Divide. Write each quotient in simplest form. Please help I can't figure this out.

Natasha2012 [34]

Simply mean to take it apart. The most basic way to decompose a fraction is to break into unit fractions

3 0

4 years ago

A group of consumers were randomly selected and asked whether they were planning to buy a new TV screen within the next year. 56

kodGreya [7K]

Answer:

The approximate probability is 0.1921.

Step-by-step explanation:

The p-value is defined as the probability, [under the null-hypothesis (H₀)], of attaining a result equivalent to or greater than what was the truly observed value of the test statistic.

The hypothesis to test whether the proportion of consumers that plan to buy a new TV screen within the next year is 0.22, is defined as:

H₀: The proportion of consumers that plan to buy a new TV screen within the next year is 0.22, i.e. p = 0.22.

Hₐ: The proportion of consumers that plan to buy a new TV screen within the next year is 0.22, i.e. p > 0.22.

The information provided is:

n = 230

X = 56

Compute the value of sample proportion as follows:

$\hat p=\frac{X}{n}=\frac{56}{230}=0.2435$

Compute the test statistic as follows:

$z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.2435-0.22}{\sqrt{\frac{0.22(1-0.22)}{230}}}=0.87$

The test statistic value is, 0.87.

Compute the p-value as follows:

$p-value=P(Z>0.87)=1-P(Z$

*Use a z-table for the probability.

The p-value of the test is 0.1921.

Thus, the approximate probability of obtaining a sample proportion equal to or larger than the one obtained here is 0.1921.

7 0

3 years ago

What does a zero slope line look like

nikdorinn [45]

It's a horizontal line

4 0

3 years ago

A common assumption in modeling drug assimilation is that the blood volume in a person is a single compartment that behaves like

mixas84 [53]

Answer:

a) $\mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}$

b) $\mathbf{x = 2000 - 2000e^{-0.015t}}$

c) the steady state mass of the drug is 2000 mg

d) t ≅ 153.51 minutes

Step-by-step explanation:

From the given information;

At time t= 0

an intravenous line is inserted into a vein (into the tank) that carries a drug solution with a concentration of 500

The inflow rate is 0.06 L/min.

Assume the drug is quickly mixed thoroughly in the blood and that the volume of blood remains constant.

The objective of the question is to calculate the following :

a) Write an initial value problem that models the mass of the drug in the blood for t ≥ 0.

From above information given :

$Rate _{(in)}= 500 \ mg/L \times 0.06 \ L/min = 30 mg/min$

$Rate _{(out)}=\dfrac{x}{4} \ mg/L \times 0.06 \ L/min = 0.015x \ mg/min$

Therefore;

$\dfrac{dx}{dt} = Rate_{(in)} - Rate_{(out)}$

with respect to x(0) = 0

$\mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}$

b) Solve the initial value problem and graph both the mass of the drug and the concentration of the drug.

$\dfrac{dx}{dt} = -0.015(x - 2000)$

$\dfrac{dx}{(x - 2000)} = -0.015 \times dt$

By Using Integration Method:

$ln(x - 2000) = -0.015t + C$

$x -2000 = Ce^{(-0.015t)$

$x = 2000 + Ce^{(-0.015t)}$

However; if x(0) = 0 ;

Then

C = -2000

Therefore

$\mathbf{x = 2000 - 2000e^{-0.015t}}$

c) What is the steady-state mass of the drug in the blood?

the steady-state mass of the drug in the blood when t = infinity

$\mathbf{x = 2000 - 2000e^{-0.015 \times \infty }}$

x = 2000 - 0

x = 2000

Thus; the steady state mass of the drug is 2000 mg

d) After how many minutes does the drug mass reach 90% of its stead-state level?

After 90% of its steady state level; the mas of the drug is 90% × 2000

= 0.9 × 2000

= 1800

Hence;

$\mathbf{1800 = 2000 - 2000e^{(-0.015t)}}$

$0.1 = e^{(-0.015t)$

$ln(0.1) = -0.015t$

$t = -\dfrac{In(0.1)}{0.015}$

t = 153.5056729

t ≅ 153.51 minutes

4 0

3 years ago

In 1902, the yearly attendance at a major league baseball park was 3.4×10^5 people. One hundred years later, the yearly attendan

Eduardwww [97]

Answer:

1,360,000

Step-by-step explanation:

1,700,000-340,000 = 1,360,000 or 1.36×10^5

5 0

3 years ago