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3 years ago

What is the area of the trapezoid?

Mathematics

1 answer:

MA_775_DIABLO [31]3 years ago

7 0

Area of Trapezoid

$A = \dfrac{(B + b)h}{2}$

where B = larger base, b = smaller base, and h = height.

For this trapezoid, you have
B = 10 + 7 + 7 = 24
b = 10
h = 9

$A = \dfrac{(24 + 10) \times 9}{2}$

$A = 153$

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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.

7 0

3 years ago

How many two-digit numbers are there in which the tens digit is greater than<br> the ones digit?

Dovator [93]

there are 46 2-digit numbers in which the tens digit is greater than the ones digit.

<h3>How many 2-digit numbers are there in which the tens digit is larger than the ones digit?</h3>

A 2-digit number is written as:

a*10 + b

Where a and b are single-digit numbers.

Tens digit and ones digit refers to place values:

a is the tens digit, and b is the ones digit. The numbers with a larger tens digit than the ones digit are:

98, 97, 96, 95, 94, 93, 92, 91, 90, 87, 86, 85, 84, 83, 82, 81, 80, 76, 75, 75, 74, 73, 72, 71, 70, 65, 64, 63, 62, 61, 60, 54, 53, 52, 51, 50, 43, 42, 41, 40, 32, 31, 30, 21, 20, 10.

We can see that there are 46 numbers in that list, so we conclude that there are 46 2-digit numbers in which the tens digit is greater than the ones digit.

If you want to learn more about place values:

brainly.com/question/12386995

#SPJ1

7 0

1 year ago

Choose the equation for the graph

DIA [1.3K]

Look up on desmos it will give the correct answer

6 0

2 years ago

How to solve. C=3p+5 solve for c

Elenna [48]