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Natalija [7]
3 years ago
14

6th grade Geometry, please help it's due soon!

Mathematics
2 answers:
Bogdan [553]3 years ago
7 0
OCD and OAB
since they use the same angles they are congruent

:))
Eddi Din [679]3 years ago
6 0

Answer:

1. ABO, 2. OCD

Step-by-step explanation:

Since they have the same degrees of angles.

EX: <OA, <BO, <CO, <DO.

(IF I AM NOT RIGHT OR I AM LOSING THE DETAIL WANTED, PLEASE LIST OUT IN THE COMMENT BELOW THE ANSWER AND I WILL CHANGE IT.)

: ) Just to make it look bolder.

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Solve the inequality y − 6&gt;10
fredd [130]

if y-6 > 10, and 16-6 = 10, then y > 16 since 17-6 = 11, which is greater than 10. Any number greater than 16 can fulfil this inequality.

8 0
3 years ago
Name a point that is 2 away from (-1,5).
pashok25 [27]

Answer:[7,2]

Step-by-step explanation:

3 0
3 years ago
Urrent
Naily [24]

Answer:

The cost of 1000 bushels is $9790

Step-by-step explanation:

Given

Cost = \$9.79 per bushel

Required

Determine the cost of 1000 bushels

If 1 bushel costs $9.79,

Then 1000 costs:

Total = \$9.79 * 1000

Total = \$9790

3 0
3 years ago
Suppose small aircraft arrive at a certain airport according to a Poisson process with rate a 5 8 per hour, so that the number o
timurjin [86]

Answer:

(a) P (X = 6) = 0.12214, P (X ≥ 6) = 0.8088, P (X ≥ 10) = 0.2834.

(b) The expected value of the number of small aircraft that arrive during a 90-min period is 12 and standard deviation is 3.464.

(c) P (X ≥ 20) = 0.5298 and P (X ≤ 10) = 0.0108.

Step-by-step explanation:

Let the random variable <em>X</em> = number of aircraft arrive at a certain airport during 1-hour period.

The arrival rate is, <em>λ</em>t = 8 per hour.

(a)

For <em>t</em> = 1 the average number of aircraft arrival is:

\lambda t=8\times 1=8

The probability distribution of a Poisson distribution is:

P(X=x)=\frac{e^{-8}(8)^{x}}{x!}

Compute the value of P (X = 6) as follows:

P(X=6)=\frac{e^{-8}(8)^{6}}{6!}\\=\frac{0.00034\times262144}{720}\\ =0.12214

Thus, the probability that exactly 6 small aircraft arrive during a 1-hour period is 0.12214.

Compute the value of P (X ≥ 6) as follows:

P(X\geq 6)=1-P(X

Thus, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8088.

Compute the value of P (X ≥ 10) as follows:

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

Thus, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2834.

(b)

For <em>t</em> = 90 minutes = 1.5 hour, the value of <em>λ</em>, the average number of aircraft arrival is:

\lambda t=8\times 1.5=12

The expected value of the number of small aircraft that arrive during a 90-min period is 12.

The standard deviation is:

SD=\sqrt{\lambda t}=\sqrt{12}=3.464

The standard deviation of the number of small aircraft that arrive during a 90-min period is 3.464.

(c)

For <em>t</em> = 2.5 the value of <em>λ</em>, the average number of aircraft arrival is:

\lambda t=8\times 2.5=20

Compute the value of P (X ≥ 20) as follows:

P(X\geq 20)=1-P(X

Thus, the probability that at least 20 small aircraft arrive during a 2.5-hour period is 0.5298.

Compute the value of P (X ≤ 10) as follows:

P(X\leq 10)=\sum\limits^{10}_{x=0}(\frac{e^{-20}(20)^{x}}{x!})\\=0.01081\\\approx0.0108

Thus, the probability that at most 10 small aircraft arrive during a 2.5-hour period is 0.0108.

8 0
3 years ago
I need help please, i’m stuck i don’t get it
Mumz [18]

Answer:

I am not sure but I think it is 3

Step-by-step explanation:

2x+29=5x+20

5x-2x+29=20

3x=-9

1x=3

3 0
3 years ago
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