Write 3.01 (1 repeating) as a mixed fraction?

A liquid dietary product implies in its advertising that use of the product for one month results in an average weight loss of a

BigorU [14]

Answer:

Following are the responses to the given question:

Step-by-step explanation:

Please find the table in the attached file.

mean and standard deviation difference: $\bar{d}=\frac{\Sigma d}{n} =\frac{-4-6-.......-4-4}{8}=-4.125 \\\\S_d=\sqrt{\frac{\Sigma (d-\bar{d})^2 }{n-1}}=\sqrt{\frac{(-4 + 4.125)^2 +.......+(-4 +4.125)^2 }{8-1}}= 1.246$

For point a:

hypotheses are:

$H_0 : \mu_d \geq -3\\\\H_a : \mu_d < -3\\\\$

degree of freedom:

$df=n-1=8-1=7$

From t table, at $\alpha = 0.05$ , reject null hypothesis if $t$ .

test statistic:

$t=\frac{\bar{d}-\mu_d }{\frac{s_d}{\sqrt{d}}}=\frac{ -4.125- (-3)}{\frac{1.246}{ \sqrt{8}}} =-2.55$

because the $t=-2.553$ , removing the null assumption. Data promotes a food product manufacturer's assertion with a likelihood of Type 1 error of 0.05.

For point b:

From t table, at $\alpha =0.01$ , removing the null hypothesis if $t$ .

because $t=-2.553 >-2.908$ , fail to removing the null hypothesis.

The data do not help the foodstuff producer's point with the likelihood of a .01-type mistake.

For point c:

Hypotheses are:

$H_0: \mu_d \geq -5\\\\H_a: \mu_d < -5$

Degree of freedom:

$df=n-1=8-1=7$

From t table, at $\alpha =0.05$ , removing the null hypothesis if $t$ .

test statistic: $t=\frac{\bar{d}-\mu_d}{\frac{s_d}{\sqrt{n}}} =\frac{-4.125-(-5)}{\frac{1.246}{\sqrt{8}}}=1.986$

Since $t-1.986 >-1.895$ , The null hypothesis fails to reject. The results do not support the packaged food producer's claim with a Type 1 error probability of 0,05.

From t table, at $\alpha= 0.01$ , reject null hypothesis if $t$ .

Since $t=1.986>-2.998$ , fail to reject null hypothesis.

Data do not support the claim of the producer of the dietary product with the probability of Type 1 error of .01.

5 0

1 year ago

In a sample of 100 steel canisters, the mean wall thickness was 8.1 mm with a standard deviation of 0.5 mm. Someone says that th

Travka [436]

Answer:

This statement can be made with a level of confidence of 97.72%.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 8.1 mm

Standard Deviation, σ = 0.5 mm

Sample size, n = 100

We are given that the distribution of thickness is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling:

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.5}{\sqrt{100}} = 0.05$

P(mean thickness is less than 8.2 mm)

P(x < 8.2)

$P( x < 8.2)\\\\ = P( z < \displaystyle\frac{8.2 - 8.1}{0.05})\\\\ = P(z < 2)$

Calculation the value from standard normal z table, we have,

$P(x < 8.2) =0.9772 = 97.72\%$

This statement can be made with a level of confidence of 97.72%.

8 0

2 years ago

Jennifer can run 12 miles in 2 hours, and she can bike 33 miles in 3 hours. What is Jennifer’s biking speed?

anastassius [24]

Answer:

11 miles per hour

Step-by-step explanation:

The biking speed is found by taking the miles and dividing by the hours

33 miles/ 3 hours

11 miles per hour

3 0

2 years ago

SAT Writing scores are normally distributed with a mean of 491 and a standard deviation of 113.A university plans to send letter

Sholpan [36]

Answer:

$z=1.405$

And if we solve for a we got

$a=491 +1.405*113=649.765$

So the value of height that separates the bottom 92% of data from the top 8% is 649.765.

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

$X \sim N(491,113)$

Where $\mu=491$ and $\sigma=113$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.08$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.92 of the area on the left and 0.08 of the area on the right it's z=1.405. On this case P(Z<1.405)=0.92 and P(z>0.92)=0.08

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.405$

And if we solve for a we got

$a=491 +1.405*113=649.765$

So the value of height that separates the bottom 92% of data from the top 8% is 649.765.

6 0

1 year ago

Complete the square to rewrite y = x2 - 6x + 15 in vertex form. Then state

love history [14]

Since $x^2$ is the square of x and 6x is twice the product between x and 3, the second square must be 3 squared, i.e. 9.

So, if we think of 15 as 9+6, we have

$x^2-6x+9+6 = (x-3)^2+6$

Which is the required vertex form. This form tells us imediately that the vertex is the point (3,6).

Since the leading coefficient is 1, the parabola is facing upwards (it's U shaped), so the vertex is a minimum.

5 0

1 year ago