All categories

2 years ago

I need help can someone help me plz

Mathematics

1 answer:

Mice21 [21]2 years ago

7 0

Answer:

y = 250x + 1500

Step-by-step explanation:

Slope is 250

y-intercept = 3000 - 6 × slope

3000 - 1500 = 1500

Please Mark as Brainliest!

You might be interested in

Translate to an algebraic expression: 15 less than six times a number, y.

Natalija [7]

Answer:

Step-by-step explanation:

6 0

3 years ago

Which expression is equivalent to 2x2 – 2x + 7?

diamong [38]

11-2x would be the answer u are looking for. Hope this helps!

4 0

3 years ago

Read 2 more answers

Suppose you pay a dollar to roll two dice. if you roll 5 or a 6 you Get your dollar back +2 more just like it the goal will be t

LiRa [457]

Answer:

(a)$67

(b)You are expected to win 56 Times

(c)You are expected to lose 44 Times

Step-by-step explanation:

The sample space for the event of rolling two dice is presented below

$(1,1), (2,1), (3,1), (4,1), (5,1), (6,1)\\(1,2), (2,2), (3,2), (4,2), (5,2), (6,2)\\(1,3), (2,3), (3,3), (4,3), (5,3), (6,3)\\(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)\\(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\\(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)$

Total number of outcomes =36

The event of rolling a 5 or a 6 are:

$(5,1), (6,1)\\ (5,2), (6,2)\\( (5,3), (6,3)\\ (5,4), (6,4)\\(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\\(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)$

Number of outcomes =20

Therefore:

P(rolling a 5 or a 6) $=\dfrac{20}{36}$

The probability distribution of this event is given as follows.

$\left|\begin{array}{c|c|c}$Amount Won(x)&-\$1&\$2\\&\\P(x)&\dfrac{16}{36}&\dfrac{20}{36}\end{array}\right|$

First, we determine the expected Value of this event.

Expected Value

$=(-\$1\times \frac{16}{36})+ (\$2\times \frac{20}{36})\\=\$0.67$

Therefore, if the game is played 100 times,

Expected Profit =$0.67 X 100 =$67

If you play the game 100 times, you can expect to win $67.

(b)

Probability of Winning $=\dfrac{20}{36}$

If the game is played 100 times

Number of times expected to win

$=\dfrac{20}{36} \times 100\\=56$ times$

Therefore, number of times expected to loose

= 100-56

=44 times

8 0

3 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

2 years ago

An amusement park has discovered that the brace that provides stability to the ferris

antiseptic1488 [7]

Answer:

Hence the correct answer is option (A) 4.708.

Step-by-step explanation:

Solution:-

Now, we need to find the angle of the sector by using sector formula,

$A = 28.25 ft^{2} \\r = 12ft.\\\\28.25 = \frac{\theta}{360^{o} } \times \pi \times 12^{2} \\\\\theta=\frac{28.25\times 360}{12 \times12 \times \pi} \\\\\theta =( \frac{70.625}{\\pi} )^{o}$

We now convert the angle to radian

$\theta = \frac{70.625}{\pi} \times \frac{\pi}{180}\\\\\theta = 0.39236 radian$

Now we calculate the length of arc by arc formula

$\theta=\frac{I}{12} \\\\0.39236 =\frac{I}{12}\\\\I = 4.708 feet.$

So, The length of steel that must be replaced is 4.708 ft.

4 0

2 years ago