Ken spends $2.50 for 3 pens. At this rate,what will he spend for 5 times as many pens?

What is the slope of the line? a. -5/2 b. -2/5 c. 2/5 d. 5/2

Sladkaya [172]

Change in y: -2, goes down 2
Change in x: 5, goes right 5
slope: change in y/ change is x
which is -2/5, and that's B

5 0

3 years ago

The graph shows the relationship between the number of cars horns Sally heard and the amount of time that has passed since midni

MariettaO [177]

She heard 22 car horns in 11 hours.

3 0

3 years ago

Read 2 more answers

Data on pull-off force (pounds) for connectors used in an automobile engine application are as follows:

netineya [11]

Answer:

a. A point estimate of the mean pull-force of all connectors in the population is approximately 75.7385

The point estimate for the mean used is the sample mean

b. The point estimate of the pull force that separates the weakest 50% of the connectors from the strongest 50% is 74.3131 N

c. The point estimate of the population variance is approximately 2.8577

The point estimate for the population standard deviation is approximately 1.6905

d. The standard error of the mean is approximately 0.3315

e. The point estimate of a proportion of the connectors are;

(72.7, 0.0385) , (73.8, 0.0769) , (73.9, 0.0385) , (74, 0.0385) , (74.1, 0.0385) ,(74.2, 0.0385), (74.6, 0.0385) , (74.7, 0.0385) , (74.9, 0.0385) , (75.1, 0.0769) , (75.3, 0.0385) , (75.4, 0.0385) , (75.5, 0.0385) , (75.6, 0.0385) , (75.8, 0.0385) , (76.2, 0.0385) , (76.3, 0.0385) , (76.8, 0.0385) , (77.3, 0.0385) , (77.6, 0.0385) , (78, 0.0385) , (78.1, 0.0385) , (78.2, 0.0385) , (79.6, 0.0385)

Step-by-step explanation:

The given data for the pull-force (pounds) for connectors used in an automobile engine are presented as follows;

Pull-force (pounds); 79.6, 75.1, 78.2, 74.1, 73.9, 75.6, 77.6, 77.3, 73.8, 74.6, 75.5, 74.0, 74.7, 75.8, 72.7, 73.8, 74.2, 78.1, 75.4, 76.3, 75.3, 76.2, 74.9, 78.0, 75.1, 76.8

a. A point estimate of the mean pull-force of all connectors in the population is the sample mean given as follows;

$Mean, \ \overline x = \dfrac{\Sigma x_i}{n}$

$\Sigma x_i$ = The sum of the data = 1966.6

n = The sample size = 26

Therefore, the sample mean, $\overline x$ = 1966.6/26 ≈ 75.7385

The point estimate for the mean is approximately 75.7385

The sample mean was used as the point estimate for the mean because it is simple and representative of the sample

b. The weakest 50% of the connectors are;

72.7, 73.8, 73.8, 73.9, 74, 74.1, 74.2, 74.6, 74.7, 74.9, 75.1, 75.1, 75.3

The sum of forces that separates the weakest 50%, $\Sigma x_{i}_{50 \%}$ = 966.2

The point estimate of the pull force that separates the weakest 50% of the connectors from the strongest 50% = $\Sigma x_{i}_{50 \%}$ /13 = 966.2/13 = 74.3131 N

c. The estimate of the population variance is the sample variance, given as follows;

$s^2 =\dfrac{\sum \left (x_i-\overline x \right )^{2} }{n - 1}}$

Where;

${\sum \left (x_i-\overline x \right )^{2} }$ ≈ 71.4415

Therefore;

$s^2 =\dfrac{71.4415 }{25}} \approx 2.8577$

The point estimate of the population variance, s², is 2.8577

The point estimate for the population standard deviation, σ, is tha sample standard deviation, 's', given as follows;

s = √s² = √2.8577 ≈ 1.6905

The point estimate for the population standard deviation ≈ 1.6905

d. The standard error of the mean is given as follows;

$SE_{\mu_x} = \dfrac{s}{\sqrt{n} }$

Therefore, we have;

$SE_{\mu_x}$ = 1.6905/√(26) ≈ 0.3315

The standard error indicates the likely hood of the difference between the sample mean and the population mean

e. The point estimate of a proportion of the connectors are;

(Number of sample with a given pull-force value)/(Sample size (26))

Therefore, using Microsoft Excel, we have

(72.7, 0.0385) , (73.8, 0.0769) , (73.9, 0.0385) , (74, 0.0385) , (74.1, 0.0385) ,(74.2, 0.0385), (74.6, 0.0385) , (74.7, 0.0385) , (74.9, 0.0385) , (75.1, 0.0769) , (75.3, 0.0385) , (75.4, 0.0385) , (75.5, 0.0385) , (75.6, 0.0385) , (75.8, 0.0385) , (76.2, 0.0385) , (76.3, 0.0385) , (76.8, 0.0385) , (77.3, 0.0385) , (77.6, 0.0385) , (78, 0.0385) , (78.1, 0.0385) , (78.2, 0.0385) , (79.6, 0.0385)

8 0

3 years ago

A bag contains red marbles, white marbles, and blue marbles. Randomly choose two marbles, one at a time, and without replacement

dsp73

Answer:

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

$P(Same) = \frac{67}{210}$

Step-by-step explanation:

Given (Omitted from the question)

$Red = 7$

$White = 9$

$Blue = 5$

Solving (a): $P(First\ White\ and\ Second\ Blue)$

This is calculated using:

$P(First\ White\ and\ Second\ Blue) = P(White) * P(Blue)$

$P(First\ White\ and\ Second\ Blue) = \frac{n(White)}{Total} * \frac{n(Blue)}{Total - 1}$

We used Total - 1 because it is a probability without replacement

So, we have:

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{21 - 1}$

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{20}$

$P(First\ White\ and\ Second\ Blue) = \frac{9*5}{21*20}$

$P(First\ White\ and\ Second\ Blue) = \frac{45}{420}$

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

Solving (b) $P(Same)$

This is calculated as:

$P(Same) = P(First\ Blue\ and Second\ Blue)\$ $or\ P(First\ Red\ and Second\ Red)\ or\ P(First\ White\ and Second\ White)$

$P(Same) = (\frac{n(Blue)}{Total} * \frac{n(Blue)-1}{Total-1})+(\frac{n(Red)}{Total} * \frac{n(Red)-1}{Total-1})+(\frac{n(White)}{Total} * \frac{n(White)-1}{Total-1})$