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alex41 [277]
3 years ago
7

The probability that a battery will last 10 hr or more is 0.9, and the probability that it will last 15 hr or more is 0.12. Give

n that a battery has lasted 10 hr, find the probability that it will last 15 hr or more. (Round your answer to three decimal places.)
Mathematics
1 answer:
Dahasolnce [82]3 years ago
4 0

Answer:

Answer is 13.3%

Step-by-step explanation:

  • 90 percent of the batteries will last 10 hours or exceed.
  • 12 percent  of the batteries will last 15 hours or more.

If the likelihood of these batteries is to hold for the given time period, then, out of one thousand (1000) batteries produced, nine hundred (900)  of the batteries will have long lasted 10 hours or longer and 120 of them will have lasted 15 hours or more.

The 120 which have lasted 15 hours or more must have lasted at most 10 hours or more, so they are part of the bunch that lasted 10 hours of more.

They are 12% of the total of 1000 batteries produced, but they are 15% of the 900 batteries that lasted 10 hours or more, Why? view the following:

  • 120 / 900 = 0.1333 = 13.3%

The solution which seems to be that 13.3 percent of the batteries that prolonged 10 hours or more lasted 15 hours or longer.

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snow_tiger [21]

The equation of line in standard form is 5x - 4y = -38

<em><u>Solution:</u></em>

Given that we have to find the equation of line perpendicular to 4x + 5y = 40 that includes the point (-10, -3)

<em><u>The equation of line in slope intercept form is given as:</u></em>

y = mx + c ------ eqn 1

Where "m" is the slope of line and "c" is the y - intercept

Given equation of line is:

4x + 5y = 40

Rearranging to slope intercept form, we get

5y = -4x + 40

y = \frac{-4}{5}x + \frac{40}{5}\\\\y = \frac{-4}{5}x + 8

On comparing the above equation with eqn 1,

m = \frac{-4}{5}

We know that product of slope of line and slope of line perpendicular to given line is equal to -1

Therefore,

\frac{-4}{5} \times \text{ slope of line perpendicular to it } = -1\\\\\text{ slope of line perpendicular to it } = \frac{5}{4}

Now find the equation of line with slope 5/4 and passing through (-10, -3)

Substitute m = \frac{5}{4} and (x, y) = (-10, -3) in eqn 1

-3 = \frac{5}{4}(-10) + c\\\\-3 = \frac{-25}{2} + c\\\\c = -3 + \frac{25}{2}\\\\c = \frac{-6 + 25}{2}\\\\c = \frac{19}{2}

Substitute c = \frac{19}{2} and m = \frac{5}{4} in eqn 1

y = \frac{5}{4}x + \frac{19}{2}

The standard form of an equation is Ax + By = C

Therefore,

\frac{5}{4}x - y = -\frac{19}{2}\\\\\frac{5x - 4y}{4} = \frac{-19}{2}\\\\5x - 4y = -38

Thus the equation of line in standard form is found

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What is the solution of the equation (x – 5)^2 + 3(x – 5) + 9 = 0? Use u substitution and the quadratic formula to solve.
xz_007 [3.2K]

Answer:

x = \dfrac{7 \pm 3i\sqrt{3}}{2}

Step-by-step explanation:

(x – 5)^2 + 3(x – 5) + 9 = 0

This is a quadratic equation in x - 5.

Let u = x - 5, then the quadratic equation becomes:

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We can use the quadratic formula to solve for u.

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u = \dfrac{-3 \pm \sqrt{3^2 - 4(1)(9)}}{2(1)}

u = \dfrac{-3 \pm \sqrt{9 - 36}}{2}

u = \dfrac{-3 \pm \sqrt{-27}}{2}

u = \dfrac{-3 \pm 3i\sqrt{3}}{2}

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x - 5 = \dfrac{-3 \pm 3i\sqrt{3}}{2}

x = \dfrac{-3 \pm 3i\sqrt{3}}{2} + 5

x = \dfrac{-3 \pm 3i\sqrt{3}}{2} + \dfrac{10}{2}

x = \dfrac{7 \pm 3i\sqrt{3}}{2}

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3 years ago
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