What is the value of x?

What is 548.6814 rounded to the nearest thousandth

tankabanditka [31]

5 486 814 rounded to the nearest thousandth so here this is 6th digit:
because 7th digit is 4 we round down:
548.6814≈548,681

5 0

3 years ago

Read 2 more answers

In recent commercials Toyota claimed that 80% of its vehicles sold in the past 20 years are still on the road. You work for an a

Anvisha [2.4K]

Answer:

Step-by-step explanation:

We have to perform one sample proportion test. To do so we have to proceed through following steps.

(1) We have to collect data about vehicle sales in last 20 years.

(2) Using random number table or random number generator or otherwise we have to select sample of vehicles from the list of sale. In this step, we have to keep an eye on the fact that sample is to be selected from all years (all of 20). Using stratified random sampling by dividing vehicles sold over different years is an effective way to do so. That means, we may divide vehicles sold into 20 homogeneous strata over 20 years and select sample from each strata.

(3) After selecting sample of vehicles we have to communicate to corresponding vehicle owners to gather information about that particular vehicle.

(4) We have to note number of vehicles which are still on the road.

Suppose, we have selected n vehicles as sample and after communicating we found that \tiny r of those are still on the road.

We have to test for null hypothesis $H_0:p=0.80$

against the alternative hypothesis $H_1 \neq 0.80$

Our test statistic is given

$z=\frac{\hat p - p_0}{\sqrt{p_0(1-p_0)/n} }$

here,

$\hat p =\frac{r}{n}$

$p_0 =0.80$

$z=\frac{r/n-0.80\sqrt{n} }{0.40} \\\\=2.5(\frac{r}{n} -0.8)\sqrt{n}$

We then calculate corresponding p-value.

We reject our null hypothesis if $\text{p-value}$ , level of significance.

According to our obtained p-value and level of significance we proceed to certain conclusion.

8 0

3 years ago

The table below shows the velocity y, in miles per minute, of a toy car at different times x, in minutes: Time (x) (minutes) 10

MrMuchimi

A. If you plot the points in a graph, it would look like that shown in the picture attached. If we use linear regression, the correlation is very poor. The coefficient of correlation (r2) is only 0.0017. There is no linear relationship between time and velocity.

B. The slope of the graph is equal to y2-y1/x2-x1, In this case, it would specifically be v2-v1/t2-t1

Slope = 0.8-0.2/20-10 = 0.06 miles/s^2

The slope represents the acceleration at time 10 to 20 minutes.

C. The table in the graph shows causation rather than correlation. The points in the data occur in a sequential manner.

8 0

3 years ago

I don't understand this factorisation <br> a2+ 4a+3

Tomtit [17]

Answer:

$\boxed{\sf (a + 3)(a + 1)}$

Step-by-step explanation:

$\sf Factor \: the \: following: \\ \sf \implies {a}^{2} + 4a + 3 \\ \\ \sf The \: factors \: of \: 3 \: that \: sum \: to \: 4 \\ \sf are \: 3 \: and \: 1. \\ \\ \sf So, \\ \sf \implies {a}^{2} + (3 + 1)a + 3 \\ \\ \sf \implies {a}^{2} + 3a + a + 3 \\ \\ \sf \implies a(a + 3) + 1(a + 3) \\ \\ \sf \implies (a + 3)(a + 1)$

7 0

2 years ago

A simple random sample from a population with a normal distribution of 99 body temperatures has xbar = 99.10°F and s = 0.64°F. C

motikmotik

The general formula for the margin of error is e = (zs)/√n, when the sample size is 30 or more (otherwise, we'd need to to a t-interval).

Since we're needing a 99% confidence interval, we need to know what the positive z-score associated with this two-tailed area under the normal curve is. This can be a little tricky if you're using a standard normal table. What you want is a two-tailed interval that has an area of .99. What this means is that the remaining 0.01 is divided into 0.005 on each end. This then means that, to the right of the mean (which is 0), the area is 0.005 less than the total right-half area of 0.5, or 0.495. The total cumulative area, then, includes this plus the left half of .5, or 0.5 + 0.495 = .0995.

Then, if all we have to go on is a cumulative standard normal table, then we need to find the area closest to 0.995 and find the corresponding z-score. This z-score is 2.58.

Now that we have the z-score, we can now plug in all the values into the formula.

e = (2.58 · 0.66)/√102 = 0.1686.

So the 99% confidence interval will be that far above and below the mean (which is 98.8). So the lower limit is 98.8 - 0.1686 = 98.631, and the upper limit is 98.8 + 0.1686 = 98.967.

The TI-84 method:

This is much easier, if you're allowed to use this technology. You begin with the STAT button, and then move over to the TESTS menu. Then select ZInterval. We have summary stats rather than data in this problem, so we select Stats as the input. For the standard deviation, we put 0.66, for x-bar 98.8, for n 102, and for C-level 0.99. Then select Calculate, and it will provide the interval at the top of the screen (98.632, 98.968). Note that the lower limit is slightly different by this method. This is because, when doing it by hand, we had to approximate the z-score. This created a rounding error, albeit a small one.

4 0

3 years ago