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2 years ago

Can someone help me out???

Mathematics

1 answer:

fenix001 [56]2 years ago

8 0

Answer:

to create this figure, connect the three sets of opiate points of the hexagon. Construction angle bisectors for each new angle, creating six new intersections with the circle, for a total of 12 points. Connect the points to make a regular 12-sided polygon.

Step-by-step explanation:

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Select the example of an inequality. o 14y = 2 0 32 - 4 044 = 15 4y < 64

faltersainse [42]

The last one is the answer

4 0

2 years ago

The common ratio in a geometric series is 0.50.50, point, 5 and the first term is 256256256.

V125BC [204]

Answer:

504

Step-by-step explanation:

I think the correct question is like:The common ratio in a geometric series is 0.50 ( point 5) and the first term is 256.

Find the sum of the first 6 terms in the series.

If it's right, then

Sₙ = (a₁ * (1 - rⁿ)) / (1 - r)

S₆ = (256 * ( 1 - 0.5⁶)) / (1 - 0.5) = (256 * 0.984375) / 0.5 = 504

6 0

2 years ago

Simplify 10 • 4 • 4 • 50 what is it lol

shusha [124]

0.0125 is your answer.
Hope this helps~!
~{Dunsforhands}

8 0

2 years ago

Read 2 more answers

Exactly 120 tickets were sold for a concert. The tickets cost $12 each for adults, $10 each for seniors, and $6 each for childre

Gemiola [76]

A=adult tickets
c=chlderenticekts
s=sinior tickets

a=c

a+c+s=120
2a+s=120

total made is 1100
12a+6c+10s=1100
sub
a=c
12a+6a+10s=1100
add
18a+10s=1100

multiply first equation by -9 and add to other equation

-18a-9s=-1080
18a+10s=1100 +
0a+s=20

s=20
20 senior tickets were sold

answer is B

5 0

3 years ago

Ninety-one percent of products come off the line within product specifications. Your quality control department selects 15 produ

Allisa [31]

Answer:

Probability of stopping the machine when $X < 9$ is 0.0002

Probability of stopping the machine when $X < 10$ is 0.0013

Probability of stopping the machine when $X < 11$ is 0.0082

Probability of stopping the machine when $X < 12$ is 0.0399

Step-by-step explanation:

There is a random binomial variable $X$ that represents the number of units come off the line within product specifications in a review of $n$ Bernoulli-type trials with probability of success $0.91$ . Therefore, the model is ${15 \choose x} (0.91) ^ {x} (0.09) ^ {(15-x)}$ . So:

$P (X < 9) = 1 - P (X \geq 9) = 1 - [{15 \choose 9} (0.91)^{9}(0.09)^{6}+...+{ 15 \choose 15}(0.91)^{15}(0.09)^{0}] = 0.0002$

$P (X < 10) = 1 - P (X \geq 10) = 1 - [{15 \choose 10}(0.91)^{10}(0.09)^{5}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0013$

$P (X < 11) = 1 - P (X \geq 11) = 1 - [{15 \choose 11}(0.91)^{11}(0.09)^{4}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0082$

$P (X < 12) = 1- P (X \geq 12) = 1 - [{15 \choose 12}(0.91)^{12}(0.09)^{3}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0399$

Probability of stopping the machine when $X < 9$ is 0.0002

Probability of stopping the machine when $X < 10$ is 0.0013

Probability of stopping the machine when $X < 11$ is 0.0082

Probability of stopping the machine when $X < 12$ is 0.0399

8 0

2 years ago