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2 years ago

Out of 2500 light bulbs that were tested, 135 were defective. Based on this information calculate the following

Mathematics

1 answer:

snow_lady [41]2 years ago

7 0

Answer:

P(defective)= 135

P(not defective)= 2365

Step-by-step explanation:

2500-135=2365

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Given the function, g(x)=5x^5/x^3-2x+1, choose the correct horizontal asymptote. none y = 0 y = 1 y = 3

viktelen [127]

Answer: NONE

<u>Step-by-step explanation:</u>

Consider that m is the degree of the numerator and n is the degree of the denominator.

The rules for horizontal asymptote (H.A.) are as follows:

If m > n then no H.A. (use long division to find the slant asymptote)

If m = n then H.A. is y = leading coefficient of numerator/leading coefficient of denominator

If m < n then H.A. is y = 0

Given: g(x) = 5x⁵/(x³ - 2x + 1)

--> m = 5, n = 3

Since m > n then there is no H.A.

8 0

3 years ago

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True [87]

Answer:

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Step-by-step explanation:

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3 years ago

The region bounded by y=(3x)^(1/2), y=3x-6, y=0

Ganezh [65]

Answer:

4.5 sq. units.

Step-by-step explanation:

The given curve is $y = (3x)^{\frac{1}{2} }$

⇒ $y^{2} = 3x$ ...... (1)

This curve passes through (0,0) point.

Now, the straight line is y = 3x - 6 ....... (2)

Now, solving (1) and (2) we get,

$y^{2} - y - 6 = 0$

⇒ (y - 3)(y + 2) = 0

⇒ y = 3 or y = -2

We will consider y = 3.

Now, y = 3x - 6 has zero at x = 2.

Therefor, the required are = $\int\limits^3_0 {(3x)^{\frac{1}{2} } } \, dx - \int\limits^3_2 {(3x - 6)} \, dx$

= $\sqrt{3} [{\frac{x^{\frac{3}{2} } }{\frac{3}{2} } }]^{3} _{0} - [\frac{3x^{2} }{2} - 6x ]^{3} _{2}$

= $[\frac{\sqrt{3}\times 2 \times 3^{\frac{3}{2} } }{3}] - [13.5 - 18 - 6 + 12]$

= 6 - 1.5

= 4.5 sq. units. (Answer)

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3 years ago

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dezoksy [38]

I don’t know spanish sorry

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4 years ago

Clara scored a 79 on her history test what grade does she have to get on her next test so she will have a average of a 83

Airida [17]

79 is 4 below 83 so her next grade needs to be 4 above in order to get an average of 83. The answer is 87.

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4 years ago