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ANTONII [103]
3 years ago
8

Part two once again... what a waste of my points

Mathematics
2 answers:
liraira [26]3 years ago
7 0
Answer: 891 miles on each of the returning days
marysya [2.9K]3 years ago
4 0

Answer: 891

Step-by-step explanation:

Multiply the miles the bus traveled per day

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A county is considering rasing the speed limit on a road because they claim that the mean speed of vehicles is greater than 30 m
DochEvi [55]

Answer:

No, at alpha equals 0.10​, we do not have enough evidence to support the​ county's claim.

Step-by-step explanation:

We are given that a county is considering raising the speed limit on a road because they claim that the mean speed of vehicles is greater than 30 miles per hour.

A random sample of 15 vehicles has a mean speed of 31 miles per hour and a standard deviation of 4.7 miles per hour.

<em><u>Let </u></em>\mu<em><u> = true mean speed of the vehicles.</u></em>

SO, <u>Null Hypothesis</u>, H_0 : \mu \leq  30 miles per hour   {means that the mean speed of vehicles is lesser than or equal to 30 miles per hour}

<u>Alternate Hypothesis,</u> H_A : \mu > 30 miles per hour   {means that the mean speed of vehicles is greater than 30 miles per hour}

The test statistics that will be used here is <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

                        T.S.  = \frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean speed of 15 vehicles = 31 mph

             s = sample standard deviation = 4.7 mph

             n = sample of vehicles = 15

So, <em><u>test statistics</u></em>  =   \frac{31-30}{\frac{4.7}{\sqrt{15} } }  ~ t_1_4

                               =  0.824

<u><em>Hence, the value of test statistics is 0.824.</em></u>

<em />

<em>Now at 0.10 significance level, the t table gives critical value of 1.345 at 14 degree of freedom for right-tailed test. Since our test statistics is less than the critical value of t as 0.824 < 1.345, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.</em>

<em />

Therefore, we conclude that the mean speed of vehicles is lesser than or equal to 30 miles per hour which means that the county's claim is not supported.

4 0
3 years ago
Equivalent fractions for 15/35
Anit [1.1K]
\frac{15}{35} /5 \\  \frac{15/5}{35/5}  \\  \frac{3}{7}

That's the simplest form of 15/35. You can use 3/7 to get other equivalent fractions by multiplying the numerator and denominator by the same number. Other examples are 9/21, 6/14, and 12/28.
Hope this helps!

5 0
3 years ago
How does an outlier affect the median?
pychu [463]

Answer:

Formulas and Procedures: Outlier An extreme value in a set of data which is much higher or lower than the other numbers. ... Outliers affect the mean value of the data but have little effect on the median or mode of a given set of data.

8 0
3 years ago
50 POINTS
larisa86 [58]

Answer:

see below

Step-by-step explanation:

f(x) = 5x^3 +1, g(x) = – 2x^2, and h(x) = - 4x^2 – 2x +5

f(-8) = 5(-8)^3 +1 = 5 *(-512) +1 =-2560+1 =-2559

g( -6) = -2 ( -6) ^2 = -2 ( 36) = -72

h(9) = -4( 9)^2 -2(9) +5 = -4 ( 81) -18+5 = -324-18+5=-337

3 0
3 years ago
Read 2 more answers
Find the midpoint of the line segment joining the points P1 and P2.
ludmilkaskok [199]

Formula for midpoint between two points is M(x,y)

x=(x1+x2)/2 and y=(y1+y2)/2

In our case (x1,y1)=(m,b) and (x2,y2)=(0,0)

x=(m+0)/2=m/2 and y=(b+0)/2=b/2     M(m/2,b/2)

Good luck!!!

7 0
3 years ago
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