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alexandr402 [8]
2 years ago
10

Can someone please help??

Mathematics
1 answer:
pshichka [43]2 years ago
5 0
Hmmm i dont dnow orry
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Keeping the properties of exponents in mind, to what power must you raise the expression 7 1/2 to get 7 as a result
zhenek [66]
For this case we have the following expression:
 7 ^ {(1/2)}

 Rewriting the expression we have:
 (7 ^ {(1/2)}) ^ x

 Where,
 x: exponent to which we must raise the expression to obtain 7 as a result.
 We have then:
 7 ^ {((1/2) x)}

 The exponent must be equal to 1:
 (1/2) x = 1

 Clearing x we have: 
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 Substituting valres:
 7 ^ {((1/2) 2)}

7 ^ 1

7
 Answer:
 
you must raise the expression to 2
7 0
3 years ago
Solve the system of linear equations by adding or subtracting. Show all work!
Thepotemich [5.8K]

Answer:

x = -2, y = -2

Step-by-step explanation:

Your goal is to try and cancel out a variable. I want to get rid of y so i subtracted the first equation from the second one.

After that I solve for x and got x=-2.

I used x=-2 and plugged it back into either of the equation to solve for y.

4 0
3 years ago
A survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 sen
tatyana61 [14]

Answer:

96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

Step-by-step explanation:

We are given that a survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 seniors, 55 was the  average desired retirement age, with a standard deviation of 3.4 years.

Firstly, the Pivotal quantity for 96% confidence interval for the population mean is given by;

                         P.Q. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample average desired retirement age = 55 years

            \sigma = sample standard deviation = 3.4 years

            n = sample of seniors = 101

            \mu = true mean retirement age of all college students

<em>Here for constructing 96% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

<u>So, 96% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-2.114 < t_1_0_0 < 2.114) = 0.96  {As the critical value of t at 100 degree

                                               of freedom are -2.114 & 2.114 with P = 2%}  

P(-2.114 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.114) = 0.96

P( -2.114 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.114 \times {\frac{s}{\sqrt{n} } } ) = 0.96

P( \bar X-2.114 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.114 \times {\frac{s}{\sqrt{n} } } ) = 0.96

<u>96% confidence interval for</u> \mu = [ \bar X-2.114 \times {\frac{s}{\sqrt{n} } } , \bar X+2.114 \times {\frac{s}{\sqrt{n} } } ]

                                           = [ 55-2.114 \times {\frac{3.4}{\sqrt{101} } } , 55+2.114 \times {\frac{3.4}{\sqrt{101} } } ]

                                           = [54.30 , 55.70]

Therefore, 96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

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