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3 years ago

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A 12 cm by 20 cm painting hangs in a museum. The museum curator wants to hang other paintings in the gallery that are proportion

al in size. The dimensions of several paintings are shown. Which paintings' dimensions are proportional to the original painting? Select all that apply. A. 1.5 cm by 2.5 cm B. 27 cm by 45 cm C. 9 cm by 15 cm D. 30 cm by 50 cm E. 6 cm by 14 cm

1 answer:

nata0808 [166]3 years ago

6 0

Answer:

A. 1.5 cm by 2.5 cm

B. 27 cm by 45 cm

C. 9 cm by 15 cm

D. 30 cm by 50 cm

Step-by-step explanation:

The dimensions of the original painting = 12 cm by 20 cm painting

Hence, the proportion is:

12/20 = 0.6

Hence, we compare the options given in the question.

A. 1.5 cm by 2.5 cm

= 1.5/2.5 = 0.6 , option A is correct

B. 27 cm by 45 cm

27cm/45 cm = 0.6 Option B is correct

C. 9 cm by 15 cm

9cm/15cm = 0.6 , Option C is correct

D. 30 cm by 50 cm

30 cm /50 cm = 0.6 Option D is correct

E. 6 cm by 14 cm

6cm/14 cm = 0.4285714286

Option E is not correct

Therefore, Options A to D are the correct options

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Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 24 students, she finds 2 w

irina1246 [14]

Answer:

A 95% confidence interval for the proportion of students who eat cauliflower on Jane's campus is [0.012, 0.270].

Step-by-step explanation:

We are given that Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 24 students, she finds 2 who eat cauliflower.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of students who eat cauliflower

n = sample of students

p = population proportion of students who eat cauliflower

<em>Here for constructing a 95% confidence interval we have used a One-sample z-test for proportions.</em>

<u>So, 95% confidence interval for the population proportion, p is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

Now, in Agresti and Coull's method; the sample size and the sample proportion is calculated as;

$n = n + Z^{2}__(\frac{_\alpha}{2})$

n = $24 + 1.96^{2}$ = 27.842

$\hat p = \frac{x+\frac{Z^{2}__(\frac{\alpha}{2}_) }{2} }{n}$ = $\hat p = \frac{2+\frac{1.96^{2} }{2} }{27.842}$ = 0.141

<u>95% confidence interval for p</u> = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.141 -1.96 \times {\sqrt{\frac{0.141(1-0.141)}{27.842} } }$ , $0.141 +1.96 \times {\sqrt{\frac{0.141(1-0.141)}{27.842} } }$ ]

= [0.012, 0.270]

Therefore, a 95% confidence interval for the proportion of students who eat cauliflower on Jane's campus [0.012, 0.270].

The interpretation of the above confidence interval is that we are 95% confident that the proportion of students who eat cauliflower on Jane's campus is between 0.012 and 0.270.

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3 years ago

What is the solution to the following system?

leva [86]

Answer:

(4,3,2)

Step-by-step explanation:

We can solve this via matrices, so the equations given can be written in matrix form as:

$\left[\begin{array}{cccc}3&2&1&20\\1&-4&-1&-10\\2&1&2&15\end{array}\right]$

Now I will shift rows to make my pivot point (top left) a 1 and so:

$\left[\begin{array}{cccc}1&-4&-1&-10\\2&1&2&15\\3&2&1&20\end{array}\right]$

Next I will come up with algorithms that can cancel out numbers where R1 means row 1, R2 means row 2 and R3 means row three therefore,

-2R1+R2=R2 , -3R1+R3=R3

$\left[\begin{array}{cccc}1&-4&-1&-10\\0&9&4&35\\0&14&4&50\end{array}\right]$

$\frac{R_2}{9}=R_2$

$\left[\begin{array}{cccc}1&-4&-1&-10\\0&1&\frac{4}{9}&\frac{35}{9}\\0&14&4&50\end{array}\right]$

4R2+R1=R1 , -14R2+R3=R3

$\left[\begin{array}{cccc}1&0&\frac{7}{9}&\frac{50}{9}\\0&1&\frac{4}{9}&\frac{35}{9}\\0&0&-\frac{20}{9}&-\frac{40}{9}\end{array}\right]$

$-\frac{9}{20}R_3=R_3$

$\left[\begin{array}{cccc}1&0&\frac{7}{9}&\frac{50}{9}\\0&1&\frac{4}{9}&\frac{35}{9}\\0&0&1&2\end{array}\right]$

$-\frac{4}{9}R_3+R_2=R2$ , $-\frac{7}{9}R_3+R_1=R_1$

$\left[\begin{array}{cccc}1&0&0&4\\0&1&0&3\\0&0&1&2\end{array}\right]$

Therefore the solution to the system of equations are (x,y,z) = (4,3,2)

Note: If answer choices are given, plug them in and see if you get what is "equal to". Meaning plug in 4 for x, 3 for y and 2 for z in the first equation and you should get 20, second equation -10 and third 15.

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3 years ago

Find the distance between points P(8, 2) and Q(3, 8) to the nearest tenth.

riadik2000 [5.3K]

<span>First of all to calculate the distance between two points we can use distance formula

d=Square Root [(x2-x1)^2 + (y2-y1)^2]

Now substitute the given points p(x1,y1) and q(x2,y2)in above distance formula

The values are X2=3, X1=8and Y2=8and Y1=2.

After Substituting the values

d=Square Root[(-5)^2+(6)^2]

d=Square Root(25+36]

d=Square Root[61]

d=7.8

7.8 is the distance between points P(8, 2) and Q(3, 8) to the nearest tenth.</span>

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Answer:

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Step-by-step explanation:

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