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earnstyle [38]
3 years ago
6

A group of friends wants to go to the amusement park. They have no more than $415 to spend on parking and admission. Parking is

$19, and tickets cost $36 per person, including tax. Use the drop-down menu below to write an inequality representing pp, the number of people who can go to the amusement park.
Mathematics
1 answer:
svetoff [14.1K]3 years ago
6 0

Answer:

p≤11

that's all i got from the delta math thing

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The Department of Commerce in a particular state has determined that the number of small businesses that declare bankruptcy per
skad [1K]

Answer:

B) 0.1487

Step-by-step explanation:

Let X be the discrete random variable that represents the number of events observed over a given time period.  If X follows a Poisson distribution, then the probability of observing k events over the time period is:

P(X=k)=\frac{\lambda^{k} *e^{-\lambda} }{k!}

Where:

\lambda=Mean\\k=number\hspace{3}of\hspace{3}events\\e=Euler's\hspace{3}number

So, the probability that exactly 5 bankruptcies occur next month is:

P(X=5)=\frac{6.4^{5} *e^{-6.4} }{5!} =\frac{17.84083537}{120} =0.1486736281\approx0.1487

4 0
3 years ago
What is the sign of 3x y3xy3, x, y when x>0x>0x, is greater than, 0 and y<0y<0y, is less than, 0?
kaheart [24]

Answer:

B ) Negative

Step-by-step explanation:

Given the expression 3xy

when x>0 and y<0

The value of the expression 3xy will be negative.

This is because the product of a positive and negative number will always be negative.

For illustration,

Let x=5>0 and y=-5<0

3xy=3(5)(-5)=-75

NOTE:

-X-=+\\-X+=-\\+X+=+

5 0
3 years ago
PRECAL:<br> Having trouble on this review, need some help.
ra1l [238]

1. As you can tell from the function definition and plot, there's a discontinuity at x = -2. But in the limit from either side of x = -2, f(x) is approaching the value at the empty circle:

\displaystyle \lim_{x\to-2}f(x) = \lim_{x\to-2}(x-2) = -2-2 = \boxed{-4}

Basically, since x is approaching -2, we are talking about values of x such x ≠ 2. Then we can compute the limit by taking the expression from the definition of f(x) using that x ≠ 2.

2. f(x) is continuous at x = -1, so the limit can be computed directly again:

\displaystyle \lim_{x\to-1} f(x) = \lim_{x\to-1}(x-2) = -1-2=\boxed{-3}

3. Using the same reasoning as in (1), the limit would be the value of f(x) at the empty circle in the graph. So

\displaystyle \lim_{x\to-2}f(x) = \boxed{-1}

4. Your answer is correct; the limit doesn't exist because there is a jump discontinuity. f(x) approaches two different values depending on which direction x is approaching 2.

5. It's a bit difficult to see, but it looks like x is approaching 2 from above/from the right, in which case

\displaystyle \lim_{x\to2^+}f(x) = \boxed{0}

When x approaches 2 from above, we assume x > 2. And according to the plot, we have f(x) = 0 whenever x > 2.

6. It should be rather clear from the plot that

\displaystyle \lim_{x\to0}f(x) = \lim_{x\to0}(\sin(x)+3) = \sin(0) + 3 = \boxed{3}

because sin(x) + 3 is continuous at x = 0. On the other hand, the limit at infinity doesn't exist because sin(x) oscillates between -1 and 1 forever, never landing on a single finite value.

For 7-8, divide through each term by the largest power of x in the expression:

7. Divide through by x². Every remaining rational term will converge to 0.

\displaystyle \lim_{x\to\infty}\frac{x^2+x-12}{2x^2-5x-3} = \lim_{x\to\infty}\frac{1+\frac1x-\frac{12}{x^2}}{2-\frac5x-\frac3{x^2}}=\boxed{\frac12}

8. Divide through by x² again:

\displaystyle \lim_{x\to-\infty}\frac{x+3}{x^2+x-12} = \lim_{x\to-\infty}\frac{\frac1x+\frac3{x^2}}{1+\frac1x-\frac{12}{x^2}} = \frac01 = \boxed{0}

9. Factorize the numerator and denominator. Then bearing in mind that "x is approaching 6" means x ≠ 6, we can cancel a factor of x - 6:

\displaystyle \lim_{x\to6}\frac{2x^2-12x}{x^2-4x-12}=\lim_{x\to6}\frac{2x(x-6)}{(x+2)(x-6)} = \lim_{x\to6}\frac{2x}{x+2} = \frac{2\times6}{6+2}=\boxed{\frac32}

10. Factorize the numerator and simplify:

\dfrac{-2x^2+2}{x+1} = -2 \times \dfrac{x^2-1}{x+1} = -2 \times \dfrac{(x+1)(x-1)}{x+1} = -2(x-1) = -2x+2

where the last equality holds because x is approaching +∞, so we can assume x ≠ -1. Then the limit is

\displaystyle \lim_{x\to\infty} \frac{-2x^2+2}{x+1} = \lim_{x\to\infty} (-2x+2) = \boxed{-\infty}

6 0
2 years ago
Find the scale factor abc def
Vesna [10]

Answer:

The scale factor is 1.2.

Step-by-step explanation:

18/15 = 1.2

30/25 = 1.2

24/20 = 1.2

Hope that helps!

8 0
3 years ago
Free question
jarptica [38.1K]

Answer:

3/4 I think its that

Step-by-step explanation:

its a guess

5 0
3 years ago
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