Write an equation in slope-intercept form of the line that passes through (−5, 19) and (5, 13)

A store had 3 packs of paper $0.93. How much would it cost if you were to buy 4 packs?

sesenic [268]

Hey you!

If I understood your question correctly, here's your answer:

0.93/3 = 0.31

0.31 × 4 = 1.24

So, it would cost $1.24 for 4 packs.

I Really Hope You Found This Helpful!

3 0

3 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

Angela's Uber cab fare is $12. A tax of 3% is added to the fare. she gives the drivers 20% of the original fare as a tip. How mu

Lisa [10]

Answer:

$14.80

Step-by-step explanation:

Tax:

3% of 12 is .4. So she would pay $0.40 for the tax.

Tip:

20% of 12 is 2.4. So she would pay $2.40 for the tip.

Sum:

Then you would add them all together, 12 + 0.4 + 2.4 = 14.8. So she payed $14.80 in all.

4 0

3 years ago

What is the rate of change and initial value for the linear relation that includes the points shown in the table?

uysha [10]

X 1 2 3
y 10 8 6
So find the slope m: (6-10)/(3-1) = -4/2 = -2
Going backwards when x = 0, y will be 12 (you can see this)
So
Initial value: 12, rate of change: −2

4 0

3 years ago

Which of the following equations is of a parabola with a vertex at (0, -5)

Harlamova29_29 [7]

The x, which is zero in this case will always be the number that goes on the inside of the equation, so far you have y=(x+0)^2 or y=x^2.

The y, which is negative will go on the end of the equation. So you get y=x^2-5

The equation is y=x^2-5 or C if it is multiple choice

4 0

3 years ago

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