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Leviafan [203]
3 years ago
10

You need to solve a system of equations. You decide to use the elimination method. Which of these is not allowed?

Mathematics
1 answer:
Serggg [28]3 years ago
8 0

Answer:

B

Step-by-step explanation:

The <u>Elimination Method</u> is the method for solving a pair of linear equations which reduces one equation to one that has only a single variable.

  • If the coefficients of one variable are opposites, you add the equations to eliminate a variable, and then solve.
  • If the coefficients are not opposites, then we multiply one or both equations by a number to create opposite coefficients, and then add the equations to eliminate a variable and solve.

When multoplying the equation by a coefficient, we multiply both sides of the equation (multiplying both sides of the equation by some nonzero number does not change the solution).

So, option B is not allowed (it is not allowed to multiply only one part of equation)

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The proprietor of a boutique in New York wanted to determine the average age of his customers. A random sample of 25 customers r
jolli1 [7]

Answer:

28-2.064\frac{10}{\sqrt{25}}=23.872    

28+2.064\frac{10}{\sqrt{25}}=32.128    

So on this case the 95% confidence interval would be given by (23.9;32.1)  

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=28 represent the sample mean

\mu population mean (variable of interest)

s=10 represent the sample standard deviation

n=25 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=25-1=24

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,24)".And we see that t_{\alpha/2}=2.064

Now we have everything in order to replace into formula (1):

28-2.064\frac{10}{\sqrt{25}}=23.872    

28+2.064\frac{10}{\sqrt{25}}=32.128    

So on this case the 95% confidence interval would be given by (23.9;32.1)    

7 0
3 years ago
Units of Length
grin007 [14]

the answer is:

78.7 (i rounded it to the nearest tenth)

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A company that manufactures cell phones has done research to determine how the pricing of thier phones affects how many they sel
pantera1 [17]

B.Less than $200 or more than $500

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Question:<br><br> $9.27 for 48 oz
AnnZ [28]

Answer:

what all does it ask

Step-by-step explanation:

6 0
3 years ago
Assume the acceleration of the object is a(t) = −32 feet per second per second. (Neglect air resistance.) A balloon, rising vert
Liula [17]

Answer:

(a) 2.79 seconds after its release the bag will strike the ground.

(b) At a velocity of 73.28 ft/second it will hit the ground.

Step-by-step explanation:

We are given that a balloon, rising vertically with a velocity of 16 feet per second, releases a sandbag at the instant when the balloon is 80 feet above the ground.

Assume the acceleration of the object is a(t) = −32 feet per second.

(a) For finding the time it will take the bag to strike the ground after its release, we will use the following formula;

                        s=ut+\frac{1}{2} at^{2}

Here, s = distance of the balloon above the ground = - 80 feet

          u = intital velocity = 16 feet per second

          a = acceleration of the object = -32 feet per second

          t = required time

So,  s=ut+\frac{1}{2} at^{2}

      -80=(16\times t)+(\frac{1}{2} \times -32 \times t^{2})

       -80=16t-16 t^{2}

       16 t^{2} -16t -80 =0

          t^{2} -t -5 =0

Now, we will use the quadratic D formula for finding the value of t, i.e;

           t = \frac{-b\pm \sqrt{D } }{2a}

Here, a = 1, b = -1, and c = -5

Also, D = b^{2} -4ac = (-1)^{2} -(4 \times 1 \times -5) = 21

So,  t = \frac{-(-1)\pm \sqrt{21 } }{2(1)}

      t = \frac{1\pm \sqrt{21 } }{2}

We will neglect the negative value of t as time can't be negative, so;

t = \frac{1+ \sqrt{21 } }{2} = 2.79 ≈ 3 seconds.

Hence, after 3 seconds of its release, the bag will strike the ground.

(b) For finding the velocity at which it hit the ground, we will use the formula;

                       v=u+at

Here, v = final velocity

So,  v=16+(-32 \times 2.79)

      v = 16 - 89.28 = -73.28 feet per second.

Hence, the bag will hit the ground at a velocity of -73.28 ft/second.

8 0
3 years ago
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