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Daniel [21]
3 years ago
5

This is the question ​

Mathematics
2 answers:
frutty [35]3 years ago
8 0
C ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎
Finger [1]3 years ago
7 0

Answer:

C

Step-by-step explanation:

over the interval [-1,1], f(x) is greater than 0

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I need you to answer with a, b, c, d
solong [7]

To find the zeros of a quadratic fiunction given the equation you can use the next quadratic formula after equal the function to 0:

\begin{gathered} ax^2+bx+c=0 \\  \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}

For the given function:

f(x)=2x^2-10x-3x=\frac{-(-10)\pm\sqrt[]{(-10)^2-4(2)(-3)}}{2(2)}x=\frac{10\pm\sqrt[]{100+24}}{4}\begin{gathered} x=\frac{10\pm\sqrt[]{124}}{4} \\  \\ x=\frac{10\pm\sqrt[]{2\cdot2\cdot31}}{4} \\  \\ x=\frac{10\pm\sqrt[]{2^2\cdot31}}{4} \\  \\ x=\frac{10\pm2\sqrt[]{31}}{4} \\  \end{gathered}\begin{gathered} x_1=\frac{10}{4}+\frac{2\sqrt[]{31}}{4} \\  \\ x_1=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \end{gathered}\begin{gathered} x_2=\frac{10}{4}-\frac{2\sqrt[]{31}}{4} \\  \\ x_2=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}

Then, the zeros of the given quadratic function are:

\begin{gathered} x=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \\  \\ x_{}=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}

Answer: Third option

8 0
1 year ago
What's sixty nine thousand one hundred eight written in standred form
Vadim26 [7]
The answer is 69,108
3 0
3 years ago
Can someone answer this !
rodikova [14]

Answer:

11253

Step-by-step explanation:

the steps are described in the picture above

6 0
3 years ago
In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the
earnstyle [38]

Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

<em>A</em> = video components need repair within 1 year

<em>B</em> = electronic components need repair within 1 year

<em>C</em> = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events <em>A</em>, <em>B</em> and <em>C</em> are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

7 0
3 years ago
There is a 60% chance that your flight will leave on time
OLEGan [10]
A 40% chance





Unless!
*wispers*
It’s a trick question?
6 0
3 years ago
Read 2 more answers
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