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pashok25 [27]
2 years ago
15

A population has a mean of 200 and a standard deviation of 50. Suppose a simple random sample of size 100 is selected and x is u

sed to estimate . a. What is the probability that the sample mean will be within 65 of the population mean
Mathematics
1 answer:
svlad2 [7]2 years ago
6 0

Answer:

z(195) = (195-200)/50 = -0.1

z(205) = (205-200)/50 = +0.1

P(195 < x < 205) = p(-0.1 < z < 0.1) = 0.0797

Step-by-step explanation:

step by step is in they're

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7√6 x √27 is equal to
mr Goodwill [35]

Answer:

63\sqrt{2}

Step-by-step explanation:

7\sqrt{6}* \sqrt{27} \\=7\sqrt2\sqrt{3} * \sqrt{9} \sqrt{3}\\ =7\sqrt{2}\sqrt{3} *3\sqrt{3}\\  =\sqrt{2} *3*3*7\\=63\sqrt{2}

3 0
3 years ago
2y+X-4=0 what is the answer
vodomira [7]
YX=2, I assume you need to get the 'X' and 'Y' alone. Also just for future reference, you're trying to get variables alone and all the numbers on the other side. Lastly, whatever you do to one side you do to the other. Good luck!
7 0
3 years ago
Read 2 more answers
Emma has 460 ballet stickers in her collection. Each month she adds 28 more stickers to her collection.
MArishka [77]
I believe you would multiply 28 by 7, then add that to 460
3 0
3 years ago
Please help with this
N76 [4]
VWX\cong KLJ means that the triangles are congruent, that is exactly identical.

The order of letters is very important. Form the given congruence we can write the following ratio:

\frac{VW}{KL} = \frac{WX}{LJ} = \frac{VX}{KJ} =1

(notice that we do not mix the order: the first 2 letters of VWX are compared to the first 2 letters of KLJ and so on)

the ratio is one because the corresponding sides are equal.

\frac{VW}{KL}=1\\\\ \frac{2z-1}{z+2}=1\\\\2z-1=z+2\\\\2z-z=2+1\\\\z=3

|VW|=2z-1=2.3-1=6-1=5


Answer: 5 units


8 0
3 years ago
Please help me with this! Picture is provided and only answer is you know it!
Sophie [7]

Answer:

Option A

Step-by-step explanation:

Number of components assembled by the new employee per day,

N(t) = \frac{50t}{t+4}

Number of components assembled by the experienced employee per day,

E(t) = \frac{70t}{t+3}

Difference in number of components assembled per day by experienced ane new employee

D(t)= E(t) - N(t)

D(t) = \frac{70t}{t+3}-\frac{50t}{t+4}

      = \frac{70t(t+4)-50t(t+3)}{(t+3)(t+4)}

      = \frac{70t^2+280t-50t^2-150t}{(t+3)(t+4)}

      = \frac{20t^2+130t}{(t+3)(t+4)}

      = \frac{10t(2t+13)}{(t+4)(t+3)}

Therefore, Option A will be the answer.

3 0
3 years ago
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