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Degger [83]
3 years ago
9

Evaluate 6x+24 for x=8

Mathematics
1 answer:
crimeas [40]3 years ago
7 0

Answer:

72

Explanation:

x = 8

6x + 24

Insert 8 for x

6 · 8 + 24

Multiply

6 · 8 = 48

Add

48 + 24 = 72

Therefore 6x + 24 = 72

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What is the value of x in the equation 13 x minus 2 (8 + 5 x) = 12 minus 11 x?
AnnZ [28]

The value of n in given proportion is 16

<u><em>Solution:</em></u>

We have to find the value of "n" in the proportion

<em><u>Given proportion is:</u></em>

<em><u></u></em>\frac{n}{28} = \frac{4}{7}<em><u></u></em>

We can solve the above proportion by cross-multiplying

Multiply the numerator of the left-hand fraction by the denominator of the right-hand fraction

Multiply the numerator of the right-hand fraction by the denominator of the left-hand fraction

Set the two products equal to each other

Solve for the variable

->\frac{n}{28} = \frac{4}{7}

-> 7 *  n = 4 * 28

-> n = \frac{4 * 28}{7}

-> n = 4 * 4 = 16

Thus the value of n in given proportion is 16

6 0
3 years ago
Read 2 more answers
Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m
Elenna [48]

Answer:

Part a: <em>The probability of no arrivals in a one-minute period is 0.000045.</em>

Part b: <em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>

Part c: <em>The probability of no arrivals in a 15-second is 0.0821.</em>

Part d: <em>The probability of at least one arrival in a 15-second period​ is 0.9179.</em>

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)

The probability mass function of X can be written as,

P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots

Substitute the value of λ=10 in the formula as,

P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}

​Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}

<em>The probability of no arrivals in a one-minute period is 0.000045.</em>

Part b:

The probability mass function of X can be written as,

P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}

<em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as \frac{X}{4}

\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}

That is,

Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)

So, the probability mass function of Y is,

P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}

<em>The probability of no arrivals in a 15-second is 0.0821.</em>

Part d:  

The probability that there is at least one arrival in a 15-second period is calculated as,

\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}

            \begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}

<em>The probability of at least one arrival in a 15-second period​ is 0.9179.</em>

​

​

7 0
3 years ago
In the isosceles triangle ABC, we have AB=AC=4. The altitude from B meets AC at H. If AH=3(HC) then determine BC.
Inga [223]

Answer:

BC=√7

Step-by-step explanation:

AC=4

AC=AH+HC

=3HC+HC

=4HC

HC=1/4AC=1/4×4=1

AH=3HC=3×1=3

BH⊥ AC

AB=AC=4

BC=\sqrt{AB^2-AH^2} =\sqrt{4^2-3^2} =\sqrt{16-9} =\sqrt{7}

5 0
2 years ago
Expand the logarithm. log4 (3xyz​)^2
tangare [24]
2<span>(<span>log4</span><span>(3)</span>+<span>log4</span><span>(x)</span>+<span>log4</span><span>(y)</span>+<span>log4</span><span>(z)</span><span>)</span></span>
4 0
3 years ago
Read 2 more answers
A jewelry manufacturer makes pendants with 1 and 1/2 ounces of silver in each. It has
Brilliant_brown [7]
Do repeated addition to 1 and 1/2 to get the answer
8 0
3 years ago
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