Write a number in each box to make the sentences true

Solve. 24= 7/8g A. 3/7 B. 21 C. 23 1/8 D. 27 3/7

Mrac [35]

I hope this helps you

3 0

3 years ago

Read 2 more answers

Help asap! 30 pts and brainliest!

myrzilka [38]

Answer:

July

Step-by-step explanation:

Took terst

3 0

3 years ago

A population proportion is . A sample of size will be taken and the sample proportion will be used to estimate the population pr

padilas [110]

Complete Question:

A population proportion is 0.4. A sample of size 200 will be taken and the sample proportion p will be used to estimate the population proportion. Use z- table Round your answers to four decimal places. Do not round intermediate calculations. a. What is the probability that the sample proportion will be within ±0.03 of the population proportion? b. What is the probability that the sample proportion will be within ±0.08 of the population proportion?

Answer:

A) 0.61351

Step-by-step explanation:

Sample proportion = 0.4

Sample population = 200

A.) proprobaility that sample proportion 'p' is within ±0.03 of population proportion

Statistically:

P(0.4-0.03<p<0.4+0.03)

P[((0.4-0.03)-0.4)/√((0.4)(.6))/200 < z < ((0.4+0.03)-0.4)/√((0.4)(.6))/200

P[-0.03/0.0346410 < z < 0.03/0.0346410

P(−0.866025 < z < 0.866025)

P(z < - 0.8660) - P(z < 0.8660)

0.80675 - 0.19325

= 0.61351

B) proprobaility that sample proportion 'p' is within ±0.08 of population proportion

Statistically:

P(0.4-0.08<p<0.4+0.08)

P[((0.4-0.08)-0.4)/√((0.4)(.6))/200 < z < ((0.4+0.08)-0.4)/√((0.4)(.6))/200

P[-0.08/0.0346410 < z < 0.08/0.0346410

P(−2.3094 < z < 2.3094)

P(z < -2.3094 ) - P(z < 2.3094)

0.98954 - 0.010461

= 0.97908

6 0

3 years ago

Is 1.88 rational or is it irrational

Anna11 [10]

Answer:

1.88 is a irrational number

3 0

4 years ago

The logistic equation for the population (in thousands) of a certain species is given by:

Eva8 [605]

Answer:

a.

b. 1.5

c. 1.5

d. No

Step-by-step explanation:

a. First, let's solve the differential equation:

$\frac{dp}{dt} =3p-2p^2$

Divide both sides by $3p-2p^2$ and multiply both sides by dt:

$\frac{dp}{3p-2p^2}=dt$

Integrate both sides:

$\int\ \frac{1}{3p-2p^2} dp =\int\ dt$

Evaluate the integrals and simplify:

$p(t)=\frac{3e^{3t} }{C_1+2e^{3t}}$

Where C1 is an arbitrary constant

I sketched the direction field using a computer software. You can see it in the picture that I attached you.

b. First let's find the constant C1 for the initial condition given:

$p(0)=3=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-1$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } =\frac{3}{2} =1.5$

c. As we did before, let's find the constant C1 for the initial condition given:

$p(0)=0.8=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=1.75$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2+1.75e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } =\frac{3}{2} =1.5$

d. To figure out that, we need to do the same procedure as we did before. So, let's find the constant C1 for the initial condition given:

$p(0)=2=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-\frac{1}{2} =-0.5$

Can a population of 2000 ever decline to 800? well, let's find the limit of the function when it approaches to ∞:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-0.5e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } =\frac{3}{2} =1.5$

Therefore, a population of 2000 never will decline to 800.

6 0

3 years ago