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Amiraneli [1.4K]

3 years ago

9

Adam is climbing up the stairs in a building. At a certain point, Adam starts counting the steps and when he reaches the last st

ep, he has counted 56 steps. If he knows that there are a total of 100 steps in the building, how many steps did he climb before he started counting the steps?

2 answers:

Katena32 [7]3 years ago

7 0

Answer:

The answer is he climbed 44 steps. Hope this helped :D

Step-by-step explanation:

Ksivusya [100]3 years ago

5 0

Answer:

He climed 3 steps before he started counting the steps.

Step-by-step explanation:

pls mark me as a branliest

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Please Hurry, I need an answer, quick my assignment is due today.

lys-0071 [83]

Answer:

8.25 hrs

Step-by-step explanation:

it takes 3/4 to replace all tires in a car

you then multiply 3/4 by 11

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3 years ago

The sum of two numbers is 72 while the difference between the numbers is 25. the larger of the numbers is

7nadin3 [17]

2 equations: the first one is the sum of 2 unknown numbers x and y: x + y = 72. Now the difference is 25, so x - y = 25. Solve the first equation for y: y = 72 - x. Sub that into the second equation to solve for x: x - (72 - x) = 25. x - 72 + x = 25 and 2x - 72 = 25. Therefore, 2x = 97 and x = 48.5 That's the larger of the 2 numbers; since we subtracted y from x we defined x as the larger of the 2. Just for fun, plug x in to find that y = 23.5

7 0

2 years ago

Please help! I will mark as brainliest IF answer is right. <3

EastWind [94]

Answer:

b^2 + 3b - 4 + 4/(b-2)

Step-by-step explanation:

see pic

7 0

2 years ago

Read 2 more answers

3. Which statements about the line that passes through (-2, 0) and (2, -4) are true?

diamong [38]

Answer:

B, C, and D.

Step-by-step explanation:

Let's go through each answer choice.

A: To find the slope of a line from two points, we use this formula: $(y_2-y_1)/(x_2-x_1)$ . Plugging in the values (-2,0) and (2, -4), we get:

$(-4-0)/(2+2)$

$-4/4$
$-1$

Because option A says the slope is 1, but we got -1, option A is false.

B: To find the y-intercept, we use this formula: $b = y_1 - m * x_1$

$b = 0 -(-1) * (-2)$

$b = 1 * (-2)$

$b = -2$

Since we know the value of b is the y-intercept, we could write the y-intercept as (0,-2). Therefore, option B is true.

C: The equation of a line (in slope-intercept form) is y = mx + b, where y - y-coordinate, m = slope, x = x-coordinate, and b = y-intercept. Since we already solved for the y-intercept and slope, let's plug in those values:

$y = mx +b$
$y = -x -2$

As you can see, the equation we got here matches the one in option C, meaning it's true.

D: We know that a point is written like this: (x, y). If the x-coordinate is not zero but the y-coordinate is, then it's the x-intercept. Since the question says that (-2,0) is one of the points the line passes through, and option D says that the x-intercept is (-2,0), it's true.

8 0

2 years ago

Read 2 more answers

The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β =

stich3 [128]

I'm assuming $\alpha$ is the shape parameter and $\beta$ is the scale parameter. Then the PDF is

$f_X(x)=\begin{cases}\dfrac29xe^{-x^2/9}&\text{for }x\ge0\\\\0&\text{otherwise}\end{cases}$

a. The expectation is

$E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\frac29\int_0^\infty x^2e^{-x^2/9}\,\mathrm dx$

To compute this integral, recall the definition of the Gamma function,

$\Gamma(x)=\displaystyle\int_0^\infty t^{x-1}e^{-t}\,\mathrm dt$

For this particular integral, first integrate by parts, taking

$u=x\implies\mathrm du=\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X]=\displaystyle-xe^{-x^2/9}\bigg|_0^\infty+\int_0^\infty e^{-x^2/9}\,\mathrm x$

$E[X]=\displaystyle\int_0^\infty e^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ , so that $\mathrm dx=\dfrac32y^{-1/2}\,\mathrm dy$ :

$E[X]=\displaystyle\frac32\int_0^\infty y^{-1/2}e^{-y}\,\mathrm dy$

$\boxed{E[X]=\dfrac32\Gamma\left(\dfrac12\right)=\dfrac{3\sqrt\pi}2\approx2.659}$

The variance is

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2$

The second moment is

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f_X(x)\,\mathrm dx=\frac29\int_0^\infty x^3e^{-x^2/9}\,\mathrm dx$

Integrate by parts, taking

$u=x^2\implies\mathrm du=2x\,\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X^2]=\displaystyle-x^2e^{-x^2/9}\bigg|_0^\infty+2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

$E[X^2]=\displaystyle2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ again to get

$E[X^2]=\displaystyle9\int_0^\infty e^{-y}\,\mathrm dy=9$

Then the variance is

$\mathrm{Var}[X]=9-E[X]^2$

$\boxed{\mathrm{Var}[X]=9-\dfrac94\pi\approx1.931}$

b. The probability that $X\le3$ is

$P(X\le 3)=\displaystyle\int_{-\infty}^3f_X(x)\,\mathrm dx=\frac29\int_0^3xe^{-x^2/9}\,\mathrm dx$

which can be handled with the same substitution used in part (a). We get

$\boxed{P(X\le 3)=\dfrac{e-1}e\approx0.632}$

c. Same procedure as in (b). We have

$P(1\le X\le3)=P(X\le3)-P(X\le1)$

and

$P(X\le1)=\displaystyle\int_{-\infty}^1f_X(x)\,\mathrm dx=\frac29\int_0^1xe^{-x^2/9}\,\mathrm dx=\frac{e^{1/9}-1}{e^{1/9}}$

Then

$\boxed{P(1\le X\le3)=\dfrac{e^{8/9}-1}e\approx0.527}$

7 0

2 years ago