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3 years ago

Measurement of a 18 gon of its interior and exterior

1 answer:

4 0

There will be 18 exterior angles, each of 360/18 = 20 deg. So each interior angle, which is supplementary to the exterior angle will be 180–20 - 160 deg. The sum of the 18 interior angles = 18*160 = 2880 deg.

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Find the value of x that makes the equation true. 12x=4(x+2)+20

d1i1m1o1n [39]

Answer:

7/2

Step-by-step explanation:

12x=4(x+2)+20

12x=4x+8+20

12x-4x=8+20

8x=28

x=28/8

simplify

x=7/2

3 0

3 years ago

Name one of two types of pentahedrons

Diano4ka-milaya [45]

Square piramid ,triangle prism.

4 0

3 years ago

Read 2 more answers

A supplier delivers an order for 20 electric toothbrushes to a store. By accident, three of the electric toothbrushes are defect

krek1111 [17]

Answer:

The probability that the first two electric toothbrushes sold are defective is 0.016.

Step-by-step explanation:

The probability of an event, say E occurring is:

$P (E)=\frac{n(E)}{N}$

Here,

n (E) = favorable outcomes

N = total number of outcomes

Let X = number of defective electric toothbrushes sold.

The number of electric toothbrushes that were delivered to a store is, n = 20.

Number of defective electric toothbrushes is, x = 3.

The number of ways to select two toothbrushes to sell from the 20 toothbrushes is:

${20\choose 2}=\frac{20!}{2!(20-2)!}=\frac{20!}{2!\times 18!}=\frac{20\times 19\times 18!}{2!\times 18!}=190$

The number of ways to select two defective toothbrushes to sell from the 3 defective toothbrushes is:

${3\choose 2}=\frac{3!}{2!(3-2)!}=\frac{3!}{2!\times 1!}=3$

Compute the probability that the first two electric toothbrushes sold are defective as follows:

P (Selling 2 defective toothbrushes) = Favorable outcomes ÷ Total no. of outcomes

$=\frac{3}{190}\\$

$=0.01579\\\approx0.016$

Thus, the probability that the first two electric toothbrushes sold are defective is 0.016.

8 0

3 years ago

Estimate each product 3/4 times 23

storchak [24]

We would have to convert 3/4 to a decimal of 0.75 (the equivalent)
0.75*23 = 17.3

-or-

3 * 23 = 69
4 * 23 = 92

6 0

2 years ago

The table below shows data from a survey about the amount of time students spend doing homework each week. The students were in

nadezda [96]

Answer:

The correct option is;

Both spreads are best described by the standard deviation

Step-by-step explanation:

The given information are;

, College High School

High, 20 20

Low, 6 3

Q₁, 8 5.5

Q₃, 18 16

IQR, 10 10.5

Median, 14 11

Mean, 13.3 11

σ, 5.2 5.4

Checking for outliers, we have

College

Q₁ - 1.5×IQR gives 8 - 1.5×10 = -7

Q₃ + 1.5×IQR gives 18 + 1.5×10 = 33

For high school

Q₁ - 1.5×IQR gives 5.5 - 1.5×10.5 = -10.25

Q₃ + 1.5×IQR gives 16 + 1.5×10.5 = 31.75

Therefore, there are no outliers and the data is representative of the population

From the data, for the college students, it is observed that the difference between the mean, 13.3 and Q₁, 8, and between Q₃, 18 and the mean,13.3 is approximately the standard deviation, σ, 5.2

The difference between the low and the high is also approximately 3 standard deviations

Therefore the college spread is best described by the standard deviation

Similarly for the high school students, the IQR is approximately two standard deviations, the difference between the mean, 11 and Q₁, 5.5, and between Q₃, 16 and the mean,11 is approximately the standard deviation, σ, 5.4

Therefore the high school spread is also best described by the standard deviation.

4 0

3 years ago