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3 years ago

13

Assuming a 90% confidence level, the margin of error is approximately?

1 answer:

elena-s [515]3 years ago

7 0

Answer:

0.3

Step-by-step explanation:

The margin of error is calculated as ...

(standard deviation)/√(sample size) × (z*-score)

where the z*-score is chosen based on the desired confidence level.

Here, you have ...

standard deviation = 2.7
√(sample size) = √225 = 15
z*-score for 90% confidence level = 1.645

Putting these values in the above expression for margin of error gives ...

2.7/15·1.645 = 0.2961 ≈ 0.3

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in the figure below trianglePQM and triangleQRP are right triangles. the measure of lineQM is 6 and the measure of lineQP is 8.

Kipish [7]

Answer:

Option 4 is correct. The length of PR is 6.4 units.

Step-by-step explanation:

From the given figure it is noticed that the triangle PQR and triangle MQR.

Let the length of PR be x.

Pythagoras formula

$hypotenuse^2=base^2+perpendicular^2$

Use pythagoras formula for triangle PQM.

$PM^2=QM^2+PQ^2$

$PM^2=(6)^2+(8)^2$

$PM^2=36+64$

$PM^2=100$

$PM=10$

The value of PM is 10. The length of PR is x, so the length of MR is (10-x).

Use pythagoras formula for triangle PQR.

$PQ^2=QR^2+PR^2$

$(8)^2=QR^2+x^2$

$64-x^2=QR^2$ .....(1)

Use pythagoras formula for triangle MQR.

$MQ^2=QR^2+MR^2$

$(6)^2=QR^2+(10-x)^2$

$36=QR^2+x^2-20x+100$

$36-x^2+20x-100=QR^2$ .... (2)

From equation (1) and (2) we get

$36-x^2+20x-100=64-x^2$

$20x-64=64$

$20x=128$

$x=6.4$

Therefore length of PR is 6.4 units and option 4 is correct.

3 0

3 years ago

Exercise #1: The amount of money in Nicole's bank account can be represented by the function f(x) = 32.50x + 200,

kolezko [41]

The initial amount in the account was $200 (y intercept). She adds $32.50 each day to her bank account (slope).

7 0

3 years ago

3. The curve C with equation y=f(x) is such that, dy/dx = 3x^2 + 4x +k

Andreas93 [3]

a. Given that y = f(x) and f(0) = -2, by the fundamental theorem of calculus we have

$\displaystyle \frac{dy}{dx} = 3x^2 + 4x + k \implies y = f(0) + \int_0^x (3t^2+4t+k) \, dt$

Evaluate the integral to solve for y :

$\displaystyle y = -2 + \int_0^x (3t^2+4t+k) \, dt$

$\displaystyle y = -2 + (t^3+2t^2+kt)\bigg|_0^x$

$\displaystyle y = x^3+2x^2+kx - 2$

Use the other known value, f(2) = 18, to solve for k :

$18 = 2^3 + 2\times2^2+2k - 2 \implies \boxed{k = 2}$

Then the curve C has equation

$\boxed{y = x^3 + 2x^2 + 2x - 2}$

b. Any tangent to the curve C at a point (a, f(a)) has slope equal to the derivative of y at that point:

$\dfrac{dy}{dx}\bigg|_{x=a} = 3a^2 + 4a + 2$

The slope of the given tangent line $y=x-2$ is 1. Solve for a :

$3a^2 + 4a + 2 = 1 \implies 3a^2 + 4a + 1 = (3a+1)(a+1)=0 \implies a = -\dfrac13 \text{ or }a = -1$

so we know there exists a tangent to C with slope 1. When x = -1/3, we have y = f(-1/3) = -67/27; when x = -1, we have y = f(-1) = -3. This means the tangent line must meet C at either (-1/3, -67/27) or (-1, -3).

Decide which of these points is correct:

$x - 2 = x^3 + 2x^2 + 2x - 2 \implies x^3 + 2x^2 + x = x(x+1)^2=0 \implies x=0 \text{ or } x = -1$

So, the point of contact between the tangent line and C is (-1, -3).

7 0

2 years ago

One tenths times one decimal point six

barxatty [35]

1/10 x 1.6

First make 1/10 over 100
1/10(10/10) = 10/100, which = 0.10

0.1 x 1.6 = 0.16

0.16, or 16/100, is your answer

hope this helps

6 0

3 years ago

Read 2 more answers

The radius of the circle is increasing at a rate of 2 meters per minute and the sides of the square are increasing at a rate of

Lunna [17]

Answer:

Change in area=24 $\pi$ -48

Step-by-step explanation:

Let s will be the side of square and r will be the radius of circle.

Then two given conditions are

1)dr/dt=2 m/s

2)ds/dt=1 m/s

Area enclosed=(Area of square)-(Area of circle)

Area of square= $s^{2}$

Area of circle= $\pi r^{2}$

Area enclosed= $(\pi r^{2})-s^{2}$

dA/dt=2 $\pi$ r(dr/dt)-2s(ds/dt)

At s=24,and r=6

dA/dt=2( $\pi$ )(6)(2)-2(24)(1)

Change in area=24 $\pi$ -48

6 0

3 years ago