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Thepotemich [5.8K]
3 years ago
9

Can anyone help me with this question I’ve been stuck ?

Mathematics
1 answer:
iren [92.7K]3 years ago
6 0
I was about to answer but I’m really bad at math and didn’t want to give you the wrong answer. Hope you find what your looking for!
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I will mark brainliest for best answer!
UNO [17]

Answer:

a) 8 b)9 c)1 d) 3

Step-by-step explanation:

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2 years ago
Find a vector parametric equation r⃗(t) for the line through the points P=(1,0,−2) and Q=(1,5,1) for each of the given condition
const2013 [10]
So the question ask to find and calculate the vector parametric equation r(t) for the line through the points P=(1,0,-2) and Q(1,5,1) for each given condition. And the possible vector parametric equation is <1,2,-2>+t/4<1,5,3>. I hope you are satisfied with my answer and feel free to ask for more 
7 0
3 years ago
Please help me I need help
Luba_88 [7]

Answer:

|28 - x| = 0  \\ 28 - x = 0 \\  - x =  - 28 \\ x = 28

3 0
3 years ago
Read 2 more answers
The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 7.8 cm.
devlian [24]

Answer:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z\leq 1.72). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15)  from 1. Therefore, we have 1- P(Z\leq -1.09). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

Step-by-step explanation:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z\leq 1.72). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15)  from 1. Therefore, we have 1- P(Z\leq -1.09). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

4 0
3 years ago
If the profits in your consulting business increase by 6​% one year and decrease by 3​% the following​ year, your profits are up
Naddika [18.5K]

Answer:

its 1.388 p

Step-by-step explanation:

i hope this helps

3 0
3 years ago
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