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MrRissso [65]
2 years ago
13

The probability of event A is x, and the probability of event B is y.

Mathematics
1 answer:
Yuki888 [10]2 years ago
7 0

Answer:

A.  P(A l B) = x

Step-by-step explanation:

For independent events A and B,

P(B | A) = P(B)

P(A | B) = P(A)

P(A ∩ B) = P(A)P(B)

If P(A) = x   and   P(B) = y  then:

⇒ P(B | A) = y

⇒ P(A | B) = x

⇒ P(A ∩ B) = xy

Therefore, the correct solution from the options given is:

A.  P(A l B) = x

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find the two angle measures from the diagram 3x+10 3x-28 there's 3 answer's 71,71 109 ,109 109,71 last question brainiest u will
andriy [413]

Answer:

<h2>109 and 71</h2>

Step-by-step explanation:

3x + 10 + 3x - 28 = 180

6x = 180 - 10 + 28

x = 198/6

x = 33

substitute the value of x = 33 back into the equation to solve for each angle

= 3x + 10

= 3(33) + 10

= 109

= 3x - 28

= 3(33) - 28

= 71

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3 years ago
How do you write the l<br> Algebraic expression 12 times the quantity 15 minus a number d
daser333 [38]
12 times the quantity 15 minus a number (d)....

12(15 - d) <== ur expression
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3 years ago
Statistics professors believe the average number of headaches per semester for all students is more than 18. From a random sampl
kompoz [17]

Complete Question

Statistics professors believe the average number of headaches per semester for all students is more than 18. From a random sample of 15 students, the professors find the mean number of headaches is 19 and the standard deviation is 1.7. Assume the population distribution of number of headaches is normal.the correct conclusion at \alpha =0.001 is?

Answer:

There is no sufficient evidence to support the professor believe

Step-by-step explanation:

From the question we are told that

     The population mean is  \mu =  18

     The sample size is  n  =  15

      The sample mean is  \=  x  =  19

      The standard deviation is  \sigma =  1.7

      The level of significance is  \alpha  =  0.001

The null hypothesis is  H_o:  \mu = 18

The  alternative hypothesis is  H_a :  \mu > 18

 The critical value of the level of significance from the normal distribution table is    

         Z_{\alpha } =  3.290527

The test hypothesis is mathematically represented as

           t =  \frac{\= x  -  \mu }{ \frac{\sigma}{ \sqrt{n} } }

substituting values  

         t =  \frac{ 19  - 18}{ \frac{1.7}{ \sqrt{15} } }

         t =  2.28

Looking at the value of  t and  Z_{\alpha } we can see that t <  Z_{\alpha } so we fail to reject the null hypothesis.

This mean that there is no sufficient evidence to support the professor believe

     

       

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Factors less than 1<br> -<br> 2
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