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sertanlavr [38]
2 years ago
5

This is hard lets see if you well be able to do it

Mathematics
1 answer:
erica [24]2 years ago
8 0

Answer:

I did  the math  the answer is 90  

Step-by-step explanation:

if  you multiply 15 by 6 you get 90

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Solve (x-2<5) u (x+7>6)
grin007 [14]

x= | 1 < x < 3

x + 2 < 5 and x - 7 > -6

x < 5 - 2 and x > -6 + 7

I don't know if I am 100% right.

4 0
3 years ago
Will mark brainliest. will be reported if only for the points
tia_tia [17]

Answer:

d=~7.6

Step-by-step explanation:

7^2+3^2=d^2

49+9=d^2

58=d^2

Square root of 58=~7.6

d=~7.6

6 0
3 years ago
Read 2 more answers
Brian’s school locker has a three-digit combination lock that can be set using the numbers 5 to 9 (including 5 and 9), without r
andreev551 [17]
The possible digits are: 5, 6, 7, 8 and 9. Let's mark the case when the locker code begins with a prime number as A and the case when <span>the locker code is an odd number as B. We have 5 different digits in total, 2 of which are prime (5 and 7).

First propability:
</span>P_A=\frac{2}{5}=40\%
<span>
By knowing that digits don't repeat we can say that code is an odd number in case it ends with 5, 7 or 9 (three of five digits).

Second probability:
</span>P_B=\frac{3}{5}=60\%
5 0
3 years ago
Which of the following numbers could not possibly be a​ probability? Justify your answer. 0.123 negative 0.577 0 What must be tr
dusya [7]

Answer:

First question: -0.577 cannot be a probability.

Second question: The number must be between 0 and 1, inclusive (D)

Third question: -0.577 cannot be a probability.

Step-by-step explanation:

A probability is a real number that is in the interval [0, 1]. Therefore, any number outside this range cannot be considered a probability. Under this premise:

First question: -0.577 cannot be a probability.

Second question: The number must be between 0 and 1, inclusive (D)

Third question: -0.577 cannot be a probability.

3 0
3 years ago
For two events and , the probability that occurs is 0.8, the probability that occurs is 0.4, and the probability that both occur
sergey [27]

Answer:

P(B|A)=0.25  , P(A|B) =0.5

Step-by-step explanation:

The question provides the following data:

P(A)= 0.8

P(B)= 0.4

P(A∩B) = 0.2

Since the question does not mention which of the conditional probabilities need to be found out, I will show the working to calculate both of them.

To calculate the probability that event B will occur given that A has already occurred (P(B|A) is read as the probability of event B given A) can be calculated as:

P(B|A) = P(A∩B)/P(A)

      = (0.2) / (0.8)  

P(B|A)=0.25

To calculate the probability that event A will occur given that B has already occurred (P(A|B) is read as the probability of event A given B) can be calculated as:

P(A|B) = P(A∩B)/P(B)

          = (0.2)/(0.4)

P(A|B) =0.5

7 0
3 years ago
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