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2 years ago

??????????????????????????????????

Mathematics

1 answer:

tiny-mole [99]2 years ago

4 0

Answer:

I don't understand this image

Step-by-step explanation:

what is the question

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Does anyone know this? please help

natali 33 [55]

Answer:

3rd answer

Step-by-step explanation:

it would be the third answer

6 0

2 years ago

Please help me so stressed

mr Goodwill [35]

Answer:

Meron ka nang loptap bakit humihingi ka pa nang answer

8 0

3 years ago

Chen kept track of how many baskets were made in a basketball game. After 6 minutes, 5 baskets were made. How long did it take u

Mekhanik [1.2K]

Answer:

Assuming the same pace was kept, it would take 9.6 minutes.

Step-by-step explanation:

We can find this by setting up a proportion and then cross multiplying.

5 baskets/6 mins = 8 baskets/x mins

5 * x = 6 * 8

5x = 48

x = 9.6

5 0

3 years ago

Translate these statements into English, where R(x) is "x is a rabbit "and H(x) is "x hops" and the domain consists of all anima

Hatshy [7]

Answer:

A- for every animal, if the animal is a rabbit, the animal hops.

B- every animal is a rabbit and it hops.

C-there are animals that, if they are rabbits, they hop.

D-there are animals that are rabbits and they hop

8 0

3 years ago

Read 2 more answers

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$