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2 years ago

Can someone help me with this question? Thank you!

Mathematics

2 answers:

Harman [31]2 years ago

6 0

Answer:

122°

Step-by-step explanation:

The Angle Sum Property of a triangle gives that the sum of the internal angles of a triangle is equal to 180°.

Hence,

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + 43° + 15° = 180°

⇒ ∠A + 58° = 180°

⇒ ∠A = 122°

emmasim [6.3K]2 years ago

4 0

Answer:

122°

Step-by-step explanation:

triangle = 180°

∴180 - 43 - 15

=122°

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Calculate the length of the circumference of a circle with radius of 2.8cm

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3 years ago

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Two exponential functions are shown in the table.

sammy [17]

Based on the table, a conclusion which can be drawn about f(x) and g(x) is that: B. the functions f(x) and g(x) are reflections over the y-axis.

<h3>How to compare the functions f(x) and g(x)?</h3>

In Mathematics, two functions are considered to be reflections over the y-axis under the following condition:

If, f(-x) = g(x).

Evaluating the given functions, we have:

f(x) = 2ˣ

f(-x) = 2⁻ˣ = ½ˣ = g(x).

Similarly, two functions are considered to be reflections over the x-axis under the following condition:

If, -f(x) = g(x).

Evaluating the given functions, we have:

f(x) = 2ˣ

-f(x) = -2ˣ ≠ g(x).

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Read more on reflections here: brainly.com/question/2702511

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2 years ago

Patricia and Joe Payne are divorced. The divorce settlement stipulated that Joe pay $485 a month for their daughter Suzanne unti

zhuklara [117]

$\bf \qquad \qquad \textit{Future Value of an ordinary annuity}
\\\\
A=pymnt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right]$

$\bf \begin{cases}
A=
\begin{array}{llll}
\textit{original amount}\\
\textit{already compounded}
\end{array} &
\begin{array}{llll}

\end{array}\\
pymnt=\textit{periodic payments}\to &
\begin{array}{llll}
485\cdot 12\\
\underline{5280}
\end{array}\\
r=rate\to 6\%\to \frac{6}{100}\to &0.06\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{a year, thus once}
\end{array}\to &1\\

t=years\to &4
\end{cases}
\\\\\\
$

$\bf A=5280\left[ \cfrac{\left( 1+\frac{0.06}{1} \right)^{1\cdot 4}-1}{\frac{0.06}{1}} \right]$

Joe is making $485 payments monthly, but the amount gets interest on a yearly basis, not monthly, so the amount that yields interest is 485*12

also, keep in mind, we're assuming is compound interest, as opposed to simple interest

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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

Which graph correctly matches the equation y = 2-x?

nalin [4]

Answer: The answer is attached in the figure.

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We have the following points on this equation.

(0,2), (1,1), (2,0), etc.

When we try to match these points with the given graphs, the we see that these points does not match with the first, third and fourth graph. The only graph containing these points is second one.

Thus, the second option is correct. Also see the attached figure for the location of the three mentioned points.

4 0

3 years ago

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