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2 years ago

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Sally drives from Turlock to Los Angeles at an average speed of 60 miles per hour. Turlock is approximately 300 miles from Lo

s Angeles. Which of the following graphs could give Sally’s approximate distance d, in miles from Los Angeles, in terms of time t,in hours?

1 answer:

docker41 [41]2 years ago

7 0

Answer:

5

Step-by-step explanation:

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The graph of y = f(x) is the solid black graph below. Which function represents the dotted graph?

nevsk [136]

Answer:

C (The lower left one)

Step-by-step explanation:

I saw that the vertex was equal to (-1,1) and that is the only answer that makes sense

6 0

2 years ago

Robert climbed 775775775 steps in 12\dfrac1212 2 1 12, start fraction, 1, divided by, 2, end fraction minutes. How many steps

Pepsi [2]

Answer:

Since the question requires you to give the answer in terms of steps/minute, the first step would be to convert hours to minutes:

12.5 hours = 750 minutes

Next, if he took 775 steps in 750 minutes, how many would he take in 1 minute?

This requires you to set up the question as a proportion:

(775 steps/ 750 minutes) = (x steps/1 minute)

(1 * 775)/750 = 31/30 = 1.033

he took ~1 step per minute

7 0

2 years ago

Read 2 more answers

F(x, y) = x2 + y2 + 4x − 4y, x2 + y2 ≤ 81 find the extreme values of f on the region described by the inequality.

Whitepunk [10]

Take partial derivatives and set them equal to 0:

$\nabla F(x,y)=\langle2x+4,2y-4\rangle=\mathbf 0$

We find one critical point within the boundary of the disk at $(x,y)=(-2,2)$ . The Hessian matrix for this function is

$\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2&0\\0&2\end{bmatrix}$

which is positive definite, and incidentally independent of $x$ and $y$ , so $F(x,y)$ attains a minimum $F(-2,2)=-8$ .

Meanwhile, we can parameterize the boundary by $\mathbf r(t)=\langle9\cos t,9\sin t\rangle$ with $0\le t\le2\pi$ , which gives

$F(x,y)=F(x(t),y(t))=f(t)=81\cos^2t+81\sin^2t+36\cos t-36\sin t=81+36(\cos t-\sin t)$

with critical points at

$f'(t)=0\implies 36(-\sin t-\cos t)=0\implies\sin t+\cos t=0\implies t=\dfrac{3\pi}4,\dfrac{7\pi}4$

At these points, we get

$f\left(\dfrac{3\pi}4\right)=81-36\sqrt2\approx30.0883$
$f\left(\dfrac{7\pi}4\right)=81+36\sqrt2\approx131.9117$

so we attain a maximum only when $t=\dfrac{7\pi}4$ , which translates to $(x,y)=\left(\dfrac9{\sqrt2},-\dfrac9{\sqrt2}\right)$ .

3 0

2 years ago

A sample with a sample proportion of 0.4 and which of the following sizes will produce the widest 95% confidence interval when e

Fed [463]

Answer:

C. 50

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

The higher the margin of error, the wider an interval is.

As the sample size increases, the margin of error decreases. If we want a widest possible interval, we should select the smallest possible confidence interval.

So the correct answer is:

C. 50

6 0

2 years ago

For quality control purposes, a company that manufactures sim chips for cell/smart phones routinely takes samples from its pro

ololo11 [35]

Answer:

There is a 22.42% probability that a sample in this size has 2 imperfections.

Step-by-step explanation:

For each chip, there are only two possible outcomes. Either they are imperfect, or they are not.

This means that we can solve this problem using binomial distribution probability concepts.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem

A success is a chip being imperfect. Suppose the average number of imperfections per 1000 sim chips is 3. So $\pi = \frac{3}{1000} = 0.003$ .

What is the probability that a sample this size (1000 chips) has 2 imperfections?

The sample has 1000 chips, wo $n = 1000$

We want P(X = 2).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{1000,2}.(0.003)^{2}.(0.997)^{998} = 0.2242$

There is a 22.42% probability that a sample in this size has 2 imperfections.

7 0

2 years ago