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Dovator [93]
3 years ago
6

There are six poles on a side of a 1 km 200 m long straight road such that there is a pole at the starting and end points of the

road. If the poles are equally spaced, then what is the distance between each consecutive pole?
Mathematics
1 answer:
Bingel [31]3 years ago
5 0

Answer:

Distance is 200m between each pole

Step-by-step explanation:

First, convert the length of the road into meters

        1km= 1000m

        1000m +200m= 1200m

There are 6 poles on the side and they're equally spaced

Divide the length of the road by the number of poles to get the distance between the poles

        Distance between poles= Length of road/ Number of poles

        Distance between poles= 1200m/ 6 poles

       Distance between poles= 200m

       

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10. A manufacturer wanted to know if more coupons would be redeemed if they were mailed to the
Mandarinka [93]

Using the t-distribution, it is found that the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if there is no difference, that is, the mean is of 0, hence:

H_0: \mu = 0

At the alternative hypothesis, it is tested if the mean number is greater for females, that is, the mean is greater than 0, hence:

H_1: \mu > 0

<h3>What is the test statistic?</h3>

The test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

The parameters are:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • s is the standard deviation of the sample.
  • n is the sample size.

In this problem, the parameters are given as follows:

\overline{x} = 1.5, \mu = 0, s = 4.75, n = 50.

Hence, the test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{1.5 - 0}{\frac{4.75}{\sqrt{50}}}

t = 2.23

<h3>What is the conclusion?</h3>

Considering a <em>right-tailed test</em>, as we are testing if the mean is greater than a value, with a <em>significance level of 0.01 and 50 - 1 = 49 df</em>, the critical value is given by t^{\ast} = 2.4.

Since the test statistic is less than the critical value for the right-tailed test, the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

More can be learned about the t-distribution at brainly.com/question/26454209

5 0
2 years ago
X²-y², x²+xy and x² + 2xy + y²<br> plz fast full formula givee​
Scorpion4ik [409]

Answer:

Step-by-step explanation:

x²- y² = (x+y) (x-y)

x² + xy = x(x+y)

x² + 2xy + y² = (x+y) (x+y)

3 0
3 years ago
I leave my family's vacation cabin at 8 a.m. and start driving home at a nice, safe 45 mph. Two hours later, my husband, who alw
Murljashka [212]

Answer:

2:30 pm

Step-by-step explanation:

In two hours (from 10am), she is 45 * 2 = 90 miles from vacation cabin.

From then on her average is still 45 mph and her husband's is 65 mph. So, husband catches her 20 miles (65-45=20) in every hour.

<em>The husband has to catch up 90 miles. How long would it take if he is catching up at 20 miles an hour?</em>

<em>90/20 = 4.5 hours</em>

<em />

<em>So husband starts at 10am and 4.5 hours form that time is 2.30pm</em>

6 0
3 years ago
Suppose each runner ran at the rate given in the table above for 3.1 miles. How much time will elapse between the first place fi
Scilla [17]
You didn't include the table but I found this table for the same statement, so I will answer you based on the next table:

Runner         distance           time

Arabella        7,299 feet        561 seconds
Bettina          3,425 yards     13 minutes, 12 seconds
Chandra       8,214 feet        0,195 hours
Divya            1,62 miles        732 seconds

To answer the question you must find the rate for each runner and then calculate the time to run 3.1 miles at each rate.

First you need to convert the data to obtain the rate in miles per second.

These are the main conversion identities:

1 mile = 5280 feet

1 mile = 1760 yards

1 hour = 3600 seconds

1 hour = 60 minutes

1 minute = 60 seconds

Arabella:

rate: 7,229 feet / 561 seconds * (1 mile / 5280 feet)  =

= 0.00244 mile/second

Time to run 3.1 miles: V = d / t => t = d / V = 3.1 miles / 0.00244 mile/second = 1270 seconds

Bettina:

13 minutes + 12 seconds = 13*60 seconds +12 seconds = 792 seconds

rate = 3425 yards / 792 seconds * 1 mile / 1760 yards = 0.00246 mile/seconds

Time to run 3.1 miles = 3.1 miles / 0.00246 mile/second = 1260 seconds

Chandra:

rate = 8214 feet / 0.195 hours * 1 mile / 5280 feet * 1hour / 3600 seconds =

= 0.00222 seconds

Time = 3.1 mile / 0.00222 seconds = 0.389 hour = 1396 seconds

Divya:

rate = 1.62 miles / 732 seconds = 0.00221 seconds

Time = 3.1 mile / 0.00221 seconds = 1403 seconds

Now you can find the difference between fhe last and the first 1403 seconds - 1260 seconds = 143 seconds

That is equivalent to 2.38 seconds.
6 0
3 years ago
What is the area of this triangle in square inches?
adelina 88 [10]

The area of a triangle is found by multiplying the base by the height, then dividing that by 2:

Area = 10 * 7 = 70

70 / 2 = 35 square yards.

3 0
3 years ago
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