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1 year ago

Pls help me out with this

Mathematics

1 answer:

vampirchik [111]1 year ago

8 0

Answer:

Question 7: 20° and 160°, Question 8: 10° and 80°

Step-by-step explanation:

Complementary angles add to 90° and supplementary angles add to 180°.

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The median and interquartile range of the a set of data is shown. Write a set of data consisting of seven values for the pair of

OleMash [197]

You use the definition of median and interquartile range. The def. of a median is the number in the middle of the data set, so the number 6 would be in the middle. Making a box and whisker plot(shown below) with the data set separates the data set into 4 "quartiles" or parts. The interquartile range is the last line in the box(Q3) minus the first line in the box(Q1). So something minus somthing must equal 5.

Data set: 1,2,3,6,8,8,10
It doesn't matter what the other numbers are as long as the median is 6 and the interquartiel range is 6.

6 0

3 years ago

Not a hard question i just don't get it. I really need some help here, its the final

Greeley [361]

Start with

$\dfrac{1}{\sin^2(x)}=4$

Invert both sides:

$\sin^2(x) = \dfrac{1}{4}$

Consider the square root of both sides with double sign:

$\sin(x) = \pm\sqrt{\dfrac{1}{4}} = \pm\dfrac{1}{2}$

We have

$\sin(x) = \dfrac{1}{2} \iff x = \dfrac{\pi}{6}\ \lor\ x = \dfrac{5\pi}{6}$

and

$\sin(x) = -\dfrac{1}{2} \iff x = \dfrac{7\pi}{6}\ \lor\ x = \dfrac{11\pi}{6}$

5 0

4 years ago

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

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

So, 95% confidence interval for the population proportion, p is ;

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

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

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

= [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.

7 0

3 years ago

Yolanda has finished 45% of the homework problems she was assigned. So far she has finished 27 problems.

sveta [45]

$p\%=\frac{p}{100}\\\\A.\ 45\%=\frac{45}{100}=\frac{45:5}{100:5}=\boxed{\frac{9}{20}}\\\\B.\ \begin{array}{ccc}45\%&-&27\\100\%&-&x\end{array}\ \ \ \ |cross\ multiply\\\\\\45\cdot x=27\cdot100\\45x=2700\ \ \ \ |divide\ both\ sides\ by\ 45\\\boxed{x=60}\\\\C.\ 60-27=\boxed{33}$

7 0

4 years ago

A sporting goods store is having a 15% off sale on all items. Which functions can be used to find the sale price of an item that

hammer [34]

F(x) = x - 0.15x
sale = original - 0.15(original)
y = 0.85x

4 0

3 years ago