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Korvikt [17]
3 years ago
11

Men absolute deviation of 14, 8, 7,6,5,10,11,8,8,6

Mathematics
1 answer:
maks197457 [2]3 years ago
4 0
I don't know much about this but as far as I can tell the mean is about 8, that would make the deviation value of about 2
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5.4 = a - (-1.8) solve for a.
Agata [3.3K]
Let's solve your equation step-by-step.<span>5.4=<span>a−<span>(<span>−1.8</span>)</span></span></span>Step 1: Simplify both sides of the equation.<span>5.4=<span>a+1.8</span></span>Step 2: Flip the equation.<span><span>a+1.8</span>=5.4</span>Step 3: Subtract 1.8 from both sides.<span><span><span>a+1.8</span>−1.8</span>=<span>5.4−1.8</span></span><span>a=3.6</span>
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Match each system of equations to its solution represented by an augmented matrix.
SCORPION-xisa [38]

Answer:

Step-by-step explanation:

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The owner of a bike shop sells unicycle and bicycles and keeps inventory by counting seats and wheels. One day she counts 20 sea
Mashcka [7]
A unicycle has 1 wheel. u unicycles have u wheels.

A bicycle has 2 wheels. b bicycles have 2b wheels.

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in the ambrose family the ages of the three children are three consecutive even integers . if the age of the youngest child is r
klemol [59]

The ages are consecutive even numbers.
The youngest child is  (x+3) years old.
Every even number is 2 more than the one before it.
So the three ages are

           (x + 3),  (x + 5),  and  (x + 7).

Their sum is    x+3  +  x+5  +  x+7  =  3x + 15

But we're told that their sum is  42, so we can write

                                              3x + 15 = 42  <== the equation

Subtract 15 from each side:  3x         =  27

Divide each side by  3 :          x         =   9    <== the solution

So the three ages are  12,  14,  and  16.

The oldest child is  16  years old.

3 0
3 years ago
United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the n
Ivanshal [37]

Answer:

<u><em>a) The probability that exactly 4 flights are on time is equal to 0.0313</em></u>

<u><em></em></u>

<u><em>b) The probability that at most 3 flights are on time is equal to 0.0293</em></u>

<u><em></em></u>

<u><em>c) The probability that at least 8 flights are on time is equal to 0.00586</em></u>

Step-by-step explanation:

The question posted is incomplete. This is the complete question:

<em>United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the number on-time flights is recorded. Round answers to 3 significant figures. </em>

<em>a) The probability that exactly 4 flights are on time is = </em>

<em>b) The probability that at most 3 flights are on time is = </em>

<em>c)The probability that at least 8 flights are on time is =</em>

<h2>Solution to the problem</h2>

<u><em>a) Probability that exactly 4 flights are on time</em></u>

Since there are two possible outcomes, being on time or not being on time, whose probabilities do not change, this is a binomial experiment.

The probability of success (being on time) is p = 0.5.

The probability of fail (note being on time) is q = 1 -p = 1 - 0.5 = 0.5.

You need to find the probability of exactly 4 success on 9 trials: X = 4, n = 9.

The general equation to find the probability of x success in n trials is:

           P(X=x)=_nC_x\cdot p^x\cdot (1-p)^{(n-x)}

Where _nC_x is the number of different combinations of x success in n trials.

            _nC_x=\frac{x!}{n!(n-x)!}

Hence,

            P(X=4)=_9C_4\cdot (0.5)^4\cdot (0.5)^{5}

                                _9C_4=\frac{4!}{9!(9-4)!}=126

            P(X=4)=126\cdot (0.5)^4\cdot (0.5)^{5}=0.03125

<em><u>b) Probability that at most 3 flights are on time</u></em>

The probability that at most 3 flights are on time is equal to the probabiity that exactly 0 or exactly 1 or exactly 2 or exactly 3 are on time:

         P(X\leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)

P(X=0)=(0.5)^9=0.00195313 . . . (the probability that all are not on time)

P(X=1)=_9C_1(0.5)^1(0.5)^8=9(0.5)^1(0.5)^8=0.00390625

P(X=2)=_9C_2(0.5)^2(0.5)^7=36(0.5)^2(0.5)^7=0.0078125

P(X=3)= _9C_3(0.5)^3(0.5)^6=84(0.5)^3(0.5)^6=0.015625

P(X\leq 3)=0.00195313+0.00390625+0.0078125+0.015625=0.02929688\\\\  P(X\leq 3) \approx 0.0293

<em><u>c) Probability that at least 8 flights are on time </u></em>

That at least 8 flights are on time is the same that at most 1 is not on time.

That is, 1 or 0 flights are not on time.

Then, it is easier to change the successful event to not being on time, so I will change the name of the variable to Y.

          P(Y=0)=_0C_9(0.5)^0(0.5)^9=0.00195313\\ \\ P(Y=1)=_1C_9(0.5)^1(0.5)^8=0.0039065\\ \\ P(Y=0)+P(Y=1)=0.00585938\approx 0.00586

6 0
3 years ago
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