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ikadub [295]
3 years ago
12

Helpppppp I NEED THIS HELP

Mathematics
2 answers:
ohaa [14]3 years ago
8 0

Answer:

A

Step-by-step explanation:

padilas [110]3 years ago
4 0

Answer:

A

Step-by-step explanation:

we need to focus on getting X alone so we remove things attached by symbols first by adding the opposite to both sides, so we need to add -2 to both sides

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2 decreased by the sum of m and 8
Elina [12.6K]

Answer:

(m+8) - 2

Step-by-step explanation:

6 0
2 years ago
Help plsss Show me the work plsss
velikii [3]

Answer:

a) y = 6x - 3

b) 1/3y = 2x -1

The first thing you need to do is isolate (y) in the second equation

3 x (1/3y) = 3 x (2x - 1)

y =6x - 3

After isolating (y) in equation b they end up being the same.

Graphing:

In order to graph this, you have to make the first point at (0, -3) since this is the Y-intercept of the equation.

In order to graph the other points, you must move 6 units up and 1 unit to the right. Or vise versa If you need a visual I'll gladly link one.

5 0
3 years ago
In the production of a plant, a treatment is being evaluated to germinate seed. From a total of 60 seed it was observed that 37
djyliett [7]

Answer:

More than 50% would germinate

Step-by-step explanation:

Given that in the production of a plant, a treatment is being evaluated to germinate seed. From a total of 60 seed it was observed that 37 of them germinated

Let us check whether more than 50% will germinate using hypothesis test

H_0: p = 0.50\\H_a: p>0.50\\

(right tailed test)

Sample proportion p =\frac{37}{60} =0.617\\q = 0.383\\Std error = \sqrt{\frac{pq}{n} } =0.0628

p difference = 0.117

Test statistic Z = p difference/std error = 1.864

p value =0.0312

Since p value <0.05 our significance level of 5% we reject null hypothesis

It is  possible to claim that most of the seed will germinate (i.e. more than 50%)

7 0
3 years ago
Determine the number of degrees of freedom for the two-sample t test or CI in each of the following situations. (Round your answ
gulaghasi [49]

Answer:

Part a ) The degrees of freedom for the given two sample non-pooled t test is 24

Part b ) The degrees of freedom for the given two sample non-pooled t test is 30

Part c ) The degrees of freedom for the given two sample non-pooled t test is 30

Part d ) The degrees of freedom for the given two sample non-pooled t test is 25

Step-by-step explanation:

Degrees of freedom for a non-pooled two sample t-test is given by;

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Now given the information;

a) :- m = 12, n = 15, s₁ = 4.0, s₂ = 6.0

we substitute

Δf =  {[ 4²/12 + 6²/15 ]²} / {[( 4²/12)²/12-1] + [(6²/15)²/15-1]}

Δf  = 30184 / 1241

Δf  = 24.3223 ≈ 24 (down to the nearest whole number)

b) :- m = 12, n = 21, s₁ = 4.0, s₂ = 6.0

we substitute using same formula

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Δf = {[ 4²/12 + 6²/21 ]²} / {[( 4²/12)²/12-1] + [(6²/21)²/21-1]}

Δf = 56320 / 1871

Δf = 30.1015 ≈ 30 (down to the nearest whole number)

c) :- m = 12, n = 21, s₁ = 3.0, s₂ = 6.0

we substitute using same formula

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Δf = {[ 3²/12 + 6²/21 ]²} / {[( 3²/12)²/12-1] + [(6²/21)²/21-1]}

Δf = 29095 / 949

Δf = 30.6585 ≈ 30 (down to the nearest whole number)

d) :- m = 10, n = 24, s₁ = 4.0, s₂ = 6.0

we substitute using same formula

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Δf = {[ 4²/10 + 6²/24 ]²} / {[( 4²/10)²/10-1] + [(6²/24)²/24-1]}

Δf = 1044 / 41  

Δf = 25.4634 ≈ 25 (down to the nearest whole number).

6 0
3 years ago
Particle P moves along the y-axis so that its position at time t is given by y(t)=4t−23 for all times t. A second particle, part
sergey [27]

a) The limit of the position of particle Q when time approaches 2 is -\pi.

b) The velocity of particle Q is v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}} for all t \ne 2.

c) The rate of change of the distance between particle P and particle Q at time t = \frac{1}{2} is \frac{4\sqrt{82}}{9}.

<h3>How to apply limits and derivatives to the study of particle motion</h3>

a) To determine the limit for t = 2, we need to apply the following two <em>algebraic</em> substitutions:

u = \pi t (1)

k = 2\pi - u (2)

Then, the limit is written as follows:

x(t) =  \lim_{t \to 2} \frac{\sin \pi t}{2-t}

x(t) =  \lim_{t \to 2} \frac{\pi\cdot \sin \pi t}{2\pi - \pi t}

x(u) =  \lim_{u \to 2\pi} \frac{\pi\cdot \sin u}{2\pi - u}

x(k) =  \lim_{k \to 0} \frac{\pi\cdot \sin (2\pi-k)}{k}

x(k) =  -\pi\cdot  \lim_{k \to 0} \frac{\sin k}{k}

x(k) = -\pi

The limit of the position of particle Q when time approaches 2 is -\pi. \blacksquare

b) The function velocity of particle Q is determined by the <em>derivative</em> formula for the division between two functions, that is:

v_{Q}(t) = \frac{f'(t)\cdot g(t)-f(t)\cdot g'(t)}{g(t)^{2}} (3)

Where:

  • f(t) - Function numerator.
  • g(t) - Function denominator.
  • f'(t) - First derivative of the function numerator.
  • g'(x) - First derivative of the function denominator.

If we know that f(t) = \sin \pi t, g(t) = 2 - t, f'(t) = \pi \cdot \cos \pi t and g'(x) = -1, then the function velocity of the particle is:

v_{Q}(t) = \frac{\pi \cdot \cos \pi t \cdot (2-t)-\sin \pi t}{(2-t)^{2}}

v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}}

The velocity of particle Q is v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}} for all t \ne 2. \blacksquare

c) The vector <em>rate of change</em> of the distance between particle P and particle Q (\dot r_{Q/P} (t)) is equal to the <em>vectorial</em> difference between respective vectors <em>velocity</em>:

\dot r_{Q/P}(t) = \vec v_{Q}(t) - \vec v_{P}(t) (4)

Where \vec v_{P}(t) is the vector <em>velocity</em> of particle P.

If we know that \vec v_{P}(t) = (0, 4), \vec v_{Q}(t) = \left(\frac{2\pi\cdot \cos \pi t - \pi\cdot t \cdot \cos \pi t + \sin \pi t}{(2-t)^{2}}, 0 \right) and t = \frac{1}{2}, then the vector rate of change of the distance between the two particles:

\dot r_{P/Q}(t) = \left(\frac{2\pi \cdot \cos \pi t - \pi\cdot t \cdot \cos \pi t + \sin \pi t}{(2-t)^{2}}, -4 \right)

\dot r_{Q/P}\left(\frac{1}{2} \right) = \left(\frac{2\pi\cdot \cos \frac{\pi}{2}-\frac{\pi}{2}\cdot \cos \frac{\pi}{2} +\sin \frac{\pi}{2}}{\frac{3}{2} ^{2}}, -4 \right)

\dot r_{Q/P} \left(\frac{1}{2} \right) = \left(\frac{4}{9}, -4 \right)

The magnitude of the vector <em>rate of change</em> is determined by Pythagorean theorem:

|\dot r_{Q/P}| = \sqrt{\left(\frac{4}{9} \right)^{2}+(-4)^{2}}

|\dot r_{Q/P}| = \frac{4\sqrt{82}}{9}

The rate of change of the distance between particle P and particle Q at time t = \frac{1}{2} is \frac{4\sqrt{82}}{9}. \blacksquare

<h3>Remark</h3>

The statement is incomplete and poorly formatted. Correct form is shown below:

<em>Particle </em>P<em> moves along the y-axis so that its position at time </em>t<em> is given by </em>y(t) = 4\cdot t - 23<em> for all times </em>t<em>. A second particle, </em>Q<em>, moves along the x-axis so that its position at time </em>t<em> is given by </em>x(t) = \frac{\sin \pi t}{2-t}<em> for all times </em>t \ne 2<em>. </em>

<em />

<em>a)</em><em> As times approaches 2, what is the limit of the position of particle </em>Q?<em> Show the work that leads to your answer. </em>

<em />

<em>b) </em><em>Show that the velocity of particle </em>Q<em> is given by </em>v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t +\sin \pi t}{(2-t)^{2}}<em>.</em>

<em />

<em>c)</em><em> Find the rate of change of the distance between particle </em>P<em> and particle </em>Q<em> at time </em>t = \frac{1}{2}<em>. Show the work that leads to your answer.</em>

To learn more on derivatives, we kindly invite to check this verified question: brainly.com/question/2788760

3 0
2 years ago
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