All categories

3 years ago

Suppose the function f(x) = 2x + 12

1 answer:

5 0

First figure out how much it costs to rent seven movies a month. f(7)=2(7) + 12–> 14 + 12–> $26 Now we need to find the difference between 26 and 10. 26 - 10= 16. So Casey needs $16 more to rent 7 movies a month

You might be interested in

I need help please thanks!

Natasha_Volkova [10]

Answer:

f=39

Step-by-step explanation:

hi! first, we can multiply by 12 on both sides to eliminate the 12 in the denominator.

f+3= (7/2)*12

f+3=84/2

f+3=42

now, subtract 3 on both sides.

f=39

7 0

3 years ago

Mr. Cobb is carpeting a 17.7 m long hallway. He has a piece of carpet 22 m long. How much does Mr. Cobb need to cut off the leng

Y_Kistochka [10]

Answer:5.7

Step-by-step explanation:

5 0

3 years ago

In the equation y=-x 4 what is the unit rate?

ioda

The unit rate is -1. because the unit rate has to have a denominator of 1.

4 0

3 years ago

A market research company conducted a survey to find the level of affluence in a city. They defined the category "affluence" for

malfutka [58]

Answer:

A 95% confidence interval for this population proportion is [0.081, 0.159].

Step-by-step explanation:

We are given that a market research company conducted a survey to find the level of affluence in a city.

Out of 267 persons who replied to their survey, 32 are considered affluent.

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 people who are considered affluent = $\frac{32}{267}$ = 0.12

n = sample of persons = 267

p = population proportion

Here for constructing a 95% confidence interval we have used 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

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

= [ $0.12-1.96 \times {\sqrt{\frac{0.12(1-0.12)}{267} } }$ , $0.12+1.96 \times {\sqrt{\frac{0.12(1-0.12)}{267} } }$ ]

= [0.081, 0.159]

Therefore, a 95% confidence interval for this population proportion is [0.081, 0.159].

4 0

3 years ago

What is the value of the expression 2x2 + 3x-y – 4y2 when x = 2 and y = –4?

sdas [7]

Write the problem: 2x^2+3x-y-4y^2
Substitute x and y: 2(2)^2+3(2)-(-4)-4(-4)^2
Exponents and Multiplication: 8+6+4-64
Add and Subtract: -46

Your Answer is -46

7 0

3 years ago