The coordinates of the cameras stationed half way down each road are;

For central avenue;

Camera location = ((6 + 2)/2, (4 + 0)/2) = (4, 2)

For Broadway;

Camera location = ((6 + 9.941)/2, (0 + 4.970)/2) = (7.97, 2.49).

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