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4 years ago

All rational numbers are whole numbers

Mathematics

1 answer:

Ilia_Sergeevich [38]4 years ago

7 0

A rational no. can be expressed as a fraction/quotient
eg. 2/5 or 2/1
thus not all rational no. are whole no.s

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How much will 25 pounds of bananas cost if they are $0.59/pound

Nadusha1986 [10]

If you times 25 by .59 you'll get 14.75 so that is probs the answer

5 0

4 years ago

Read 2 more answers

use a(t)=-32 feet per second squared as the acceleration due to gravity. A ball is thrown vertically upward from the ground with

liberstina [14]

Answer:

144 ft

Step-by-step explanation:

The speed decreases linearly by 32 ft/sec/sec, so will be zero after ...

(96 ft/s)/(32 ft/s²) = 3 s

The average speed over that period is half the initial speed, so the distance covered between the ground and the maximum height is ...

(3 s)(96 ft/s)/2 = 144 ft

_____

<em>Alternate solution</em>

The relationship between velocity and distance for a constant acceleration is ...

$v_f^{2}- v_i^{2} = 2ad$

For an initial velocity of 96 ft/s, a final velocity (at the top of travel) of 0, and the given acceleration, we have ...

$d = \dfrac{v_f^{2}- v_i^{2}}{2a} = \dfrac{0-96^{2}}{2(-32)}= 144\,\dots\ \text{ft}$

4 0

3 years ago

Read 2 more answers

Find the sum of

Bad White [126]

Answer:

Step-by-step explanation:

(2x^2-x)+(x^2+3x-1)

2x^2-x+x^2+3x-1

x^2+3x-1

5x^2+2x-1+(4x^2-x+2)

5x^2+2x-1+4x^2+x-2

9x^2+6x-3

(3x^2+2x+1)+(x^2-1)

3x^2+2x+1+x^2-1

4x^2+2x

-x^2-1+(4x^2-x)

-x^2-1+4x^2-x

3x^2-1-x

5 0

3 years ago

(a) A lamp has two bulbs of a type with an average lifetime of 1600 hours. Assuming that we can model the probability of failure

lara [203]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability is $P_T= 0.4560$

The probability is $P_F= 0.0013$

Step-by-step explanation:

From the question we are told that

The mean for the exponential density function of bulbs failure is $\mu = 1600 \ hours$

Generally the cumulative distribution for exponential distribution is mathematically represented as

$1 - e^{- \lambda x}$

The objective is to obtain the p=probability of the bulbs failure within 1800 hours

So for the first bulb the probability will be

$P_1(x < 1800)$

And for the second bulb the probability will be

$P_2 (x< 1800)$

So from our probability that we are to determine the area to the left of 1800 on the distribution curve

Now the rate parameter $\lambda$ is mathematically represented as

$\lambda$ $= \frac{1}{\mu}$

$\lambda = \frac{1}{1600}$

The probability of the first bulb failing with 1800 hours is mathematically evaluated as

$P_1(x < 1800) = 1 - e^{\frac{1}{1600} * 1800 }$

$= 0.6753$

Now the probability of both bulbs failing would be

$P_T=P_1(x < 1800) * P_2(x < 1800)$

$= 0.6375 * 06375$

$P_T= 0.4560$

Let assume that one bulb failed at time $T_a$ and the second bulb failed at time $T_b$ then

$T_a + T_b = 1800\ hours$

The mathematical expression to obtain the probability that the first bulb failed within between zero and $T_a$ and the second bulb failed between $T_a \ and \ 1800$ is represented as

$P_F=\int_{0}^{1800}\int_{0}^{1800-x} \f{\lambda }^{2}e^{-\lambda x}* e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\lambda }e^{-\lambda x}\int_{0}^{1800-x} {\lambda } e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\lambda x}\int_{0}^{1800-x} \frac{1}{1600 } e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\lambda x}[e^{- \lambda y}]\left {1800-x} \atop {0}} \right. dx$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\frac{x}{1600} }[e^{- \frac{1800 -x}{1600} }-1] dx$

$=[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{- \frac{x}{1600} }] \left {1800} \atop {0}} \right.$

$=[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{- \frac{1800}{1600} }] -[[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{-0}]$