You drop a rock off a bridge. The rock’s height, h (in feet above the water), after t seconds is modeled by h = -16t^2 + 541. What is the height of the rock after 2 seconds?

Answers

Answer 1

Answer: For this case we have the following equation:
$h = -16t ^ 2 + 541$
To answer the question, what we must do is evaluate the function for t = 2.
We have then:
$h = -16 (2) ^ 2 + 541$
Rewriting we have:
$h = -16 (4) + 541h = -64 + 541h = 477 feet$
Answer:
The height of the rock after 2 seconds is:
$h = 477 feet$

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discrete random variable X has the following probability distribution: x 13 18 20 24 27 P ( x ) 0.22 0.25 0.20 0.17 0.16 Compute each of the following quantities. P ( 18 ) . P(X > 18). P(X ≤ 18). The mean μ of X. The variance σ 2 of X. The standard deviation σ of X.

Answers

Answer:

(a) P(X = 18) = 0.25

(b) P(X > 18) = 0.53

(c) P(X ≤ 18) = 0.47

(d) Mean = 19.76

(e) Variance = 22.2824

(f) Standard deviation = 4.7204

Step-by-step explanation:

We are given that discrete random variable X has the following probability distribution:

X P (x) X * P(x) $X^(2)$ $X^(2)$ * P(x)

13 0.22 2.86 169 37.18

18 0.25 4.5 324 81

20 0.20 4 400 80

24 0.17 4.08 576 97.92

27 0.16 4.32 729 116.64

(a) P ( X = 18) = P(x) corresponding to X = 18 i.e. 0.25

Therefore, P(X = 18) = 0.25

(b) P(X > 18) = 1 - P(X = 18) - P(X = 13) = 1 - 0.25 - 0.22 = 0.53

(d) Mean of X, $\mu$ = ∑X * P(x) ÷ ∑P(x) = (2.86 + 4.5 + 4 + 4.08 + 4.32) ÷ 1

= 19.76

(e) Variance of X, $\sigma^(2)$ = ∑ $X^(2)$ * P(x) - $(\sum X * P(x))^(2)$

= 412.74 - $19.76^(2)$ = 22.2824

(f) Standard deviation of X, $\sigma$ = $√(variance)$ = $√(22.2824)$ = 4.7204 .

Final answer:

The probabilities for the given X values are calculated by summing the relevant given probabilities. The mean of X is computed as a weighted average, and the variance and standard deviation are calculated using formula involving the mean and the individual probabilities.

Explanation:

The probability P(18) is given as 0.25 according to the distribution. The probability P(X > 18) is the sum of the probabilities for all x > 18, so we add the probabilities for x=20, x=24, and x=27, giving us 0.20 + 0.17 + 0.16 = 0.53. The probability P(X ≤ 18) includes x=18 and any values less than 18. As 18 is the lowest value given, P(X ≤ 18) is just P(18), or 0.25.

The mean μ of X is the expected value of X, computed as Σ(xP(x)). That gives us (13*0.22) + (18*0.25) + (20*0.20) + (24*0.17) + (27*0.16) = 2.86 + 4.5 + 4 + 4.08 + 4.32 = 19.76.

The variance σ 2 of X is computed as Σ [ (x - μ)^2 * P(x) ]. That gives us [(13-19.76)^2 * 0.22] + [(18-19.76)^2 * 0.25] + [(20-19.76)^2 * 0.20] + [(24-19.76)^2 * 0.17] + [(27-19.76)^2 * 0.16] = 21.61. The standard deviation σ of X is the sqrt(σ^2) = sqrt(21.61) = 4.65.

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Salmon Weights: Assume that the weights of spawning Chinook salmon in the Columbia river are normally distributed. You randomly catch and weigh 17 such salmon. The mean weight from your sample is 19.2pounds with a standard deviation of 4.4 pounds. You want to construct a 90% confidence interval for the mean weight of all spawning Chinook salmon in the Columbia River.(a) What is the point estimate for the mean weight of all spawning Chinook salmon in the Columbia River?
pounds

(b) Construct the 90% confidence interval for the mean weight of all spawning Chinook salmon in the Columbia River. Round your answers to 1 decimal place.
< ? <

(c) Are you 90% confident that the mean weight of all spawning Chinook salmon in the Columbia River is greater than 18 pounds and why?

No, because 18 is above the lower limit of the confidence interval.

Yes, because 18 is below the lower limit of the confidence interval.

No, because 18 is below the lower limit of the confidence interval.

Yes, because 18 is above the lower limit of the confidence interval.

(d) Recognizing the sample size is less than 30, why could we use the above method to find the confidence interval?

Because the sample size is greater than 10.

Because we do not know the distribution of the parent population.

Because the parent population is assumed to be normally distributed.

Because the sample size is less than 100.

Answers

Answer:

a) $\bar X= 19.2$ represent the sample mean. And that represent the best estimator for the population mean since $\hat \mu =\bar X=19.2$ b) The 90% confidence interval is given by (17.3;21.1)

c) No, because 18 is above the lower limit of the confidence interval.

d) Because the parent population is assumed to be normally distributed.

The reason of this is because the t distribution is an special case of the normal distribution when the degrees of freedom increase.

Step-by-step explanation:

1) Notation and definitions

n=17 represent the sample size

Part a

$\bar X= 19.2$ represent the sample mean. And that represent the best estimator for the population mean since $\hat \mu =\bar X=19.2$ Part b

$s=4.4$ represent the sample standard deviation

m represent the margin of error

Confidence =90% or 0.90

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".

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".

2) Calculate the critical value tc

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$ . The degrees of freedom are given by:

$df=n-1=17-1=16$

We can find the critical values in excel using the following formulas:

"=T.INV(0.05,16)" for $t_(\alpha/2)=-1.75$

"=T.INV(1-0.05,16)" for $t_(1-\alpha/2)=1.75$

The critical value $tc=\pm 1.75$

3) Calculate the margin of error (m)

The margin of error for the sample mean is given by this formula:

$m=t_c (s)/(√(n))$

$m=1.75 (4.4)/(√(17))=1.868$

4) Calculate the confidence interval

The interval for the mean is given by this formula:

$\bar X \pm t_(c) (s)/(√(n))$

And calculating the limits we got:

$19.2 - 1.75 (4.4)/(√(17))=17.332$

$19.2 + 1.75 (4.4)/(√(17))=21.068$

The 90% confidence interval is given by (17.332;21.068) and rounded would be: (17.3;21.1)

Part c

No, because 18 is above the lower limit of the confidence interval.

Part d

Because the parent population is assumed to be normally distributed.

The reason of this is because the t distribution is an special case of the normal distribution when the degrees of freedom increase.

5 (8 4) = (5 × 8) + (5 × 4)

Answers

Answer: it think its 60

Step-by-step explanation:

Answer:...

Step-by-step explanation:

The number of minutes that Samantha waits to catch the bus is uniformly distributed between 0 and 15 minutes. What is the probability that Samantha has to wait less than 4.5 minutes to catch the bus? a. 10%
b. 3%
c. 20%
d. 30%

Answers

Answer:

option d. 30%

Step-by-step explanation:

Data provided in the question:

$a=0$

$b=15$

Now,

The probability density function of the uniform distribution is given as:

$f(x)=(1)/(15-0)$

$f(x)=(1)/(15), 0<x<15$

Therefore,

The required probability will be

$P(X<4.5)=\int_(0)^(4.5) f(x) d x$

$P(X<4.5)=\int_(0)^(4.5) (1)/(15) d x$

$P(X<4.5)=\left((x)/(15)\right)_(0)^(4.5)$

$P(X<4.5)=\left((4.5)/(15)-(0)/(15)\right)$

$P(X<4.5)=0.3$

= 0.3 × 100%

= 30%

Hence,

The answer is option d. 30%

Final answer:

In a uniform distribution, all values have an equal chance of occurring. To find the probability of Samantha waiting less than 4.5 minutes, you divide 4.5 by the total time interval of 15 minutes. The answer is 30%.

Explanation:

The question is about the concept of uniform distribution in probability, which means that each value within the given range has an equal chance of being drawn. In this case, Samantha can wait anywhere between 0 and 15 minutes, so each minute has an equally likely chance of being the waiting time. To find the probability that Samantha has to wait less than 4.5 minutes, we simply divide this time interval by the total time interval. Hence, the calculation is 4.5 / 15 = 0.30 or 30%. So, the answer to the question of the probability that Samantha has to wait less than 4.5 minutes to catch the bus is 30%.

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a new recipe calls for 1/3 cups sugar and you have 2 1/3 cups in the pantry if you have unlimited supplies of the other ingredients how many recipes could you make? PLS HELP