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Natali [406]
4 years ago
13

Lisa tied a piece of rope to a tree branch. She is standing 5 feet away from the base of the tree, and the branch is 12 feet off

the ground. The length of the rope from the tree branch to Lisa’s feet is r feet long.
Mathematics
1 answer:
Ksenya-84 [330]4 years ago
8 0
When you draw this out, what you have is a right triangle problem.  The height of the rope from the ground is 12 (a side), the distance she is from the base of the tree is 5 (a side), and the length of the rope is the hypotenuse.  Use Pythagorean's Theorem to find the length of the hypotenuse: 5^2+12^2 = c^2 or
25+144 = c^2 and c^2 = 169, which makes the length of the rope come out to be 13 feet long.
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Compare the fractions. Use &lt;, &gt;, or =.<br> 6<br> 1<br> 7<br> 7
AURORKA [14]

Answer:

-6/7 < -1/7

because it is more than negative one by seven

8 0
3 years ago
Which number can each term of the equation be multiplied by to eliminate the fractions before solving?
Lynna [10]

Answer:

would multiply everything by the common denominator of 4

-3/4m - 1/2 = 2 + 1/4m....multiply everything by 4 and u get :

-3m - 2 = 8 + m

that will get rid of the fractions....but u could multiply by -4 ...giving u :

3m + 2 = -8 - m

Step-by-step explanation:

7 0
3 years ago
Brenda’s bank offers car financing for 3, 4 or 5 years. If brenda chooses 5-year financing, how many monthly payments will she h
klemol [59]
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7 0
3 years ago
HELP ME WITH THIS QUESTION ASAP
Stels [109]
Factor out 12x^3y^2

12x^3y^2(3y^2+5x^ 2 y z^2 -1)
8 0
3 years ago
A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the
Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

             P.Q. = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = sample average age = 25 years

            \sigma = population standard deviation = 1.8 years

            n = sample of students = 36

            \mu = population average age

So, 98% confidence interval for the average age, \mu is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < 2.3263) = 0.98

P( -2.3263 \times {\frac{\sigma}{\sqrt{n} } < {\bar X - \mu} < 2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

P( \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } < \mu < \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

98% confidence interval for \mu = [ \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } , \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ]

                                                  = [ 25 - 2.3263 \times {\frac{1.8}{\sqrt{36} } , 25 + 2.3263 \times {\frac{1.8}{\sqrt{36} } ]

                                                  = [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0
3 years ago
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