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RUDIKE [14]
3 years ago
5

Which of the following statements istrue for every

Mathematics
1 answer:
Gelneren [198K]3 years ago
8 0

Answer:

Please see attachment

Step-by-step explanation:

Please see attachment

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Calculate the value of x.
Nikitich [7]

Answer:

x = 17deg

Step-by-step explanation:

Let angle y be the unknown angle IN THE TRIANGLE.

Angle y = 180 - 90 - 73

= 17deg

Angle x = Angle y (Alternate angles on 2 parallel lines)

Angle x = 17 deg

8 0
2 years ago
Read 2 more answers
Use multiplication in the distributive property to find the quotient 252 divided by 9
ira [324]

252 divided by 9 is 28

Firstly, I muplity 20 and 9.

I have 180.

Then I also muplity 8 and 9.

I got 72.

Then I add 72 and 180 which got me 252.

(I hope this helped!)

8 0
3 years ago
Tina placed a 15 meter rope along one side of the bicycle path. She hung a ribbon on each end of the rope and every 3 meters in-
dusya [7]
Hello!

She placed 6 ribbons

2 at the ends and 5 every 3 meters
7 0
3 years ago
20.
Alik [6]

Answer:

After 3 seconds

Step-by-step explanation:

Given

h(t) = -16t^2 + 96t

Required

Seconds to attain maximum height

The maximum of a quadratic function y = ax^2 + bx + c is calculated using:

x = -\frac{b}{2a}

So, we have:

t = -\frac{b}{2a}

Where:

a = -16 and b = 96

So:

t = -\frac{96}{2 * -16}

t = -\frac{96}{-32}

Cancel out negatives

t = \frac{96}{32}

t = 3

5 0
2 years ago
An instructor who taught two sections of engineering statistics last term, the first with 20 students and the second with 30, de
Lelechka [254]

Answer:

(a) P(X=10) = 0.2070

(b) P(X\geq 10) = 0.3798

Step-by-step explanation:

Note that in this problem we have an initial population N = 50, of which 30 fulfill a certain characteristic "m" (belong to the second section). Then, from the population N, a sample of size n = 15 is selected and it is desired to know how many comply with the desired characteristic (second section).

So

Let X be the number of projects in the second section, then X is a discrete random variable that can be modeled by a hypergeometric distribution.

(a)

Therefore, to answer question (a) we use the following equation presented in the attached image:

Where:

N = 50\\m = 30\\n = 15\\X = 10

Then:

P (X = 10) = \frac{\frac{30!}{10!(30-10)!}\frac{20!}{5!(20-5)!}}{\frac{50!}{15!(50-15)!}}\\\\\\P(X=10) = 0.2070

(b)

For part (b) we have:

P(X\geq10) = 1-P(X

P(X\geq 10) = 1-0.6202 \\\\P(X\geq 10) = 0.3798

4 0
3 years ago
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