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scoundrel [369]
2 years ago
12

6x2-30=0

Mathematics
2 answers:
katrin2010 [14]2 years ago
7 0

Step-by-step explanation:

6 {x}^{2}  - 30  = 0 \\  \therefore \: 6( {x}^{2}  - 5) = 0 \\ \therefore \: {x}^{2}  - 5 = 0 \\ \therefore \: {x}^{2}  =  5\\ \therefore \: x = \pm \:  \sqrt{5}\\  \\

agasfer [191]2 years ago
6 0

The quadratic form is ax^2 + bx + c

The given equation is 6x^2 -30 = 0

a = 6 because it is with x^2.

There is no x term in the equation so b is 0

c is the numerical term with no variable so c = -30

a = 6

b = 0

c =-30

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If the zeros of a quadratic equation are<br> 7 and 4, what would be the factors?
Snowcat [4.5K]

Answer:

The factors are (x-7) and (x-4)

Step-by-step explanation:

We are finding the factors of a function from the zeros

The zeros are 7 and 4

We can use the zero product property which is

(x-b)(x-c) = 0  where b and c are the zeros

(x-7) (x-4) =0

The factors are (x-7) and (x-4)

4 0
1 year ago
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Which of the following expressions is equivalent to the expression?
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Answer should be C! Have a great day!
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3 years ago
A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l
Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = <u><em>the length of the calls, in minutes.</em></u>

So, X ~ Normal(\mu=4.5,\sigma^{2} =0.70^{2})

The z-score probability distribution for the normal distribution is given by;

                           Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time = 4.5 minutes

           \sigma = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X \leq 4.50 min)

    P(X < 5.30 min) = P( \frac{X-\mu}{\sigma} < \frac{5.30-4.5}{0.7} ) = P(Z < 1.14) = 0.8729

    P(X \leq 4.50 min) = P( \frac{X-\mu}{\sigma} \leq \frac{4.5-4.5}{0.7} ) = P(Z \leq 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = <u>0.3729</u>.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

    P(X > 5.30 min) = P( \frac{X-\mu}{\sigma} > \frac{5.30-4.5}{0.7} ) = P(Z > 1.14) = 1 - P(Z \leq 1.14)

                                                              = 1 - 0.8729 = <u>0.1271</u>

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X \leq 5.30 min)

    P(X < 6.00 min) = P( \frac{X-\mu}{\sigma} < \frac{6-4.5}{0.7} ) = P(Z < 2.14) = 0.9838

    P(X \leq 5.30 min) = P( \frac{X-\mu}{\sigma} \leq \frac{5.30-4.5}{0.7} ) = P(Z \leq 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = <u>0.1109</u>.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X \leq 4.00 min)

    P(X < 6.00 min) = P( \frac{X-\mu}{\sigma} < \frac{6-4.5}{0.7} ) = P(Z < 2.14) = 0.9838

    P(X \leq 4.00 min) = P( \frac{X-\mu}{\sigma} \leq \frac{4.0-4.5}{0.7} ) = P(Z \leq -0.71) = 1 - P(Z < 0.71)

                                                              = 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = <u>0.745</u>.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

            P(X > x) = 0.05            {where x is the required time}

            P( \frac{X-\mu}{\sigma} > \frac{x-4.5}{0.7} ) = 0.05

            P(Z > \frac{x-4.5}{0.7} ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

                      \frac{x-4.5}{0.7}=1.645

                      {x-4.5}{}=1.645 \times 0.7

                       x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0
3 years ago
What is sin B ?
Neporo4naja [7]

Answer:

The answer to your question is sin B = \frac{8}{17}

Step-by-step explanation:

Sine is the trigonometric function that relates the opposite side and the hypotenuse.

In the picture, we have the hypotenuse and the adjacent side, so we must calculate the opposite side using the Pythagorean theorem.

                b² = c² - a²

                b² = 17² - 15²

                b² = 289 - 225

                b² = 64

               b = 8

Now, we can calculate the sine

               sin B = \frac{opposite side}{hypotenuse}

               sin B = \frac{8}{17}

3 0
2 years ago
The graph of the function f is shown. f(0) =<br> A) f(-1) <br> B) f(2) <br> C) f(3) <br> D) f(4)
aalyn [17]

f(0)  means x = 0

so when x = 0, y = -5

and when x = 4, f(4) is also equal -5 (y = -5)

so f(0) = f(4) = -5

Answer

D) f(4)

5 0
2 years ago
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