What percent increase is 1/3 to 1/6

Leokris [45]

I think it is 10 because 18×10=180 but it is close hope this helps

7 0

2 years ago

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x overbar, is found

joja [24]

Answer:

(a) 80% confidence interval for the population mean is [109.24 , 116.76].

(b) 80% confidence interval for the population mean is [109.86 , 116.14].

(c) 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed.

Step-by-step explanation:

We are given that a simple random sample of size n is drawn from a population that is normally distributed.

The sample mean is found to be 113 and the sample standard deviation is found to be 10.

(a) The sample size given is n = 13.

Firstly, the pivotal quantity for 80% 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 = 113

s = sample standard deviation = 10

n = sample size = 13

$\mu$ = population mean

Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 80% confidence interval for the population mean, $\mu$ is ;

P(-1.356 < $t_1_2$ < 1.356) = 0.80 {As the critical value of t at 12 degree

of freedom are -1.356 & 1.356 with P = 10%}

P(-1.356 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.356) = 0.80

P( $-1.356 \times }{\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.356 \times }{\frac{10}{\sqrt{13} } }$ , $113+1.356 \times }{\frac{10}{\sqrt{13} } }$ ]

= [109.24 , 116.76]

Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].

(b) Now, the sample size has been changed to 18, i.e; n = 18.

So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.

80% confidence interval for $\mu$ = [ $\bar X-1.333 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.333 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.333 \times }{\frac{10}{\sqrt{18} } }$ , $113+1.333 \times }{\frac{10}{\sqrt{18} } }$ ]

= [109.86 , 116.14]

Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].

(c) Now, we have to construct 98% confidence interval with sample size, n = 13.

So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.

98% confidence interval for $\mu$ = [ $\bar X-2.681 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+2.681 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-2.681 \times }{\frac{10}{\sqrt{13} } }$ , $113+2.681 \times }{\frac{10}{\sqrt{13} } }$ ]

= [105.56 , 120.44]

Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed because t test statistics is used only when the data follows normal distribution.

6 0

2 years ago

Need help Asap thank you in advance

julia-pushkina [17]

Answer:

$12x^4-20x^2+32x=4x(3x^3-5x+8)$

Step-by-step explanation:

Common Factors

An algebraic expression that is formed by sums or subtractions of terms can be factored provided there are numeric or variable common factors in all the terms.

The following expression

$Z=12x^4-20x^2+32x$

Can be factored in the constants and in the variable x.

1. To find the common factor of the variable, we must locate if the variable is present in all terms. If so, we take the common factor as the variable with an exponent which is the lowest of all the exponents found throughout the different terms. In this case, the lowest exponent is x (exponent 1).

2. To find the common factors of the constants, we take all the coefficients:

12 - 20 - 32

and find the greatest common divisor of them, i.e. the greatest number all the given numbers can be divided by. This number is 4, since 12/4=3, 20/4=5 and 32/4=8

3. The factored expression is

$Z=12x^4-20x^2+32x=4x(3x^3-5x+8)$

$\boxed{12x^4-20x^2+32x=4x(3x^3-5x+8)}$

3 0

2 years ago

In the triangle shown below, what is the approximate value of x?

Verdich [7]

Answer:

the answer the answer is the answer is C

Step-by-step explanation:

21 Unity

3 0

2 years ago

Read 2 more answers

Help me wegnerkolmp2741o

Pie

Answer:

Step-by-step explanation:

All these amount of money are based on Dan's initial amount, which is x. Ben had 7 more than Dan, so Ben's amount of money is x + 7. Kaden has 4 more than Ben, so Kaden's amount of money is 4 + (x + 7) which is x + 11. So here's the breakdown so far:

Dan: x

Ben: x + 7

Kaden: x + 11

Now it looks like Kaden is the favorite and got another 20 from his mom, so his amount became x + 31. After Dan spends a dollar, making his amount x - 1, Kaden had 9 times what Dan had. Now we have finally:

x + 31 = 9(x - 1) and

x + 31 = 9x - 9 and

40 = 8x so

x = 5.

Dan originally had $5, Ben originally had $12, and Kaden originally had $16

8 0

2 years ago