One half of a number is 3 more than one-sixth the same number. what is the number?

The percentage of body fat of a random sample of 36 men aged 20 to 29 found a sample mean of 14.42. Find a 95% confidence interv

Rina8888 [55]

Answer:

$14.42-1.96\frac{6.95}{\sqrt{36}}=12.150$

$14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690$

So on this case the 95% confidence interval would be given by (12.150;16.690)

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".

$\bar X=14.42$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=6.95$ represent the population standard deviation

n=36 represent the sample size

Solution to the problem

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

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

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: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

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

$14.42-1.96\frac{6.95}{\sqrt{36}}=12.150$

$14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690$

So on this case the 95% confidence interval would be given by (12.150;16.690)

5 0

4 years ago

I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be

Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

$P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}$

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: $g'(-1)\,(1)$ since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: $g'(-1)= 7$ as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1

6 0

3 years ago

These are the first six terms of a sequence with a = 2:

SpyIntel [72]

Answer:

Step-by-step explanation:

98÷14

=7

686÷98

=7

4 802÷686

=7

33 614÷4 802

=7

Then the common ratio q for this sequence is 7

recursive formula : an+1 = q×an = ?

an= a1 × qⁿ⁻¹

=2×7ⁿ⁻¹

an+1 = q×an

= 7×(2×7ⁿ⁻¹)

= 2×7ⁿ

3 0

3 years ago

Read 2 more answers

Write an inequality that represents the graph.

Anika [276]

Answer:

$y - 2x - 1 = 0$

Step-by-step explanation:

the slope is :

$\tan( \alpha ) = 4 \div 2 = 2$

at a pt on graph , when

$x = 1 \\ y = 3 \\$

therefore, coordinates are (1,3)

The equation is :

$(y - 3) = slope \times (x - 1) \\ (y - 3) = 2 \times (x - 1) \\ y - 3 = 2x - 2 \\ y - 2x - 1 = 0$

5 0

3 years ago

For the formula F = 95K – 455.67, what is the value of K when F = 0?

Inessa [10]

Answer:

Step-by-step explanation:

<u>Equation Solving</u>

We are given the equation:

F = 95K - 455.67

We are required to find the value of K when F=0:

95K - 455.67 = 0

Adding 455.67:

95K = 455.67

Dividing by 95;

K = 4.8

4 0

3 years ago