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2 years ago

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Use the position function s(t) = −16t² + 400, which gives the height (in feet) of an object that has fallen for t seconds from a

height of 400 feet. The velocity at time t = a seconds is given by $\underset{(t \rightarrow a)}{lim} \frac{s(a) - s(t)}{a-t}$ . A construction worker drops a full paint can from a height of 500 feet. When will the paint can hit the ground? At what velocity will the paint can impact the ground?

1 answer:

skad [1K]2 years ago

7 0

Answer:

160m/s

Step-by-step explanation:

The object can hit the ground when t = a; meaning that s(a) = s(t) = 0

So, 0 = -16a² + 400

16a² = 400

a² = 25

a = √25

a = 5 (positive 5 only because that's the only physical solution)

The instantaneous velocity is

v(a) = lim(t->a) [s(t) - s(a)]/[t-a)

Where s(t) = -16t² + 400

and s(a) = -16a² + 400

v(a) = Lim(t->a) [-16t² + 400 + 16a² - 400]/(t-a)

v(a) = Lim(t->a) (-16t² + 16a²)/(t-a)

v(a) = lim (t->a) -16(t² - a²)(t-a)

v(a) = -16lim t->a (t²-a²)(t-a)

v(a) = -16lim t->a (t-a)(t+a)/(t-a)

v(a) = -16lim t->a (t+a)

But a = t

So, we have

v(a) = -16lim t->a 2a

v(a) = -32lim t->a (a)

v(a) = -32 * 5

v(a) = -160

Velocity = 160m/s

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If ABCD is a square and AB = 10, what is the measure of CE to the nearest hundredth?

andrey2020 [161]

Answer:

$AC=14.14\ units$

Step-by-step explanation:

<em><u>The correct question is </u></em>

If ABCD is a square and AB = 10, what is the measure of AC? (rounded to the nearest hundredth)

see the attached figure to better understand the problem

we know that

All four sides of the square are congruent and all four interior angles are equal to 90°.

so

AB=BC=CD=AD

In the right triangle ACD

Applying the Pythagorean Theorem

$AC^2=CD^2+AD^2$

we have that

$CD=AD=AB=10\ units$

substitute the given value

$AC^2=10^2+10^2$

$AC^2=200$

$AC=\sqrt{200}\ units$

$AC=14.14\ units$

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2 years ago

A math teacher tells her students that eating a healthy breakfast on a test day will help their brain function and perform well

kogti [31]

Answer:

Step-by-step explanation:

Hello!

To test the claim that eating a healthy breakfast improves the performance of students on their test a math teacher randomly asked 46 students what did they have for breakfast before they took the final exam and classified them as:

<u>Group 1</u>: Ate healthy breakfast

X₁: Number of students that ate a healthy breakfast before the exam and earned 80% or higher.

n₁= 26

<u>Group 2: </u>Did not eat healthy breakfast

X₂: Number of students that did not eat a healthy breakfast before the exam and earned 80% or higher.

n₂= 20

After the test she counted the number of students that got 80% or more in the test for each group obtaining the following sample proportions:

p'₁= 0.50

p'₂= 0.40

The parameters of study are the population proportions, if the claim is true then p₁ > p₂

And you can determine the hypotheses as

H₀: p₁ ≤ p₂

H₁: p₁ > p₂

α: 0.05

$Z= \frac{(p'_1-p'_2)-(p_1-p_2)}{\sqrt{p'(1-p')[\frac{1}{n_1} +\frac{1}{n_2}] } } }$ ≈N(0;1)

pooled sample proportion: $p'= \frac{x_1+x_2}{n_1+n_2} =\frac{13+8}{46} = 0.46$

$Z_{H_0}= \frac{(0.5-0.4)-0}{\sqrt{0.46(1-0.46)[\frac{1}{26} +\frac{1}{20}] } } }= 0.67$

p-value: 0.2514

The decision rule is:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

The p-value: 0.2514 is greater than the significance level 0.05, the test is not significant.

At a 5% significance level you can conclude that the population proportion of math students that obtained at least 80% in the test and had a healthy breakfast is equal or less than the population proportion of math students that obtained at least 80% in the test and didn't have a healthy breakfast.

So having a healthy breakfast doesn't seem to improve the grades of students.

I hope this helps!

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2 years ago

A website advertises job openings on its website, but job seekers have to pay to access the list of job openings. The website re

nataly862011 [7]

Answer:

The 90% confidence interval would be given by (57.006;62.994)

Step-by-step explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

$\bar X=60$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=10$ represent the population standard deviation

n=30 represent the sample size

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=10)$

The sample mean $\bar X$ is distributed on this way:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

The confidence interval on this case is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The next step would be find the value of $\z_{\alpha/2}$ , $\alpha=1-0.90=0.1$ , $\alpha/2 =0.05$ and $z_\alpha/2=1.64$

Using the normal standard table, excel or a calculator we see that:

$z_{\alpha/2}=1.64$

Since we have all the values we can replace:

$60 - 1.64\frac{10}{\sqrt{30}}=57.006$

$60 + 1.64\frac{10}{\sqrt{30}}=62.994$

So on this case the 90% confidence interval would be given by (57.006;62.994)

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2 years ago

Xavier is a salesperson who is paid a fixed amount of $455 per week. He also earns a commission of 3% on the sales he makes. If

melomori [17]

Xavier wants to earn more then 575 dollars a week, he will have to sale x<3,600 a week . So the answer is C

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2 years ago

Money Spent on Road Repair A politician wishes to compare the variances of the amount of money spent for road repair in two diff

Mkey [24]

Answer:

(A) Null Hypothesis, $H_0$ : $\sigma_1^{2} =\sigma_2^{2}$ {means that there is no significant difference in the variances of the amounts spent in the two counties}

Alternate Hypothesis, $H_A$ : $\sigma_1^{2} \neq \sigma_2^{2}$ {means that there is a significant difference in the variances of the amounts spent in the two counties}

(B) The F-table gives critical values of 0.359 and 2.781 for (14,17) degrees of freedom for the two-tailed test.

(C) The value of the F-test statistic is 0.611.

(D) We conclude that there is no significant difference in the variances of the amounts spent in the two counties.

Step-by-step explanation:

We are given that a politician wishes to compare the variances of the amount of money spent on road repair in two different counties.

The data are given here below;

County A County B

s1 = $11,596 s2 = $14,837

n1 = 15 n2 = 18

Let $\sigma_1^{2}$ = variances of the amounts spent in County A.

$\sigma_2^{2}$ = variances of the amounts spent in County B.

(A) So, Null Hypothesis, $H_0$ : $\sigma_1^{2} =\sigma_2^{2}$ {means that there is no significant difference in the variances of the amounts spent in the two counties}

Alternate Hypothesis, $H_A$ : $\sigma_1^{2} \neq \sigma_2^{2}$ {means that there is a significant difference in the variances of the amounts spent in the two counties}

The test statistics that would be used here <u>Two-sample F-test statistics </u>distribution;

T.S. = $\frac{s_1^{2} }{s_2^{2} } \times \frac{\sigma_2^{2} }{\sigma_1^{2} }$ ~ $F__n_1_-_1,_ n_2_-_1$

where, $s_1$ = sample standard deviation for County A = $11,596

$s_2$ = sample standard deviation for County B = $14,837

$n_1$ = sample size for County A = 15

$n_2$ = sample size for County B = 18

(B) Now at 0.05 level of significance, the F-table gives critical values of 0.359 and 2.781 for (14,17) degrees of freedom for the two-tailed test.

(C) So, <u><em>the test statistics</em></u> = $\frac{11,596^{2} }{14,837^{2} } \times 1$ ~ $F__1_4,_ 1_7$

= 0.611

The value of the F-test statistic is 0.611.

Now, as we can see that our test statistics lie within the range of critical values of F, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

(D) Therefore, we conclude that there is no significant difference in the variances of the amounts spent in the two counties.

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2 years ago