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joja [24]
2 years ago
14

Help pls pls!!! tyty​

Mathematics
1 answer:
blsea [12.9K]2 years ago
4 0

here's the link to asnwer:

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PLEASE ANSWER THIS: look at the pic for question. Thanks!!!
Nadya [2.5K]

Answer:

\text{D. }b^2-4ac>0

Step-by-step explanation:

The equation b^2-4ac represents the discriminant of a quadratic. It is the part taken from under the radical in the quadratic formula.

For any quadratic:

  • If the discriminant is positive, or greater than 0, the quadratic has two solutions
  • If the discriminant is equal to 0, the quadratic has one distinct real solution (the solution is repeated).
  • If the discriminant is negative, or less than 0, the quadratic has zero solutions

In the graph, we see that the equation intersects the x-axis at two distinct points. Therefore, the quadratic has two solutions and the discriminant must be positive. Thus, we have b^2-4ac>0.

4 0
2 years ago
Please answer this question
Scrat [10]

Answer:

the answer would be 1/225

4 0
3 years ago
1.) Consider the right triangle below in which a = 5 and b = 5.
patriot [66]

Answer:

c=7.07

Step-by-step explanation:

c²=a²+b²

c²=5²+5²

c²=25+25

c²=50

c=√50

c=7.07

B. c=7.07 (to nearest hundredth)

C. c=√50= 5√2 in radical form

3 0
2 years ago
PLEASE HELP WILL MARK BRAINLIEST
egoroff_w [7]
Option 4 is the correct answer
3 0
3 years ago
Read 2 more answers
Suppose that, after measuring the duration of many telephone calls, a telephone company found their data was well-approximated b
Musya8 [376]

Answer:

a) 7.79%

b) 67.03%

c) Cumulative Distribution Function

P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b

Step-by-step explanation:

We are given the following in the question:

p(x) = 0.1 e^{-0.1x}

where x is the duration of a call, in minutes.

a) P( calls last between 2 and 3 minutes)

=\displaystyle\int^3_2 p(x)~ dx\\\\= \displaystyle\int^3_20.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^3_2\\\\=-\Big[e^{-0.3}-e^{-0.2}\Big]\\\\= 0.0779\\=7.79\%

b) P(calls last 4 minutes or more)

=\displaystyle\int^{\infty}_4 p(x)~ dx\\\\= \displaystyle\int^{\infty}_40.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^{\infty}_4\\\\=-\Big[e^{\infty}-e^{-0.4}\Big]\\\\=-(0- 0.6703)\\= 0.6703\\=67.03\%

c) cumulative distribution function

P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b

6 0
3 years ago
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