Find the Value of PT. PT=4x-6 and TQ=3x+4

Suppose we have a population whose proportion of items with the desired attribute is p = 0:5. (a) If a sample of size 200 is tak

crimeas [40]

Answer:

a. If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b. If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

Step-by-step explanation:

This problem should be solved with a binomial distribution sample, but as the size of the sample is large, it can be approximated to a normal distribution.

The parameters for the normal distribution will be

$\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/200}= 0.0353$

We can calculate the z values for x1=0.47 and x2=0.51:

$z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.0353}=-0.85\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.0353}=0.28$

We can now calculate the probabilities:

$P(0.47$

If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b) If the sample size change, the standard deviation of the normal distribution changes:

$\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/100}= 0.05$

We can calculate the z values for x1=0.47 and x2=0.51:

$z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.05}=-0.6\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.05}=0.2$

We can now calculate the probabilities:

$P(0.47$

If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

8 0

3 years ago

Vladimir says that the equation of the line that passes through points (negative 5, negative 3) and (10, 9) is y = four-fifths x

love history [14]

Answer:

Vladimir and Robyn both are correct.

Step-by-step explanation:

Let us check whether the points (-5,-3) and (10,9) are on the line $y = \frac{4}{5}x + 1$ or not.

The equation of the straight line passing through the given points (-5,-3) ans (10,9) is $\frac{y - 9}{9 + 3} = \frac{x - 10}{10 + 5}$

⇒ 5(y - 9) = 4(x - 10)

⇒ 5y = 4x + 5

⇒ $y = \frac{4}{5}x + 1$ .............. (1)

So, Vladimir is correct.

Now, Robyn says that the line passes through the points (-10,-7) and (-15,-11).

Then, both of the points satisfy the equation (1).

Therefore, Vladimir and Robyn both are correct. (Answer)

4 0

3 years ago

Read 2 more answers

Рабочий изготовил за 1 час 17 деталей . А его ученик 12 деталей . Сколько деталей они изготовят за 7 часов совместной работы

Svetradugi [14.3K]

Answer:

203

Step-by-step explanation:

17*7=119

12*7=84

84+119=203

4 0

3 years ago

Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x

Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

4 0

3 years ago

Write the formula for the perimeter of a rectangle.Then substitute 12in for the length and 6in for the width solve the equation

Anastaziya [24]

Answer:

Step-by-step explanation:

Let L represent the Length of the rectangle.

Let W represent the width of the rectangle.

The formula for determining the perimeter of a rectangle is expressed as

Perimeter = 2(L + W)

Then substitute 12 in for the length and 6in for the width. It becomes.

Perimeter = 2(12 + 6).

Perimeter =2 × 18 = 36

3 0

3 years ago