Find the mean, median, and mode of the data. Which reflects the best measure of the center? 43, 38, 37, 57, 57, 58, 45

This is the call plan for Robert's cell phone:

butalik [34]

To answer this, first make an equation representing the data:

15+ 0.20X= Bill

now plug in data you know:

15 + 0.20(30)= bill

and finally solve it:
Bill= $21.00

4 0

3 years ago

Read 2 more answers

You have 6 pints of glaze. It takes 7/8 of a pint to glaze a bowl and 9/16 of a pint to glaze a plate.

son4ous [18]

Answer:

a. How many bowls could you glaze?

6 bowls

How many plates could you glaze?

10 plates

b. You want to glaze 5 bowls, and then use the rest for plates.

How many plates can you glaze?

2 plates

How much glaze will be left over?

8/9 of glaze would be left

c. How many of each object could you glaze so that there is no glaze left over? Explain how you found your answer.

4 4/23 bowls and 4 4/23 plates

Step-by-step explanation:

We are told in the question

Number of pints of glaze = 6 pints

Number of pints to glaze a bowl= 7/8 of pint

Number of pints to glaze a plate = 9/16 of a pint

a.

How many bowls could you glaze?

7/8 pint = 1 bowl

6 pints = x

Cross Multiply

x = 6 pints/ 7/8

= 6 × 8/7

= 48/7

= 6.8571428571 bowls

We can only glaze whole numbers of a bowl, hence, we can only glaze 6 bowls

How many plates could you glaze?

9/16 pint = 1 bowl

6 pints = y

y = 6 pints/ 9/16

= 6 × 16/9

= 10.666666667 plates

We can only glaze whole number of a plate, hence, we can only glaze 10 plates

b. You want to glaze 5 bowls, and then use the rest for plates.

1 bowl = 7/8 pints

5 bowls = z

Cross Multiply

z = 5 × 7/8pints

= 4 3/8 pints of glaze would be used for bowls

How many plates can you glaze?

The rest is for plates, hence:

The amount of glaze left for plates is calculated as:

6 pints - 4 3/8

1 5/8 pints of glaze would be left over to glaze plates.

So therefore,

9/16 pints = 1 plate

1 5/8 pints =

= 2 8/9

Hence we can only glaze 2 plates

How much glaze will be left over?

2 8/9 pints - 2 pints

= 8/9 pints of glaze.

c. How many of each object could you glaze so that there is no glaze left over?

We have 6 pints of glaze

Number of pints to glaze a bowl= 7/8 of pint

Number of pints to glaze a plate = 9/16 of a pint

Let the number of each objects be represented by x

7/8 × x + 9/16 × x = 6 pints

= 4 4/23 bowls and 4 4/23 plates

3 0

3 years ago

Which histogram represents a data set with a mean of 4.5 and a standard deviation of 1.7?

AVprozaik [17]

Answer:

that would be the third one

Step-by-step explanation:

5 0

3 years ago

Is 6/6 more than 3/3

hodyreva [135]

They are both equivalent to 1, so they are equal.

4 0

3 years ago

Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your

Veronika [31]

The expression of integral as a limit of Riemann sums of given integral $\int\limits^5_b {1} \, x/(2+x^{3}) dx$ is 4 $\lim_{n \to \infty}$ ∑ $n(n+4i)/2n^{3}+(n+4i)^{3}$ from i=1 to i=n.

Given an integral $\int\limits^5_b {1} \, x/(2+x^{3}) dx$ .

We are required to express the integral as a limit of Riemann sums.

An integral basically assigns numbers to functions in a way that describes displacement, area, volume, and other concepts that arise by combining infinite data.

A Riemann sum is basically a certain kind of approximation of an integral by a finite sum.

Using Riemann sums, we have :

$\int\limits^b_a {f(x)} \, dx$ = $\lim_{n \to \infty}$ ∑f(a+iΔx)Δx ,here Δx=(b-a)/n

$\int\limits^5_1 {x/(2+x^{3}) } \, dx$ =f(x)= $x/2+x^{3}$

⇒Δx=(5-1)/n=4/n

f(a+iΔx)=f(1+4i/n)

f(1+4i/n)= $[n^{2}(n+4i)]/2n^{3}+(n+4i)^{3}$

$\lim_{n \to \infty}$ ∑f(a+iΔx)Δx=

$\lim_{n \to \infty}$ ∑ $n^{2}(n+4i)/2n^{3}+(n+4i)^{3}4/n$

=4 $\lim_{n \to \infty}$ ∑ $n(n+4i)/2n^{3}+(n+4i)^{3}$

Hence the expression of integral as a limit of Riemann sums of given integral $\int\limits^5_b {1} \, x/(2+x^{3}) dx$ is 4 $\lim_{n \to \infty}$ ∑ $n(n+4i)/2n^{3}+(n+4i)^{3}$ from i=1 to i=n.

Learn more about integral at brainly.com/question/27419605

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5 0

2 years ago