Help doing test

dem82 [27]

So we see that the ratio between the two octagons is 7:1, since 28/4=7 so what we do next is multiply the values of the smaller octagon by 7. But that’s the long way. There’s actually a shortcut by multiplying the perimeter of the smaller octagon, 34, by 7. This in turn equals 238.

6 0

4 years ago

Read 2 more answers

Which of these is an example of continuous random variable?

Nana76 [90]

Answer:

B. weight of a bag of apples

Step-by-step explanation:

A continuous random variable is a variable that is measured not counted. It can be any number between integers in decimal form.

So from the options given above, only the weight of bag of apples is a continuous variable as it will be measured.

While the others cannot be expressed in the form of decimals.

So the correct answer is:

B. weight of a bag of apples ..

4 0

4 years ago

A total of 12 players consisting 6 male and 6 female badminton players are attending a training camp

abruzzese [7]

Step-by-step explanation:

"A total of 12 players consisting 6 male and 6 female badminton players are attending a training camp."

"(a) During a morning activity of the camp, these 12 players have to randomly group into six pairs of two players each."

"(i) Find the total number of possible ways that these six pairs can be formed."

The order doesn't matter (AB is the same as BA), so use combinations.

For the first pair, there are ₁₂C₂ ways to choose 2 people from 12.

For the second pair, there are ₁₀C₂ ways to choose 2 people from 10.

So on and so forth. The total number of combinations is:

₁₂C₂ × ₁₀C₂ × ₈C₂ × ₆C₂ × ₄C₂ × ₂C₂

= 66 × 45 × 28 × 15 × 6 × 1

= 7,484,400

"(ii) Find the probability that each pair contains players of the same gender only. Correct your final answer to 4 decimal places."

We need to find the number of ways that 6 boys can be grouped into 3 pairs. Using the same logic as before:

₆C₂ × ₄C₂ × ₂C₂

= 15 × 6 × 1

= 90

There are 90 ways that 6 boys can be grouped into 3 pairs, which means there's also 90 ways that 6 girls can be grouped into 3 pairs. So the probability is:

90 × 90 / 7,484,400

= 1 / 924

≈ 0.0011

"(b) During an afternoon activity of the camp, 6 players are randomly selected and 6 one-on-one matches with the coach are to be scheduled.

(i) How many different schedules are possible?"

There are ₁₂C₆ ways that 6 players can be selected from 12. From there, each possible schedule has a different order of players, so we need to use permutations.

There are 6 options for the first match. After that, there are 5 options for the second match. Then 4 options for the third match. So on and so forth. So the number of permutations is 6!.

The total number of possible schedules is:

₁₂C₆ × 6!

= 924 × 720

= 665,280

"(ii) Find the probability that the number of selected male players is higher than that of female players given that at most 4 females were selected. Correct your final answer to 4 decimal places."

If at most 4 girls are selected, that means there's either 0, 1, 2, 3, or 4 girls.

If 0 girls are selected, the number of combinations is:

₆C₆ × ₆C₀ = 1 × 1 = 1

If 1 girl is selected, the number of combinations is:

₆C₅ × ₆C₁ = 6 × 6 = 36

If 2 girls are selected, the number of combinations is:

₆C₄ × ₆C₂ = 15 × 15 = 225

If 3 girls are selected, the number of combinations is:

₆C₃ × ₆C₃ = 20 × 20 = 400

If 4 girls are selected, the number of combinations is:

₆C₂ × ₆C₄ = 15 × 15 = 225

The probability that there are more boys than girls is:

(1 + 36 + 225) / (1 + 36 + 225 + 400 + 225)

= 262 / 887

≈ 0.2954

7 0

3 years ago

50 POINTS!! Pleasee helppppp!!! A scientific journal recently reported that there are currently 2400 black rhinos in the world t

Harman [31]

Answer:

C

Step-by-step explanation:

So, we started out with 2,400 black rhinos.

And we know that they are declining at a rate of 10% per year.

In other words, each subsequent year will only have 90% of the previous year's rhinos.

Therefore, our function is:

$f(x)=2400(.9)^x$

We can do some elimination first.

First, the starting point or y-intercept must be 2400 since that's the number of rhinos we started out with.

Therefore, we can eliminate A and D.

From the remaining choices B or C, we can technically also eliminate B. This is because from our function, we know it's exponential decay, and exponential decay is not a straight line.

However, to double check, let's check the population after 10 years using our function. So:

$f(10)=2400(.9)^{10}$

Use a calculator:

$f(10)\approx837$

So, after 10 years, the population should be around 840.

B is definitely not 840 (at x=10). However, C is right around it.

Therefore, we can conclude that C is the right answer.

And we're done!

6 0

3 years ago

Multiply 60 times 7 410, 420, 67, or 430?

DochEvi [55]

Answer:

420

Step-by-step explanation:

4 0

3 years ago