Elements of {1,2,3,4,6,12} and {4,6,8,10}

Hansika paid $9.25 for 2.8 pounds of pretzels. about how much did she pay for a pound of pretzels?

alekssr [168]

9.25 per 2.8 pound
9.25/2.8=3.303/1=$3.30 per pound

4 0

4 years ago

Read 2 more answers

Find the slope of the lines graphed below. Giving brainliest to whoever answers all correct.

Rashid [163]

Answer:

1.slope is 2

2.slope is -3/5

3.slope is 1

4.vertical line slope 0 undefined

5.x=-5

6.horizontal line slope 0 undefined

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

"Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given st

11111nata11111 [884]

Answer:

The error E = ± 4.04 %

Step-by-step explanation:

Solution:-

- The sample data is used to estimate the population proportion ( p ).

- The success p^ = success percentage = 40 %

- The confidence interval CI = 98%

- The sample size n = 800

- The margin of error E:

- The margin of error "E" for estimation of population proportion ( p ) is given by:

$E = z-critical*\sqrt{\frac{p~ ( 1 - p~ )}{n} }$

Where, Z-critical value is defined by the significance level:

P ( Z < Z-critical ) = α / 2

Where, α : Significance level

α = 1 - CI

P ( Z < Z-critical ) = (1 - 0.98) / 2

P ( Z < Z-critical ) = 0.01

Z-critical = 2.33

- The error E of estimation is:

$E = 2.33*\sqrt{\frac{0.4 ( 1 - 0.4 )}{800} }\\\\E = 0.04035$

- The error E = ± 4.04 %

6 0

3 years ago

An indoor track is made up of a rectangular region with two semi-circles at the ends. The distance around the track is 400 meter

dybincka [34]

Answer:

width of rectangle = 2R = (200/π) = 400/π meters

length of rectangle = 400 - π(200/π) = 400 - 200 = 200 meters

Step-by-step explanation:

The distance around the track (400 m) has two parts: one is the circumference of the circle and the other is twice the length of the rectangle.

Let L represent the length of the rectangle, and R the radius of one of the circular ends. Then the length of the track (the distance around it) is:

Total = circumference of the circle + twice the length of the rectangle, or

= 2πR + 2L = 400 (meters)

This equation is a 'constraint.' It simplifies to πR + L = 400. This equation can be solved for R if we wish to find L first, or for L if we wish to find R first. Solving for L, we get L = 400 - πR.

We wish to maximize the area of the rectangular region. That area is represented by A = L·W, which is equivalent here to A = L·2R = 2RL. We are to maximize this area by finding the correct R and L values.

We have already solved the constraint equation for L: L = 400 - πR. We can substitute this 400 - πR for L in

the area formula given above: A = L·2R = 2RL = 2R)(400 - πR). This product has the form of a quadratic: A = 800R - 2πR². Because the coefficient of R² is negative, the graph of this parabola opens down. We need to find the vertex of this parabola to obtain the value of R that maximizes the area of the rectangle:

-b ± √(b² - 4ac)

Using the quadratic formula, we get R = ------------------------

2a

-800 ± √(6400 - 4(0)) -1600

or, in this particular case, R = ------------------------------------- = ---------------

2(-2π)

-800

or R = ----------- = 200/π

-4π

and so L = 400 - πR (see work done above)

These are the dimensions that result in max area of the rectangle:

width of rectangle = 2R = (200/π) = 400/π meters

length of rectangle = 400 - π(200/π) = 400 - 200 = 200 meters

5 0

3 years ago

Help please i dont know this thank you<33

jok3333 [9.3K]

Answer:

42
35
16

Step-by-step explanation:

Put the value where the variable is and do the arithmetic. It can save some steps to simplify the expression first.

35 -c³ +8 = 43 -c³

c = 1

43 - 1³ = 43 -1 = 42

c = 2

43 - 2³ = 43 -8 = 35

c = 3

43 -3³ = 43 -27 = 16

The values that go in the blanks are 42, 35, 16.

_____

Additional comment

It does not take long to learn how to use a spreadsheet for evaluating the same formula with a number of different values of the variable(s). Graphing calculators can do this, too. It is always appropriate to use the right tool for the job.

Familiarity with multiplication and addition facts is a very good place to start. It is also useful to memorize the squares and cubes of small integers. The latter are needed here.

3 0

3 years ago