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WITCHER [35]
3 years ago
11

Can someone please help me with this page and show work ?!?! I’m struggling with this will give brainleist!

Mathematics
1 answer:
egoroff_w [7]3 years ago
3 0
You can download this app called photo math and take a pic of it and it tells you the answer
You might be interested in
Weary of the low turnout in student elections, a college administration decides to choose an SRS of three students to form an ad
-Dominant- [34]

Answer:

P(ABC) = 0.110592

P(ABC^c) = 0.119808

P(AB^cC) = 0.119808

P(A^cBC) = 0.119808

P(AB^cC^c)  = 0.129792

P(A^cBC^c)  = 0.129792

P(A^cB^cC)  = 0.129792

P(A^cB^cC^c)  = 0.140608

Step-by-step explanation:

Given

P(A) = P(B) = P(C) = 48\%

Convert the probability to decimal

P(A) = P(B) = P(C) = 0.48

Solving (a): P(ABC)

This is calculated as:

P(ABC) = P(A) * P(B) * P(C)

This gives:

P(ABC) = 0.48*0.48*0.48

P(ABC) = 0.110592

Solving (b): P(ABC^c)

This is calculated as:

P(ABC^c) = P(A) * P(B) * P(C^c)

In probability:

P(C^c) = 1 - P(C)

So, we have:

P(ABC^c) = P(A) * P(B) * (1 - P(C))

P(ABC^c) = 0.48 * 0.48 * (1 - 0.48)

P(ABC^c) = 0.48 * 0.48 * 0.52

P(ABC^c) = 0.119808

Solving (c): P(AB^cC)

This is calculated as:

P(AB^cC) = P(A) * P(B^c) * P(C)

P(AB^cC) = P(A) * [1 - P(B)] * P(C)

P(AB^cC) = 0.48 * (1 - 0.48)* 0.48

P(AB^cC) = 0.48 * 0.52* 0.48

P(AB^cC) = 0.119808

Solving (d): P(A^cBC)

This is calculated as:

P(A^cBC) = P(A^c) * P(B) * P(C)

P(A^cBC) = [1-P(A)] *P(B) * P(C)

P(A^cBC) = (1 - 0.48)* 0.48 * 0.48

P(A^cBC) = 0.52* 0.48 * 0.48

P(A^cBC) = 0.119808

Solving (e): P(AB^cC^c)

This is calculated as:

P(AB^cC^c)  = P(A) * P(B^c) * P(C^c)

P(AB^cC^c)  = P(A) * [1-P(B)] * [1-P(C)]

P(AB^cC^c)  = 0.48 * [1-0.48] * [1-0.48]

P(AB^cC^c)  = 0.48 * 0.52*0.52

P(AB^cC^c)  = 0.129792

Solving (f): P(A^cBC^c)

This is calculated as:

P(A^cBC^c)   = P(A^c) * P(B) * P(C^c)

P(A^cBC^c)   = [1-P(A)] * P(B) * [1-P(C)]

P(A^cBC^c)   = [1-0.48] * 0.48 * [1-0.48]

P(A^cBC^c)   = 0.52 * 0.48 * 0.52

P(A^cBC^c)  = 0.129792

Solving (g): P(A^cB^cC)

This is calculated as:

P(A^cB^cC)  = P(A^c) * P(B^c) * P(C)

P(A^cB^cC)  = [1-P(A)] * [1-P(B)] * P(C)

P(A^cB^cC)  = [1-0.48] * [1-0.48] * 0.48

P(A^cB^cC)  = 0.52 * 0.52 * 0.48

P(A^cB^cC)  = 0.129792

Solving (h): P(A^cB^cC^c)

This is calculated as:

P(A^cB^cC^c)  = P(A^c) * P(B^c) * P(C^c)

P(A^cB^cC^c)  = [1-P(A)] * [1-P(B)] * [1-P(C)]

P(A^cB^cC^c)  = [1-0.48] * [1-0.48] * [1-0.48]

P(A^cB^cC^c)  = 0.52*0.52*0.52

P(A^cB^cC^c)  = 0.140608

5 0
3 years ago
Leilani practiced piano for 3/5 of an hour. Sarah practiced piano for 7/8 of an hour.
ivolga24 [154]

Answer:

The number of hours did Sarah practice more than Leilani is \dfrac{11}{40}  hours  .

Step-by-step explanation:

Given as :

The time for which Leilani practiced piano = \dfrac{3}{5} hours

The time for which Sarah practiced piano = \dfrac{7}{8} hours

Let the number of hours did Sarah practice more than Leilani = T hours

<u>Now, According to question</u>

The number of hours did Sarah practice more than Leilani = \dfrac{7}{8} hours -  \dfrac{3}{5} hours

Or, T =  \dfrac{35-24}{40} hours

Or, T =  \dfrac{11}{40} hours

So,The number of hours did Sarah practice more than Leilani = T =  \dfrac{11}{40}  hours

Hence, The number of hours did Sarah practice more than Leilani is \dfrac{11}{40}  hours  . Answer

3 0
3 years ago
ANYONE ABLE TO HELP??? ASAP!!!!
Artemon [7]
I belive it c but i mit be wrong so dont mark it

5 0
3 years ago
Question 13 PLEASE HELP find the x-intercepts of -x+2y=6 X-intercept:(____,0). Type only the value that fills one the blank
Free_Kalibri [48]

Answer:

Step-by-step explanation:

to be honest I'm not sure how

8 0
3 years ago
What option is correct: A or B
Yuliya22 [10]
The image isn’t visible. Would recommend reposting
5 0
3 years ago
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