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3 years ago

Could someone please help answer this, thank you, will give brainliest, plz no spam (home work help) thanks!

Mathematics

2 answers:

kolezko [41]3 years ago

5 0

A < -12

1/3a < -4
a < -12

Mark brainliest please

Zielflug [23.3K]3 years ago

3 0

Answer:

this is my answer

a< -12

hope it helps

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Bernie has to write a report. He will write 3/8 of the report on Monday and 1/3 of the report on Tuesday. How much of the report

natta225 [31]

Answer:

17/24

Step-by-step explanation:

3 x 3= 9

8 x 3= 24

1 x 8=8

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2 years ago

100pts: Is 999962000357 a prime number? If it is not, then give its prime factorization.

Serhud [2]

Answer:

No, it is NOT a prime number.

Step-by-step explanation:

To find out if a number is a prime number, you can divide it by two. If it is a whole number, it isn't prime. It would be a mixed number/fraction if so.

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3 years ago

Read 2 more answers

In a certain Algebra 2 class of 29 students, 7 of them play basketball and 24 of them play baseball. There are 3 students who pl

antiseptic1488 [7]

Answer:

$\frac{5}{29}$

Step-by-step explanation:

Let $n(A)$ represent students playing basketball, $n(B)$ represent students playing baseball.

Then, $n(A)=7$ , $n(B)=24$

Let $n(S)$ be the total number of students. So, $n(S)=29$ .

Now,

$P(A)=\frac{n(A)}{n(S)}=\frac{7}{29}$

$P(B)=\frac{n(B)}{n(S)}=\frac{24}{29}$

3 students play neither of the sport. So, students playing either of the two sports is given as:

$n(A\cup B)=n(S)-3\\n(A\cup B)=29-3=26$

∴ $P(A\cup B)=\frac{n(A\cup B)}{n(S)}=\frac{26}{29}$

From the probability addition theorem,

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Where, $P(A\cap B)$ is the probability that a student chosen randomly from the class plays both basketball and baseball.

Plug in all the values and solve for $P(A\cap B)$ . This gives,

$\frac{26}{29}=\frac{7}{29}+\frac{24}{29}+P(A\cap B)\\\\\frac{26}{29}=\frac{7+24}{29}+P(A\cap B)\\\\\frac{26}{29}=\frac{31}{29}+P(A\cap B)\\\\P(A\cap B=\frac{31}{29}-\frac{26}{29}\\\\P(A\cap B=\frac{31-26}{29}=\frac{5}{29}$

Therefore, the probability that a student chosen randomly from the class plays both basketball and baseball is $\frac{5}{29}$

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3 years ago

Find the area of a regular octagon with apothem K and side of 10.

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The answer is indeed 40 K.

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1 year ago