Select all of the following points that lie on the graph of f(x) = 7

Sanjit drives the 45 km distance between York and Leeds he then drives a further 42 km from Leeds to Blackpool

Leona [12618]

The average speed for his entire journey from York to Blackpool is about 61.41 km/h.

Here’s a breakdown of how we arrive at this:

$\text{ The average speed}=\frac{\text{Total distance covered}}{\text{Total time taken}}$

The distance he travelled from York to Leeds is 45 km,

and the speed during that section was 54 km/h.

Therefore, the time taken to travel from York to Leeds is 45/54 hours (since Time = Distance/Speed).

Next, the distance from Leeds to Blackpool is 42 km,

and the time for that leg of the journey is 35 minutes, which is 35/60 hours.

This leads to the total duration for his trip as

$=\frac{45}{54}+\frac{35}{60}=\frac{450+315}{540}=\frac{765}{540}=\frac{17}{12}$ hours.

The cumulative distance covered equals 45 + 42 = 87 km.

Thus, his average speed is calculated as:

$=\frac{87}{17/12}= \frac{12\times 87}{17}=\frac{1044}{17}=61.4117647\approx 61.41\text{ km per hour}$

3 0

2 months ago

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Which expression is equivalent to StartFraction 3 x Superscript negative 6 Baseline y Superscript negative 3 Baseline Over 15 x

zzz [12365]

Answer:

The appropriate expression is 1 Over 5 x Superscript minus 8 Baseline y Superscript minus 13 Baseline EndFraction

Step-by-step explanation:

In order to simplify the expression $\frac{3x^{-6}y^{-3} }{15x^{2}y^{10} }$ where x ≠ 0, y ≠ 0, we must work through the given expression.

$\frac{3x^{-6}y^{-3} }{15x^{2}y^{10} }\\= \frac{3}{15} x^{-6-2}y^{-3-10}\\ = \frac{3}{15}x^{-8}y^{-13} \\ =\frac{1}{5}x^{-8}y^{-13} \\$

The appropriate expression is 1 Over 5 x Superscript minus 8 Baseline y Superscript minus 13 Baseline EndFraction

6 0

2 months ago

The manufacturer of hardness testing equipment uses steel-ball indenters to penetrate metal that is being tested, however, the m

Inessa [12570]

Answer:

$1.667-2.31\frac{1.803}{\sqrt{9}}=0.2789$

$1.667+2.31\frac{1.803}{\sqrt{9}}=3.055$

Based on the analysis, the 95% confidence interval is specified as (0.2789;3.055)

The question regarding the 95% confidence interval's ability to ascertain potential differences in measurements between the two indenters is as follows:

Indeed, the confidence interval does not include the value 0, thus indicating that the Diamond values significantly exceed those of the Steel Ball at a 5% significance level.

Step-by-step explanation:

Here is the dataset in consideration:

specimen 1 2 3 4 5 6 7 8 9

Steel Ball 51 57 61 70 68 54 65 51 53

Diamond 53 55 63 74 69 56 68 51 56

By calculating the differences between diamond and steel ball measurements, we create the dataset:

d: 2, -2, 2, 4, 1, 2, 3, 0, 3

In the next step, we compute the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}=1.667$

Following that, we determine the standard deviation of the differences, arriving at:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =1.803$

A confidence interval refers to "a range of values that’s likely to encompass a population value with a certain level of confidence. It is typically expressed as a percentage indicating where a population mean falls within an upper and lower limit."

The margin of error represents the extent to which values diverge above and below the sample statistic in a confidence interval.

Normal distribution, is described as a "probability distribution that is symmetric about the mean, illustrating that data points close to the mean occur more frequently than those further away."

The confidence interval for the mean is derived using the following formula:

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

To calculate the critical value $t_{\alpha/2}$ , we first determine the degrees of freedom using:

$df=n-1=9-1=8$

Given a confidence level of 0.95 or 95%, the appropriate critical value can be found using Excel, a calculator, or a table. The Excel command would be: "=-T.INV(0.025,9)". Therefore, we observe that $t_{\alpha/2}=2.31$ .

Finally, we can substitute all our findings into formula (1):

$1.667-2.31\frac{1.803}{\sqrt{9}}=0.2789$

$1.667+2.31\frac{1.803}{\sqrt{9}}=3.055$

In this case, the 95% confidence interval is calculated as (0.2789;3.055)

In determining if the two indenters yield distinct measurements, we find that the confidence interval does not enclose zero, allowing us to conclude that Diamond readings greatly surpass Steel Ball readings at the 5% significance level.

4 0

1 month ago

Svet_ta [12734]

T<span>he area of a figure signifies the measure of space within a two-dimensional shape, typically expressed as square units based on the figure's dimensions.

For instance, with a shape having dimensions of k, its area can be given by $k^2$ .
</span>
<span>Consider that one similar figure possesses an area nine times that of another.

As these figures are similar, their areas correspond to the proportionality of their dimensions.

Let the smaller shape's dimensions be k, while the larger is p times the dimensions of the smaller shape. The smaller shape's area is $k^2$ and the larger shape's area is $(pk)^2$ .

Now, knowing the larger figure's area is nine times the area of the smaller figure, we have:
$\frac{(pk)^2}{k^2} = \frac{9}{1} \\ \\ \frac{p^2k^2}{k^2} =9 \\ \\ p^2=9 \\ \\ p= \sqrt{9} \\ \\ p=3$
</span>
Thus, the dimensions of the larger figure must be 3 times those of the smaller figure.

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1 month ago

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Select all of the following points that lie on the graph of f(x) = 7 - 3x.

The appropriate expression is 1 Over 5 x Superscript minus 8 Baseline y Superscript minus 13 Baseline EndFraction