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3 years ago

What is the answer to this problem 6(1-5m)

1 answer:

5 0

Apply the distributive property.6⋅1+6(−5m)

Multiply 66 by 11 to get 66.6+6(−5m)6+6(-5m)Multiply −5-5 by 66 to get −30-30.6−30m6-30mReorder 66 and −30m-30m.−30m+6
So your answer is −30m+6

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Construct a 99% confidence interval for the population mean, mu. Assume the population has a normal distribution. A group of 1

Zarrin [17]

Answer:

99% confidence interval for the population mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% 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 mean age of selected students = 22.4 years

s = sample standard deviation = 3.8 years

n = sample of students = 19

$\mu$ = population mean

Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.878 < $t_1_8$ < 2.878) = 0.99 {As the critical value of t at 18 degree of

freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.878) = 0.99

P( $-2.878 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.878 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ , $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ]

= [ $22.4 -2.878 \times {\frac{3.8}{\sqrt{19} }$ , $22.4 +2.878 \times {\frac{3.8}{\sqrt{19} }$ ]

= [19.891 , 24.909]

Therefore, 99% confidence interval for the population mean is [19.891 , 24.909].

6 0

3 years ago

Leonard uses 3 peach slices and 1 banana for his fruit parfaits. If he used 42 peach slices to make parfaits for his friends, ho

Naddika [18.5K]

Answer:ez 26

Step-by-step explanation: how do you not know this

5 0

3 years ago

Read 2 more answers

The resquest of the y'+6y=e^4t, y(0)=2

kotegsom [21]

This is a linear differential equation of first order. Solve this by integrating the coefficient of the y term and then raising e to the integrated coefficient to find the integrating factor, i.e. the integrating factor for this problem is e^(6x).
Multiplying both sides of the equation by the integrating factor: 

(y')e^(6x) + 6ye^(6x) = e^(12x) 

The left side is the derivative of ye^(6x), hence 

d/dx[ye^(6x)] = e^(12x) 

Integrating 

ye^(6x) = (1/12)e^(12x) + c where c is a constant 

y = (1/12)e^(6x) + ce^(-6x) 

Use the initial condition y(0)=-8 to find c: 

-8 = (1/12) + c 
c=-97/12 

Hence 

y = (1/12)e^(6x) - (97/12)e^(-6x)

7 0

3 years ago

Solve 4 over x minus 4 equals the quantity of x over x minus 4, minus four thirds for x and determine if the solution is extrane

Vinil7 [7]

I'm thinking this is what the problem looks like: $\frac{4}{x-4}= \frac{x}{x-4}- \frac{4}{3}$ . The first thing to do is to move the $\frac{x}{x-4}$ over to the other side because it has a common denominator with the other side. Doing that and at the same time combining them over their common denominator looks like this: $\frac{4-x}{x-4}= -\frac{4}{3}$ . The best way to solve for x now is to cross-multiply to get 3(4-x)=-4(x-4). Distributing through the parenthesis is 12 - 3x = -4x + 16. Solving for x gives us x = 4. Of course when we sub a 4 back in for x we get real problems, don't we? Dividing by zero breaks every rule in math that there ever was! So, yes, the solution is extraneous.

6 0

3 years ago

Need help for 18) 21) Solve each equation and check. Show all work please

Alexxandr [17]

18) (10)7d/10=35(10)
7d/7=350/7
d=50
21) 3/4 w=27
(4/3)3/4 w=27(4/3)
w=36
Hope this helps!

5 0

3 years ago