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kogti [31]
1 year ago
12

What is the perimeter, in centimeters, of Trian

Mathematics
1 answer:
inessss [21]1 year ago
6 0

For the given right triangle, the perimeter is 60cm, so the correct option is D.

<h3></h3><h3>How to get the perimeter of the triangle?</h3>

We can see a right triangle, by using the Pythagorean theorem we can get the missing cathetus:

a^2 + (20cm)^2 = (25cm)^2

a = \sqrt{(25cm)^2 - (20cm)^2} = 15cm

Now we know the measure of the 3 sides, then the perimeter of the given triangle is:

P = 15cm + 20cm + 25cm = 60cm

The perimeter of the triangle is 60cm.

If you want to learn more about right triangles:

brainly.com/question/2217700

#SPJ1

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The Answer is option B : 5

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Nia has numbers cards for 2,3,8,9,and 5.What are all the different ways Nia can arrange the cards to show only even numbers with
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Answer:

any order where 2 is the last number in the sequence

Step-by-step explanation:

the ten-thousand place in 12,345 is 5.

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5 0
3 years ago
What is the length of the apothem of a regular hexagon with 10-cm sides?
vodka [1.7K]

Answer:

The length of apothem of given hexagon is 8.66 cm

Step-by-step explanation:

Apothem of a polygon is given by:

apothem = \frac{s}{2\ tan\ (\frac{180}{n})}

Here

s is length of side

tan is trigonometric function

and n denotes number of sides

Given that the polygon is a hexagon

s = 10cm\\n = 6

Putting the values in the formula

apothem = \frac{10}{2\ tan\ (\frac{180}{6})}\\= \frac{10}{2\ tan\ 30}\\=\frac{10}{2 * 0.5773}\\=\frac{10}{1.1546}\\=8.6610\\Rounding\ off\ to\ nearest\ hundredth\\= 8.66\ cm

Hence,

The length of apothem of given hexagon is 8.66 cm

6 0
3 years ago
The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airp
motikmotik

Answer:

The 95% confidence interval for the population mean rating is (5.73, 6.95).

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{50}(6+4+6+. . .+6)\\\\\\M=\dfrac{317}{50}\\\\\\M=6.34\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{49}((6-6.34)^2+(4-6.34)^2+(6-6.34)^2+. . . +(6-6.34)^2)}\\\\\\s=\sqrt{\dfrac{229.22}{49}}\\\\\\s=\sqrt{4.68}=2.16\\\\\\

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=6.34.

The sample size is N=50.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.16}{\sqrt{50}}=\dfrac{2.16}{7.071}=0.305

The degrees of freedom for this sample size are:

df=n-1=50-1=49

The t-value for a 95% confidence interval and 49 degrees of freedom is t=2.01.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.01 \cdot 0.305=0.61

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 6.34-0.61=5.73\\\\UL=M+t \cdot s_M = 6.34+0.61=6.95

The 95% confidence interval for the mean is (5.73, 6.95).

4 0
3 years ago
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