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Alenkasestr [34]
2 years ago
8

It is known that driving can be difficult in regions where winter conditions involve snow-covered roads. For cars equipped with

all-season tires traveling at 90 kilometers per hour, the mean stopping time in fresh snow is known to be 215 meters, with a standard deviation of σ = 2.5 meters. It is often advocated that automobiles in such areas should be equipped with special tires to compensate for such conditions, especially with respect to stopping distance. A manufacturer of tires made for driving in fresh snow claims that vehicles equipped with their tires have a decreased stopping distance. A study was done using a random sample of nine snow tires from the manufacturer on a snow-covered test track. The tests resulted in a mean stopping distance of = 212.9 meters. What are the appropriate null and alternative hypotheses to test the manufacturer's claim?
Mathematics
1 answer:
noname [10]2 years ago
5 0

Answer:

The null and alternative hypothesis for this test are

H_0: \mu\ge 215\\\\H_1: \mu< 215

Step-by-step explanation:

If we perform a hypothesis test, we can reject or not reject the null hypothesis.

To conclude that the tires have a decreased stopping distance (μ<215), we should state the null hypothesis H_0: \mu\ge 215 and then go on with the analysis to reject it (or not).

If the null hypothesis is rejected, the claim of the manufacturer is rigth.

The alternative hypothesis would be H_1: \mu, that would turn rigth if the null hypothesis is rejected.

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or the past 50 days, daily sales of laundry detergent in a large grocery store have been recorded (to the nearest 10). Units Sol
bagirrra123 [75]

Answer, step-by-step explanation:

A. With the previous exercise we can deduce that there is the situation of a number of sales in a grocery store, the relative frequency for the number of units sold, is shown below:

units sold. relative frequency. Acumulative frequency. interval of random numbers

30. 0.16. 0.16. 0.00 <0.16

40. 0.24. 0.4. 0.16 <0.4

50. 0.3. 0.7. 0.4 <0.7

60. 0.2. 0.9. 0.7<09

70. 0.1. 1. 0.9<1

B. For the next point, they give us some random numbers and then it is compared with the simulation of 10 days in sales:

random Units

number. sold

0.12. 30

0.96. 70

0.53. 50

0.80. 60

0.95. 70

0.10. 30

0.40. 50

0.45. 50

0.77. 60

0.29. 40

the two lists are compared so that opposite each one is the result of the simulation

3 0
1 year ago
Grace works as a salesperson at an electronics store and sells phones and phone accessories, and makes a fixed commission for ea
spayn [35]

Answer:

Commissions:

Accessoriy: $2

Phone: $9

Step-by-step explanation:

18p + 21a = 204

6p + 14a = 82

21a = 42

a = 2

p = [82 - 14(2)]/6 = 9

4 0
2 years ago
Read 2 more answers
Aubree wanted to see if there is a connection between the time a given exam takes place and the average score of this exam. She
Andrews [41]

Answer:

you can divide the answers by the time duration of the exam and get your answer

Step-by-step explanation:

5 0
1 year ago
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Write an equation in point-slope form for the line that has a slope of 9/8 and contains the point (9,−5). please help I am in de
Andreyy89

Answer: y+5 = 9/8(x-9)

Step-by-step explanation: Point-slope form is written in y-y1 = m(x-x1) so when you put in (9,-5) in the y1 and x1 spots and substitute the slope 9/8 in for m you get the point-slope form.

7 0
2 years ago
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compare the graphs of the logarithmic functions f(x)=log7x and g(x)=log4x. for what values of x is f=g, and f&gt;g, and f
VMariaS [17]

This problem has several items. So, let's solve it step by step.


1. Compare the graphs of the logarithmic functions f(x)=log7x and g(x)=log4x.


In the Figure below, we have tha graph of the two functions. The graph in red is f(x)=log(7x) and the graph in blue is g(x)=log(4x). The x-intercept of f is:


y=log(7x) \\ \\ 0=log(7x) \\ \\ 10^{0}=10^{log(7x)} \\ \\ 1=7x \\ \\ x=\frac{1}{7}=0.14


On the other hand, the x-intercept of g is:


y=log(4x) \\ \\ 0=log(4x) \\ \\ 10^{0}=10^{log(4x)} \\ \\ 1=4x \\ \\ x=\frac{1}{4}=0.25


Each graph begins in the fourth quadrant and is increasing quickly. As the graph crosses the x-axis at each x-intercept, each graph does not increase as fast. The graph continues to increase slowly throughout the first quadrant.


2. For what values of x is f=g


We can find this answer by taking this equation:


f(x)=g(x) \\ \\ log(7x)=log(4x) \\ \\ 10^{log(7x)}=10^{log(4x)} \\ \\ 7x=4x \\ \\ 7=4


As you can see this is an absurd result since 7 is not equal to 4. The conclusion is that the function f is always different from g, that is, f\neq g \ always!


3. For what values f>g


From the graph, we can see that the red function is always greater than the blue function. Therefore, f>g \ Always!


5 0
2 years ago
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