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3 years ago

Plz help, asap, do not paste link for other website.

Mathematics

2 answers:

riadik2000 [5.3K]3 years ago

3 0

Answer:

D is the answer not a or b or c

Step-by-step explanation:

none

timurjin [86]3 years ago

3 0

Answer Choice D, If you graph the equation into the Desmos Graphing Calculater and graph the point (2,3) you will see that the point is right along the line of the equation.

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Chang knows one side of a triangle is 13 cm. Which set of two sides is possible for the lengths of the other two

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7 cm and 2 cm

Step-by-step explanatio

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3 years ago

Read 2 more answers

Evaluate the integral ∫2−1|x−1|dx

defon

I think you might be referring to the definite integral,

$\displaystyle \int_{-1}^2|x-1|\,\mathrm dx$

Recall the definition of absolute value:

$|x| = \begin{cases}x&\text{if }x\ge0\\-x&\text{if }x$

Then $|x-1|=x-1$ if $x\ge1$ , and $|x-1|=1-x$ is $x$ . So spliting up the integral at <em>x</em> = 1, we have

$\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \int_{-1}^1(1-x)\,\mathrm dx + \int_1^2(x-1)\,\mathrm dx$

The rest is simple:

$\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \left(x-\dfrac{x^2}2\right)\bigg|_{-1}^1 + \left(\dfrac{x^2}2-x\right)\bigg|_1^2 \\\\ = \left(\left(1-\frac12\right)-\left(-1-\frac12\right)\right) + \left(\left(2-2\right)-\left(\frac12-1\right)\right) \\\\ = \boxed{\frac52}$

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3 years ago

Example 2: Combining Use of the Multiplication and Addition Rules

bulgar [2K]

Answer:

The probability of getting 6's on both cubes is $\frac{1}{36}$ .

The probability that the total score is at least 11 is $\frac{1}{12}$ .

Step-by-step explanation:

Consider the provided information.

A red cube has faces labeled 1 through 6, and a blue cube has faces labeled in the same way.

Part (A) Both cubes show 6’s.

Probability of getting 6 on red cube is $\frac{1}{6}$

Probability of getting 6 on blue cube is $\frac{1}{6}$

Thus, the probability of getting 6's on both cubes is:

$P(\text{Both 6's})=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$

Hence, the probability of getting 6's on both cubes is $\frac{1}{36}$ .

Part (B) The total score is at least 11.

The possible number of outcomes in which total score is at least 11 is:

Red shows 6 and Blue shows 5.

Blue shows 6 and Red shows 5.

Blue shows 6 and Red shows 6.

Thus, the probability of total score is at least 11.

$P(\text{Total is at least 11})=\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}\\P(\text{Total is at least 11})=\frac{1}{12}$

Hence, the probability that the total score is at least 11 is $\frac{1}{12}$ .

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