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How many degrees does the hour hand of a clock turn between 3 p.m and 7:30 p.m ?

BlackZzzverrR [31]

A clock has 12 hours on it.

360/12 = 30

Each hour is 30 degrees.

3pm to 7:30pm is 4.5 hours

Multiply total hours by degrees per hour:

4.5 x 30 = 135 degrees

3 0

2 years ago

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Suppose I have an urn with 9 balls: 4 green, 3 yellow and 2 white ones. I draw a ball from the urn repeatedly with replacement.

Mrrafil [7]

Answer:

$E(X_n)=\frac{2(n-1)}{27}$

$E(y)=\frac{14}{9}$

Step-by-step explanation:

From the question we are told that:

Sample size n=9

Number of Green $g=4$

Number of yellow $y=4$

Number of white $w=4$

Probability of Green Followed by yellow P(GY) ball

$P(GY)=\frac{4}{9}*\frac{3}{9}$

$P(GY)=\frac{4}{27}$

Generally the equations for when n is even is mathematically given by

$Probability of success P(S)=\frac{4}{27}$

$Probability of Failure P(F)=\frac{27-4}{27}$

$Probability of Failure P(F)=\frac{23}{27}$

Therefore

$E(X_n)=\frac{n}{2}*P$

$E(X_n)=\frac{n}{2}*\frac{4}{27}$

$E(X_n)=\frac{2n}{27}$

Generally the equations for when n is odd is mathematically given by

$\frac{n-1}{2}$

$E(X_n)=\frac{n-1}{2}*\frac{4}{27}$

$E(X_n)=\frac{2(n-1)}{27}$

b)

Probability of drawing white ball

$P(w)=\frac{2}{9}$

Therefore

$E(w)=\frac{1}{p}$

$E(w)=\frac{1}{\frac{2}{9}}$

$E(w)=\frac{9}{2}$

Therefore

$E(y)=[E(w)-1]\frac{4}{9}$

$E(y)=[\frac{9}{2}-1]\frac{4}{9}$

$E(y)=\frac{14}{9}$

5 0

2 years ago

5. The superintendent of the local school district claims that the children in her district are brighter, on average, than the g

anygoal [31]

Answer:

We conclude that children in district are brighter, on average, than the general population.

Step-by-step explanation:

We are given the following data set:

105, 109, 115, 112, 124, 115, 103, 110, 125, 99

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{1117}{10} = 111.7$

Sum of squares of differences = 642.1

$S.D = \sqrt{\frac{642.1}{49}} = 8.44$

We are given the following in the question:

Population mean, μ = 106

Sample mean, $\bar{x}$ = 111.7

Sample size, n = 10

Alpha, α = 0.05

Sample standard deviation, s = 8.44

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 106\\H_A: \mu > 106$

We use one-tailed(right) t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{111.7 - 106}{\frac{8.44}{\sqrt{10}} } = 2.135$

Now,

$t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833$

Since,

$t_{stat} > t_{critical}$

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

We conclude that children in district are brighter, on average, than the general population.

4 0

2 years ago

6.

mel-nik [20]

Answer:

-2

Step-by-step explanation:

maybe?

5 0

2 years ago

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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