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maw [93]
3 years ago
15

How many are on a hexagon - 4th grader

Mathematics
1 answer:
Anna35 [415]3 years ago
7 0

Answer:

There are 6 sides to a hexagon

Step-by-step explanation:

simple questions like this can just be googled

You might be interested in
How many four-character passwords can be formed using the characters A, B, C, 1, 2 if the characters can be repeated
olga55 [171]
<h3>Answer: 625</h3>

Work Shown:

The set {A,B,C,1,2} has five items. There are four slots to fill.

So we have 5^4 = 5*5*5*5 = 625 different possible passwords where the characters can be repeated.

7 0
3 years ago
I need help ASAP! Can anyone please check my work?
STALIN [3.7K]

A = event the person got the class they wanted

B = event the person is on the honor roll

P(A) = (number who got the class they wanted)/(number total)

P(A) = 379/500

P(A) = 0.758

There's a 75.8% chance someone will get the class they want

Let's see if being on the honor roll changes the probability we just found

So we want to compute P(A | B). If it is equal to P(A), then being on the honor roll does not change P(A).

---------------

A and B = someone got the class they want and they're on the honor roll

P(A and B) = 64/500

P(A and B) = 0.128

P(B) = 144/500

P(B) = 0.288

P(A | B) = P(A and B)/P(B)

P(A | B) = 0.128/0.288

P(A | B) = 0.44 approximately

This is what you have shown in your steps. This means if we know the person is on the honor roll, then they have a 44% chance of getting the class they want.

Those on the honor roll are at a disadvantage to getting their requested class. Perhaps the thinking is that the honor roll students can handle harder or less popular teachers.

Regardless of motivations, being on the honor roll changes the probability of getting the class you want. So Alex is correct in thinking the honor roll students have a disadvantage. Everything would be fair if P(A | B) = P(A) showing that events A and B are independent. That is not the case here so the events are linked somehow.

8 0
3 years ago
I need help answering
jekas [21]

Put the values of the variables in place of the variables in the expression, then do the arithmetic.

8p+3q-18\\\\=8\left(\dfrac{1}{2}\right)+3(7)-18=\dfrac{8}{2}+21-18\\\\=4+21-18\\\\=25-18\\\\=\bf{7}

6 0
3 years ago
An insurance company has 10,000 automobile policyholders. The expected yearly claim per policyholder is $240, with a standard de
elena55 [62]

Answer:

almost 0%

Step-by-step explanation:

Given that for an insurance company with 10000 automobile policy holders, the expected yearly claim per policyholder is $240 with a standard deaviation of 800

using normal approximation, the probability that the total yearly claim exceeds $2.7 million is calculated as follows:

Sea sumatoria de x = SUMX, tenemos que:

P (SUMX \geq 2700000) = P(\frac{SUMX - 240*10000}{800 *\sqrt{10000} } \geq \frac{2700000 - 240*10000}{800 *\sqrt{10000} })

= P (z\geq \frac{2700000 - 240*10000}{800 *\sqrt{10000}})

= P (z => 3.75)

= 1 - P ( z < 3.75)

P = 1 - 0.999912

P = 0.000088

Which means that the probability is almost 0%

4 0
3 years ago
Please help me with these questions
Katarina [22]

Answer:


I can only do 1 but I hope it helps.


3 0
3 years ago
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