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3 years ago

Rick eats 8 burgers and has 563 leftover how many did he have to begin with

Mathematics

2 answers:

Alex73 [517]3 years ago

3 0

563+8=571 burgers to start with

Luba_88 [7]3 years ago

3 0

He had 571 at first because
563 + 8 = 571. :) :) :) :)

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4 years ago

Use the diagram to complete these statements. If m26 = 50°, then m2 = 50° If m 25 = 70°, then m = 70 What is the measure of 22?

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3 years ago

Sqrt(2x-6)=3-x<br> Solve for x

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Answer:

x = 3

Step-by-step explanation:

$\sqrt{2x - 6} = 3 - x$

Square both sides of the equation

$2x - 6 = (3 - x)^{2} = 9 - 6x + x^{2} \\$

$x^{2} - 8x + 15 = 0\\$

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Now, you must always check your results because a result may not satisfy the original equation.

If x = 3, then $\sqrt{2x - 6} = \sqrt{2(3) - 6} = \sqrt{6 - 6} = \sqrt{0} = 0$ and 3 - x = 3 - 3 = 0

So 3 satisfies the original.

If x = 5, then $\sqrt{2(5) - 6} = \sqrt{10 - 6} = \sqrt{4} = 2$ , but 3 - x = 3 - 5 = -2. Therefore, 5 does NOT satisfy the original equation.

That means that x = 3 is the solution to the equation.

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3 years ago

I need to know on how to do these problems

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Just remember that
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3 years ago

Read 2 more answers

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

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

7 0

3 years ago