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2 years ago

Help please I don’t know it!

Mathematics

1 answer:

Sidana [21]2 years ago

4 0

99kg of water.

Since 100kg of mass = 66kg of water, all you have to do to find a mass of 60 is divide by 2 giving you 50kg of mass, and 33 kg of water. Then add that up onto 100kg of water and 66kg of water to get 150kg of mass and 99kg of water.

Hope this helps you, and have a nice day.

-R3TR0 Z3R0

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A past study claims that adults in America spend an average of 17 hours a week on leisure activities. A researcher wanted to tes

8090 [49]

Answer:

The claim is false.

Step-by-step explanation:

Given the data :

13 24 21 37 15 25 18 22 40 32

The sample mean and standard deviation can ben calculated for the given sample.

Using calculator :

Sample mean, xbar = 24.7

Sample standard deviation, s = 9.04

Sample size, n = 10

The hypothesis :

H0 : μ = 17

H1 : μ ≠ 17

The test statistic :

(xbar - μ) ÷ (s/√(n))

(24.7 - 17) ÷ (9.04/√(10))

7.7 / 2.8586990

Test statistic = 2.694

We can obtain the Pvalue, at α = 0.05 ; df = n-1 = 9

Pvalue = 0.0246

Since Pvalue < α ; we reject the null ; Hence, there is significant evidence to conclude that an adult American does not spend average of 17 hours in leisure

5 0

3 years ago

Solve the following equation.

elena-14-01-66 [18.8K]

Answer:

x=6

Step-by-step explanation:

x + 6 = x + x original problem

x + 6 = 2x combine like terms

-x = -x subtract x from both sides to have like terms on each side

6 = x solution

8 0

2 years ago

What is 40% as a whole number

maksim [4K]

% is a number over 100, so 40% is 40/100, which is 0.40.

5 0

3 years ago

Read 2 more answers

Find 48% of 30. Please help

baherus [9]

Answer:

14.4 would be the answer hope this is helpful

Step-by-step explanation:

14.4/30= 48%

4 0

2 years ago

Read 2 more answers

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

Eva8 [605]

Answer:

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