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3 years ago

Blake can jog 5,000 feet in 10 minutes. If he jogs at the same rate, how many feet can he jog in 16 minutes?

Mathematics

1 answer:

icang [17]3 years ago

6 0

Answer:

8,000 feet

Step-by-step explanation:

10/5,000 = 0.002

so 16 divided by what equals 0.002?

16/ 8,000=0.002

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Solve for x.<br> 5/3=x-4/2<br> Simplify your answer as much as possible.

lianna [129]

Answer:

Step-by-step explanation:

5/3=x-4/2

10=x-12

x=22

8 0

3 years ago

Read 2 more answers

The function c = 100 + 0.30m represents the cost c (in dollars) of renting a car after driving m miles. What would the cost be t

aleksley [76]

Answer:

c = $130

Step-by-step explanation:

Given:

c = 100 + 0.30m

Where,

c = cost of renting a car (in dollars)

m = number of miles driven (in miles)

If m = 100 miles

c = 100 + 0.30m

= 100 + 0.30(100)

= 100 + 30

= 130

c = $130

7 0

3 years ago

Cindy shoots a bottle rocket straight up into the air such that its height h, in feet, at time t, in seconds, is given by the fu

dezoksy [38]

Answer:

t = [0.493,4.2]

Step-by-step explanation:

The height of the rocket straight up into the air is given by :

$h(t)=-16t^2+85t+5$

We need to find the interval the rocket be at least 43 feet above the ground.

$43\ge -16t^2+85t+5\\\\t\ge 0.493\ and\ t\le 4.82\ s$

So, the required interval during which the rocket is at least 43 feet above the ground is [0.493,4.2].

6 0

3 years ago

Suppose that the population of the scores of all high school seniors who took the SAT Math (SAT-M) test this year follows a Norm

Vlad1618 [11]

Answer:

Confidence = $1-\alpha=1-0.01=0.99$

And then the confidence level would be given by 99%

Step-by-step explanation:

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".

Assuming the X follows a normal distribution

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

The distribution for the sample mean is given by:

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

$\bar x =512$ represent the sample mean

$\sigma=100$ represent the population standard deviation

n= 100 sample size selected.

The confidence interval is given by this formula:

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

The marginof error for this case is given by Me=25.76. And we know that the formula for the margin of error is given by:

$Me=z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

$25.76=z_{\alpha/2} \frac{100}{\sqrt{100}}$

And we can find the critical value $z_{\alpha/2}$ like this:

$z_{\alpha/2}=\frac{25.76(\sqrt{100})}{100}=2.576$

And we know that on the right tail of the z score =2.576 we have $\alpha/2$ of the total area. We can find the area on the right of the z score using this excel code:

"=1-NORM.DIST(2.576,0,1,TRUE)" or using a table of the normal standard distribution, and we got 0.004998= $\alpha/2$ , so then $\alpha=0.00498*2=0.009995 \approx 0.01$ , and then we can find the confidence like this: