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Degger [83]
3 years ago
13

What is the volume of the square pyramid shown below?

Mathematics
1 answer:
NemiM [27]3 years ago
4 0

Answer: C) 147cm^{3}

Step-by-step explanation:

First we need to calculate the area of the square base.

This is 7*7 which equals 49 cm squared.

Plug this value into the formula of a square pyramid which will be V = 49 * \frac{h}{3}, now plug in the h, which is 9 cm.

V = 49 * \frac{9}{3}

Now solve.

V = 147 cubic cm.

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A survey was conducted in the United Kingdom, where respondents were asked if they had a university degree. One question asked,
katovenus [111]

Answer:

z=\frac{0.121-0.0892}{\sqrt{0.101(1-0.101)(\frac{1}{373}+\frac{1}{639})}}=1.620  

p_v =2*P(Z>1.620)=0.105  

If we compare the p value and using any significance level for example \alpha=0.05 always p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the two proportions are not statistically different at 5% of significance

Step-by-step explanation:

Data given and notation  

X_{1}=45 represent the number of correct answers for university degree holders

X_{2}=57 represent the number of correct answers for university non-degree holders  

n_{1}=373 sample 1 selected

n_{2}=639 sample 2 selected

p_{1}=\frac{45}{373}=0.121 represent the proportion of correct answers for university degree holders  

p_{2}=\frac{57}{639}=0.0892 represent the proportion of correct answers for university non-degree holders  

z would represent the statistic (variable of interest)  

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

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the proportions are different between the two groups, the system of hypothesis would be:  

Null hypothesis:p_{1} = p_{2}  

Alternative hypothesis:p_{1} \neq p_{2}  

We need to apply a z test to compare proportions, and the statistic is given by:  

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{45+57}{373+639}=0.101

Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

z=\frac{0.121-0.0892}{\sqrt{0.101(1-0.101)(\frac{1}{373}+\frac{1}{639})}}=1.620  

Statistical decision

For this case we don't have a significance level provided \alpha, but we can calculate the p value for this test.  

Since is a one side test the p value would be:  

p_v =2*P(Z>1.620)=0.105  

If we compare the p value and using any significance level for example \alpha=0.05 always p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the two proportions are not statistically different at 5% of significance

7 0
2 years ago
An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect
Serggg [28]

Answer:

k = 1

P(x > 3y) = \frac{2}{3}

Step-by-step explanation:

Given

f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y  \leq x }  & { \text 0, { elsewhere. } } \end{array} \right.

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1

Where

{ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y  \leq x }

Substitute values for the interval of x and y respectively

So, we have:

\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1

Isolate k

k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1

Integrate y, leave x:

k \int\limits^2_{0} y {dx} \, [0,x/2]= 1

Substitute 0 and x/2 for y

k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1

k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1

Integrate x

k * \frac{x^2}{2*2} [0,2]= 1

k * \frac{x^2}{4} [0,2]= 1

Substitute 0 and 2 for x

k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1

k *[ \frac{4}{4} - \frac{0}{4} ]= 1

k *[ 1-0 ]= 1

k *[ 1]= 1

k = 1

Solving (b): P(x > 3y)

We have:

f(x,y) = k

Where k = 1

f(x,y) = 1

To find P(x > 3y), we use:

\int\limits^a_b \int\limits^a_b {f(x,y)}

So, we have:

P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy

P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy

P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0  dxdy

Integrate x leave y

P(x > 3y) = \int\limits^2_0  x [0,y/3]dy

Substitute 0 and y/3 for x

P(x > 3y) = \int\limits^2_0  [y/3 - 0]dy

P(x > 3y) = \int\limits^2_0  y/3\ dy

Integrate

P(x > 3y) = \frac{y^2}{2*3} [0,2]

P(x > 3y) = \frac{y^2}{6} [0,2]\\

Substitute 0 and 2 for y

P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}

P(x > 3y) = \frac{4}{6} -\frac{0}{6}

P(x > 3y) = \frac{4}{6}

P(x > 3y) = \frac{2}{3}

8 0
2 years ago
You go to the post office to mail 6 letters. The postage for each letter is 39¢. How much money will the post office charge you
Alborosie
The answer is $2.34 hope this helps

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3 years ago
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Answer:

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Step-by-step explanation:

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Answer:

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Decimal form-2.88443731

Step-by-step explanation:

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