Bag A contains 3 red and 2 white marbles

find the area of a circle with a diameter of four. Either enter an exact answer in terms of Pi or use 3.14 for pi as a decimal

Serggg [28]

Answer:

4 pi

Step-by-step explanation:

diameter = 2r

4 =2r

r =2

area = pi × (r)^2

= pi × 2^2

= 4 pi

3 0

2 years ago

In selecting a sulfur concrete for roadway construction in regions that experience heavy frost, it is important that the chosen

Lesechka [4]

Step-by-step explanation:

Given - In selecting a sulfur concrete for roadway construction in regions that experience heavy frost, it is important that the chosen concrete has a low value of thermal conductivity in order to minimize subsequent damage due to changing temperatures. Suppose two types of concrete, a graded aggregate and a no-fines aggregate, are being considered for a certain road. The table below summarizes data on thermal conductivity from an experiment carried out to compare the two types of concrete.

Type ni xi si

Graded 42 0.486 0.187

No-fines 42 0.359 0.158

To find - a. Formulate the above in terms of a hypothesis testing problem.

b. Give the test statistic and its reference distribution (under the null hypothesis).

c. Report the p-value of the test statistic and use it to assess the evidence that this sample provides on the scientific question of difference in mean conductivity of the two materials at the 5% level of significance.

Proof -

a.)

Hypothesis testing problem :

H0 : There is significant difference between mean conductivity for the graded concrete and mean conductivity for the no fines concrete.

H1 : There is no significant difference between mean conductivity for the graded concrete and mean conductivity for the no fines concrete.

b)

Test statistic :

$Z = \frac{x_{1} - x_{2} }{\sqrt{\frac{s_{1} ^{2} }{n_{1}} + \frac{s_{2}^{2} }{n_{2}} } }$

$Z = \frac{0.486 - 0.359 }{\sqrt{\frac{0.03496 }{42} + \frac{0.02496 }{42} } }$

$Z = \frac{0.127 }{\sqrt{0.001468}}$

$Z = \frac{0.127 }{0.0377}$

⇒Z(cal) = 3.3687

Z(tab) = 1.96

As Z (cal) > Z(tab)

So, we reject H0 at 5% Level of significance

p-value = 0.99962

Hence

There is significant difference in mean conductivity at the two materials.

5 0

2 years ago

Sarah claims that the thickness of the spearmint gum she produces is 7.5 one-hundredths of an inch. A quality control specialist

satela [25.4K]

Answer:

$t=\frac{7.55-7.5}{\frac{0.103}{\sqrt{10}}}=1.539$

$p_v =2*P(t_{(9)}>1.539)=0.158$

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis.

Step-by-step explanation:

Data given and notation

Data: 7.65, 7.60, 7.65, 7.7, 7.55, 7.55, 7.4, 7.4, 7.5, 7.5

We can begin calculating the sample mean given by:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And the sample deviation given by:

$s=\sart{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And we got:

$\bar X=7.55$ represent the sample mean

$s=0.103$ represent the sample standard deviation

$n=10$ sample size

$\mu_o =7.5$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal to 7.5 or no, the system of hypothesis would be:

Null hypothesis: $\mu = 7.5$

Alternative hypothesis: $\mu \neq 7.5$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{7.55-7.5}{\frac{0.103}{\sqrt{10}}}=1.539$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=10-1=9$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(9)}>1.539)=0.158$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis.

3 0

3 years ago

PLEEEASEE HELP <8888

4vir4ik [10]

Answer:

D. 56

Step-by-step explanation:

4 0

2 years ago

the bases of a frustum of a cone have radii 1 1/3 and 2. The height of the frustum is 2. What were the dimension of the original

Sedbober [7]

One of the major advantage of the two-condition experiment has to do with interpreting the results of the study. Correct scientific methodology does not often allow an investigator to use previously acquired population data when conducting an experiment. For example, in the illustrative problem involving early speaking in children, we used a population mean value of 13.0 months. How do we really know the mean is 13.0 months? Suppose the figures were collected 3 to 5 years before performing the experiment. How do we know that infants haven’t changed over those years? And what about the conditions under which the population data were collected? Were they the same as in the experiment? Isn’t it possible that the people collecting the population data were not as motivated as the experimenter and, hence, were not as careful in collecting the data? Just how were the data collected? By being on hand at the moment that the child spoke the first word? Quite unlikely. The data probably were collected by asking parents when their children first spoke. How accurate, then, is the population mean?

7 0

2 years ago

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