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I am Lyosha [343]
3 years ago
12

Please help me with this question. I am very confused.

Mathematics
1 answer:
Oksanka [162]3 years ago
7 0
I think u do the ones in the parenthesis first bc of PEMDAS since it's both negative it become positive and u can cross out both the 2's and that leaves u with 1/7 and then u just times 1/7 with 8 and then u get 8/7

hoped this helped! :)
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A seventh-grade teacher asks her students a survey question. Which questions are appropriate for the teacher to ask her
zzz [600]

Answer:

How many hours of sleep do you get per night?

On average, how long does it take you to complete your homework?

What time do you leave for school in the morning?

Step-by-step explanation:

5 0
4 years ago
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A ball is draw at random form a box containing 6 red balls, 4 white balls and 5 blue balls. what is the probability that the bal
makkiz [27]
1 = 6 out of 15
2 = 4 out of 15
3 = 5 out of 15
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hope this helps you
3 0
3 years ago
Just #35 please piecewise functions
saveliy_v [14]

Answer: No, the friend is correct. In any function, each input value can only lead to one output value. When you input 3 for the x-values, you would get two output values because 3 is included in both equations. To fix this, you need to have the 3 not included in one of the equations.

For example, you could say y=\left \{ {{2x-2, x\leq 3} \atop {-3, x > 3}} \right. or y=\left \{ {{2x-2, x < 3} \atop {-3, x\geq 3}} \right. because the input value of 3 would not be included twice.

If you look at the attached screenshot, you will see that if you keep your friend's function, inputting 3 will result in two outputs of 4 and -3, so therefore, y=\left \{ {{2x-2,x\leq 3} \atop {-3,x\geq 3}} \right. cannot represent a piecewise function.

3 0
1 year ago
Delta Airlines' flights from Boston to Seattle are on time 70 % of the time. Suppose 6 flights are randomly selected, and the nu
emmainna [20.7K]

Answer:

P(at least 4)  = 0.74431

Step-by-step explanation:

Here

Probability of the flights on time is = 70% = 70/100= 0.7

q= 0.3

n= 6

As the probability of success and failure is constant and the n is also fixed then the binomail probability distribution can be used to solve this.

P (x= 4) = 6c4 *(0.7)^4 (0.3)^2= 15*0.2401* 0.09= 0.324135

P (x= 5) = 6c5 *(0.7)^5(0.3)^1= 6*0.16807* 0.3=0.302526

P (x= 6) = 6c6 *(0.7)^6 (0.3)^0= 0.117649P

P (x= 0) = 6c0 *(0.7)^0 (0.3)^6= 0.000729

P (x= 1) = 6c1 *(0.7)^1 (0.3)^5=0.010206

P (x= 2) = 6c 2*(0.7)^2 (0.3)^4=0.059535

P (x= 3) = 6c3 *(0.7)^3 (0.3)^3=0.18522

Probability of at least 4 means that probability of greater than 4 inclusive of 4.

This can be obtained in two ways.

First is by subtracting the probabilities of less than 4.

Second is by adding probabilities greater than 4 including 4.

We get the same answer with any of the two methods used.

P(at least 4) = 1- P(x<4)

                     = 1- [ 0.000729+ 0.010206 +0.059535+ 0.18522]

                     = 1- 0.25569

                       = 0.74431

P (at least 4) = P(x=4) +P(X=5) + P(x=6)

                        =0.324135+ 0.302526+ 0.117649

                        =0.74431

7 0
3 years ago
Help meeeeeeeeeeeeeee ​
Travka [436]

Answer:

b. 4 min   c. 5 min

Step-by-step explanation:

6 0
3 years ago
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