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stira [4]
3 years ago
6

15 Points and Brainliest :)

Mathematics
1 answer:
kodGreya [7K]3 years ago
6 0

Answer:

Step-by-step explanation:

Hello, please consider the following.

Option A. First week we got $200.

Week 2, we got $200+$50=$250

Week 3, we got $250+$50=$300

Week 4, we got $300+$50=$350

Week 5, we got $350+$50=$400

Week 6, we got $400+$50=$450

\begin{array}{c |c |c |c |c |c |c|}Week & 1 & 2 & 3 & 4 & 5 & 6\\----&---&---&---&---&---&---\\Amount &200 &250 &300 &350 & 400 &450\end{array}

Option B. First week we got $200.

Week 2, we got $200+$200*10%=$200+$20=$220

Week 3, we got $220(1+10%)=$220(1.10)=$242

Week 4, we got $242(1.10)=$266.2

Week 5, we got $266.2(1.10)=$292.82

Week 6, we got $292.82(1.10)=$322.102

\begin{array}{c |c |c |c |c |c |c|}Week & 1 & 2 & 3 & 4 & 5 & 6\\----&---&---&---&---&---&---\\Amount &200 &220 &242 &266.2 & 292.82 &322.102\end{array}

Thank you.

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You have 3/4 of a pie leftover after your party. Each guest takes home 1/6 of the leftover
Anit [1.1K]

Answer:

3/24 of the original pie.

Step-by-step explanation:

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1/6*3/4=3/24

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8 0
3 years ago
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>

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