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NikAS [45]
3 years ago
6

What is the sum of interior angels of an octagonal box?

Mathematics
2 answers:
Gala2k [10]3 years ago
8 0
8-2=6
180×6=1080 degrees
sum of interior angle of an octagonal box = 1080 degrees
Elodia [21]3 years ago
7 0
Just like any regular polygon, to find the interior angle we use the formula 180(n–2) For an octagon, <span>n=8. (n=the number of sides of the polygon; a polygon is a shape with 3 or more sides) T</span><span>he sum of interior angles of an octagon=(n-2)*180 degrees </span><span>=(8-2)*180 degrees </span><span>= 6*180 degrees </span><span>=1080 degrees.

Hope this helps xox :)</span>
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Rectangle JKLM is congruent to a 4 inch by 7 inch rectangle. How many different values are possible for the length
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The volume of a sphere is increased by 6%. Calculate the corresponding percentage increased in it area.
lions [1.4K]

Answer:

The new volume will be V + 0.06V or 1.06V

Step-by-step explanation:

The volume of the sphere increased by 6% or 0.06V. This means the total volume of the sphere will be V + 0.06V or 1.06V.


3 0
3 years ago
Select the correct answer from each drop-down menu. Trapezoid PQRS be inscribed in a circle because the .
prisoha [69]

Trapezoid PQRS be inscribed in a circle because its opposite angles are supplementary.

According to the statement

we have given that the Trapezoid PQRS be inscribed in a circle and we have to find the correct answer from the given data.

So, For this purpose, we know that the

The inscribed quadrilateral conjecture states that the opposite angle of any inscribed quadrilateral are supplementary to each other. That is, they have a sum of 180 degrees.

From the diagram given,

the opposite angles in the trapezoid are 115 and 65 degree.

So, after adding it become

115 + 65 = 180 degrees.

Therefore, we can conclude that: trapezoid QPRS can be inscribed in a circle because its opposite angles are supplementary.

So, Trapezoid PQRS be inscribed in a circle because its opposite angles are supplementary.

Learn more about inscribed quadrilateral conjecture here

brainly.com/question/12238046

Disclaimer: This question was incomplete. Please find the full content below.

Question:

Select the correct answer from each drop-down menu.

Trapezoid PQRS

4650

be inscribed in a circle because the

Reset

115⁰

Next

65°

For more understanding please see the image below.

#SPJ4

3 0
1 year ago
What is the perimeter of a square that has and area equal to 64??
egoroff_w [7]

Answer:12

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
A market research firm supplies manufacturers with estimates of the retail sales of their products from samples of retail stores
EastWind [94]

Answer:

(52-49) -4.300= -1.300

(52-49) +4.300= 7.300

And the 95% confidence would be :

-1.300 \leq \mu_1 -\mu_2 \leq 7.300

Step-by-step explanation:

We have the following info given from the problem

\bar X_1 = 52 sample mean for this year

s_1= 13 sample deviation for this year

n_1 = 75 random sample selected for this year

\bar X_2 = 49 sample mean for last year

s_2= 11 sample deviation for last year

n_1 = 53 random sample selected for last year

And we want to construct a 95% confidence interval estimate of the difference μ1−μ2, where μ1 is the mean of this year's sales and μ2 is the mean of last year's sales

For this case the formula that we need to use is:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

The degrees of freedom are given by:

df= n_1 +n_2 -2 = 75+53-2= 126

The confidence level is 0.95 and the significance would be \alpha=0.05 and \alpha/2 =0.025 so then the critical value for this case is :

t_{\alpha/2}= 1.979

The margin of error would be:

ME = 1.979 \sqrt{\frac{13^2}{75} +\frac{11^2}{49}}= 4.300

And the confidence interval would be given by:

(52-49) -4.300= -1.300

(52-49) +4.300= 7.300

And the 95% confidence would be :

-1.300 \leq \mu_1 -\mu_2 \leq 7.300

6 0
3 years ago
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