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3 years ago

The perimeter of an equilateral triangle is 48 centimeters, what’s the height to the nearest whole number?

Mathematics

2 answers:

Lisa [10]3 years ago

6 0

Answer:

Step-by-step explanation:

Because if its all equal sides then you just divide 48 by 3 which gives 16 since there is 3 sides is how you got the 3

Lerok [7]3 years ago

5 0

Answer:

Height = 13.86 cm

Step-by-step explanation:

1. Find the length of each side

48/3 = 16

2. cut the triangle in half.

Base = 8 Hypotenuse = 16

3. use the formula A^2 + B^2 = C^2

(16^2 - 8^2)^(1/2) = 13.86 cm

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A population of 400 beetles grows by 5% each week. How can the beetle population be determined after a number of weeks, w?

Helga [31]

For this case we have an equation of the form:
f (w) = A * (b) ^ w
Where,
A: initial amount
b: growth rate
w: number of weeks
Substituting values we have:
f (w) = 400 * (1.05) ^ w
Answer:
the beetle population can be determined after a number of weeks, w, with the following function:
f (w) = 400 * (1.05) ^ w

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: 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

4 years ago

Please help me and ty

TEA [102]

Answer:

D. 1963 square inches

Step-by-step explanation:

Area of a circle= πr^2

π= 3•14

r= 25 inches

3•14 × 25 × 25

= 1962•5

~ 1963 inches

hope it helps

6 0

3 years ago

Carl is boarding a plane. He has 222 checked bags of equal weight and a backpack that weighs 4 Write an equation to determine th

dangina [55]

Answer: 2w + 4 = 35

Step-by-step explanation:

Carl's two checked bags and his one backpack together weigh 35kg.

Carl's two checked bags have equal weight so both their weight can be denoted as 2w.

Equation is;

2w + 4 = 35

If you were to solve;

2w + 4 = 35

2w = 35 - 4

w = 31/2

w = 15.5kg

4 0

3 years ago

(5-x) (b+2) using distributive properties simplify the following expression

lyudmila [28]

Answer:

-bx+5b-2x+10

Step-by-step explanation:

https://www.mathpapa.com/algebra-calculator.html?q=%5Cleft((5-x)(b%2B2%5Cright)

8 0

3 years ago