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3 years ago

2.5d + 7.25 = 1 + 3.75d solve for d

Mathematics

1 answer:

dimulka [17.4K]3 years ago

4 0

Answer:

Step-by-step explanation:

Subtract 1 from both sides. The equation becomes 2.5d + 6.25 = 3.75d

Subtract 2.5d from both sides. The equation becomes 6.25 = 1.25d

Divide 1.25 from both sides. You get 5 = d

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5t-14=-14 solve the equation for t

ikadub [295]

Hi there!

Before we begin, let's rewrite your equation. :)

$5t - 14 = -14$

Step 1) Add 14 to both sides.

$5t - 14 + 14 = -14 + 14 5t = 0$

Step 2) Divide both sides by 5.

$\frac{5t}{5} = \frac{0}{5}$

$t = 0$

Final Answer -

$t = 0$

Hope this helps!

Message me if you need anymore help! :D

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4 years ago

What is the distance from the origin to the point P(3,2)

Alenkasestr [34]

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3 years ago

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The revenue, in dollars, of a company that makes toy cars can be modeled by the polynomial 3x2 + 4x – 60. The cost, in dollars,

Tems11 [23]

For this case we have the following quadratic functions:

Revenue: $3x ^ 2 + 4x - 60$

Cost: $3x ^ 2 - x + 200$

Then, we observe that the profit is given by the following mathematical relationship:

$Profit = Revenue - Cost$

Substituting values we have:

$Profit = (3x ^ 2 + 4x - 60) - (3x ^ 2 - x + 200)$

Making the corresponding calculations we have:

$Profit = x ^ 2 (3-3) + x (4 + 1) + (-60-200)\\Profit = 5x - 260$

Answer:

An expression that represents the profit is:

$Profit = 5x - 260$

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3 years ago

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Find the quotient of -85^2-27n+76 / -17n-19

bagirrra123 [75]

Answer:
5n-4

Explanation:
The given expression is:
$\frac{-85n^2-27n+76}{-17n-19}$

1- Take the negative as a common factor from both the numerator and the denominator. This will give us:
$\frac{-(85n^2+27n-76)}{-(17n+19)}$

2- Cancel out the negative sign (common factor) from the numerator and denominator. This will give us:
$\frac{(85n^2+27n-76)}{(17n+19)}$

3- Factor the numerator. This wil give:
$\frac{(5n-4)(17n+19)}{(17n+19)}$

4- Finally, cancel out the (17n+19) which common in both numerator and denominator. This will give us the final expression:
5n-4

Hope this helps :)

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3 years ago

The probability density function of the time you arrive at a terminal (in minutes after 8:00 A.M.) is f(x) = 0.1 exp(−0.1x) for

Blababa [14]

$f_X(x)=\begin{cases}0.1e^{-0.1x}&\text{for }x>0\\0&\text{otherwise}\end{cases}$

a. 9:00 AM is the 60 minute mark:

$f_X(60)=0.1e^{-0.1\cdot60}\approx0.000248$

b. 8:15 and 8:30 AM are the 15 and 30 minute marks, respectively. The probability of arriving at some point between them is

$\displaystyle\int_{15}^{30}f_X(x)\,\mathrm dx\approx0.173$

c. The probability of arriving on any given day before 8:40 AM (the 40 minute mark) is

$\displaystyle\int_0^{40}f_X(x)\,\mathrm dx\approx0.982$

The probability of doing so for at least 2 of 5 days is

$\displaystyle\sum_{n=2}^5\binom5n(0.982)^n(1-0.982)^{5-n}\approx1$

i.e. you're virtually guaranteed to arrive within the first 40 minutes at least twice.

d. Integrate the PDF to obtain the CDF:

$F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x$

Then the desired probability is

$F_X(30)-F_X(15)\approx0.950-0.777=0.173$

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3 years ago