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2 years ago

Sixteen is 64% of what number? A.12 B.25 C.32 D.48

Mathematics

2 answers:

zhannawk [14.2K]2 years ago

4 0

answer is B.25 because 16 divided by 64% is 25 i hope this helps

Basile [38]2 years ago

3 0

B. 25

16 divided by 0.64 (the percentage as a decimal) = 25

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The largest angle in a triangle is three times as big as the smallest angle, and the middle angle is twice as big as the smalles

PtichkaEL [24]

Let X = smallest angle
largest 3x
middle=2x. add the 3 angles of any triangle and you get ,180

x+2x+3x=180
6x=180
x= 30
smallest =30
middle=60
largest=90

8 0

3 years ago

Please solve with working much appreciated

mojhsa [17]

Answer:

x°= 55°

Step-by-step explanation:

First, find the degrees of the larger triangle.

Since every triangle equals 180° degrees, make an equation.

60°+90°+a°= 180°

Now, add the degrees together.

120°+a°= 180°

Then subtract from both sides.

120°+a°= 180°

-120 ° -120°

a°= 60°

Now that you know the sum of all the angles, subtract 25° from 60°(which is the measure of a).

60°-25°= 35°

Now that we have the measurement of the angle from the smaller triangle inside the larger one, make an equation.

35°+90°+x°= 180°

Then, add the degrees.

125°+x°= 180°

Subtract from both sides of the equation to get the answer for the variable.

125°+x°= 180°

-125° -125°

x°= 55°

6 0

3 years ago

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

Elenna [48]

Answer:

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

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

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

Part d: The probability of at least one arrival in a 15-second period is 0.9179.

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,

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

The probability of no arrivals in a one-minute period is 0.000045.

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,

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

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

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:

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

The probability of no arrivals in a 15-second is 0.0821.

Part d:

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

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

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

The probability of at least one arrival in a 15-second period is 0.9179.

7 0

3 years ago

The Tigers, a football team, must gain 10 yards in the next four plays to keep possession of the ball . The Tigers lose 12 yards

mestny [16]

No because They lose 12 sand that puts them at -12. Then they gain 5 which puts them at -7 and then they lose 8 which puts them at -15. Please mark as brainliest

5 0

3 years ago

Let w = x2 + y2 + z2, x = uv, y = u cos(v), z = u sin(v). use the chain rule to find ∂w ∂u when (u, v) = (9, 0).

jasenka [17]

By the chain rule,

$\dfrac{\partial w}{\partial u}=\dfrac{\partial w}{\partial x}\cdot\dfrac{\partial x}{\partial u}+\dfrac{\partial w}{\partial y}\cdot\dfrac{\partial y}{\partial u}+\dfrac{\partial w}{\partial z}\cdot\dfrac{\partial z}{\partial u}$
$\dfrac{\partial w}{\partial u}=2xv+2y\cos v+2z\sin v$

We also have $x(9,0)=0$ , $y(9,0)=9$ , and $z(9,0)=0$ , so at this point we get

$\dfrac{\partial w}{\partial u}(9,0)=2\cdot9\cdot\cos0=18$

7 0

3 years ago