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Sholpan [36]
3 years ago
14

If r is the midpoint of AB and AR=14, then AB=______

Mathematics
1 answer:
skelet666 [1.2K]3 years ago
4 0

Answer:

28

Step-by-step explanation:

The midpoint is halfway, so if half the length is 14 then the full length is 28 (14 × 2 = 28)

Hope this helps!

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The gym is 3.2 inches wide brainlist please!!!!!!!!!!!!!!!!!!
6 0
4 years ago
7100 dollars is placed in an account with an annual interest rate of 5%. To the nearest tenth of a year, how long will it take f
kow [346]

Answer:

After 26.91\ years the principle amount value will reach to 26400\ dollars.

Step-by-step explanation:

Given that,

                 Principle Amount (P)= 7100\ dollars

                 Rate of Interest (R)= 5%

                 Total Amount (A)= 26400\ dollars

If Principle Amount (P) =P\ dollars, Time =t\ years, Rate of Interest =R% pa

when interest is compounded annually:

Total Amount after t\ years = P(1+\frac{R}{100}) ^{t}

Now,

         A= P(1+\frac{R}{100}) ^{t}

     ⇒ 26400=7100(1+\frac{5}{100}) ^{t}

     ⇒ 3.7183=(1+0.05)^{t}

     ⇒ 3.7183=(1.05)^{t}

taking log both sides, we get

         ln(3.7183)=ln {(1.05)^{t}}

     ⇒ 1.31327=t\times ln(1.05)

     ⇒ 1.31327=t\times 0.04879

    ⇒ t=26.91

Therefore,

After 26.91\ years the principle amount value will reach to 26400\ dollars.

6 0
4 years ago
Read 2 more answers
The weights of steers in a herd are distributed normally. The standard deviation is 300lbs and the mean steer weight is 1100lbs.
gizmo_the_mogwai [7]

Answer:

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

<em> P(920≤ x≤1730) = 0.7078 </em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given mean of the Population = 1100 lbs

Standard deviation of the Population = 300 lbs

Let 'X' be the random variable in Normal distribution

Let x₁ = 920

Z = \frac{x-mean}{S.D} = \frac{920-1100}{300} = - 0.6

Let x₂ = 1730

Z = \frac{x-mean}{S.D} = \frac{1730-1100}{300} = 2.1

<u><em>Step(ii)</em></u>

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

P(x₁≤ x≤x₂) = P(Z₁≤ Z≤ Z₂)

                  = P(-0.6 ≤Z≤2.1)

                  = P(Z≤2.1) - P(Z≤-0.6)

                 = 0.5 + A(2.1) - (0.5 - A(-0.6)

                 =  A(2.1) +A(0.6)               (∵A(-0.6) = A(0.6)

                 =  0.4821 + 0.2257

                 = 0.7078

<u><em>Conclusion:-</em></u>

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

           <em> P(920≤ x≤1730) = 0.7078 </em>

5 0
3 years ago
What is the correct solution set for the following graph?
nalin [4]
Point in quadrqnt 111
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3 years ago
5. An angle bisector of a triangle divides the opposite side of the triangle into segments 6 cm and 5 cm long. A second side of
AlladinOne [14]
You have to set up 2 proportions.

If the 6.9 goes with the 5 then the proportion will look like this.
6.9/5 = x/6 Cross multiply
6 * 6.9 = 5x
41.4 / 5 = x
x = 8.28 or 8.3

If on the other hand the 6.9 goes with the 6 then the second proportion is
6.9/6 = x/5
6.9 * 5 = 6X
34.5 / 6 = x
x = 5.75

Answer: D <<<<<<<<======

6 0
4 years ago
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