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3 years ago

8

Let g be the function given by g(x)=x^2*e^(kx) , where k is a constant. For what value of k does g have a critical point at x=2/

3?

1 answer:

ArbitrLikvidat [17]3 years ago

8 0

G `( x ) = $2x * e^{kx} + k x^{2} *e^{kx} = \\ = xe^{kx}(2 + kx )$
2 + k x = 0
k x = -2
k = -2: x = - 2 : 2/3 = - 2 * 3/2
k = - 3
Answer: for k= - 3, the function g ( x ) have a critical point at x = 2/3.

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Ann is 5 feet 3 inches tall. To find the height of a lamppost, she measured her shadow to be 8 feet 9 inches and the lamppost’s

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Answer: The lamppost is 7 feet 2 inches

Step-by-step explanation: If Ann measured her own height and her shadow, then what she used is a ratio between both measurements. If she can measure the shadow of the lamppost, then she can use the same ratio of her height and it’s shadow to derive the correct measurement of the lamppost.

If Ann’s height was measured as 5 feet 3 inches, and her shadow was 8 feet 9 inches, the ratio between them can be expressed as 3:5.

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Reduce to the least fraction

Ratio = 3/5

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3 years ago

While conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modem

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Answer:

We conclude that this is an unusually high number of faulty modems.

Step-by-step explanation:

We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.

The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.

Let p = <em><u>population proportion</u></em>.

So, Null Hypothesis, $H_0$ : p = 0.013 {means that this is an unusually 0.013 proportion of faulty modems}

Alternate Hypothesis, $H_A$ : p > 0.013 {means that this is an unusually high number of faulty modems}

The test statistics that would be used here <u>One-sample z-test</u> for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion faulty modems= $\frac{10}{367}$ = 0.027

n = sample of modems = 367

So, <u><em>the test statistics</em></u> = $\frac{0.027-0.013}{\sqrt{\frac{0.013(1-0.013)}{367} } }$

= 2.367

The value of z-test statistics is 2.367.

Since, we are not given with the level of significance so we assume it to be 5%. <u>Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.</u>

Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.

Therefore, we conclude that this is an unusually high number of faulty modems.

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The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 -57, where t is the time

cupoosta [38]

A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?

Answer:

$1\leq t\geq \sqrt{2}$

Step-by-step explanation:

Given:

A stuntman jumping off a 20-m-high building is modeled by the equation

$h =20-5t^{2}$ -----------(1)

A high-speed camera is ready to making film between 15 m and 10 m above the ground

when the stuntman is 15m above the ground.

height $h = 15m$

Put height value in equation 1

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$5t^{2} =20-15$

$5t^{2} =5$

$t^{2} =1$

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We know that the time is always positive, therefore $t=1$

when the stuntman is 10m above the ground.

height $h = 10m$

Put height value in equation 1

$10 =20-5t^{2}$

$5t^{2} =20-10$

$5t^{2} =10$

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