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3 years ago

14

Consider the set {4, 5, 7, 7, 8, 8, 12, 13, 13, 15, 18, 20, 22, 24, 26, 27, 27, 37, 43}. What would be an appropriate interval t

o make a histogram with the data
A. 20
B. 10
C. 5
D. 1

1 answer:

adelina 88 [10]3 years ago

8 0

Proper interval would be 5

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A student bought a book for $7.50 and a pen.the total cost was $9.50.which of the following can be used to find the cost of the

zloy xaker [14]

D: because if you take 9.50-7.50=2 you would get p=price or the price of the pen.
Hope I helped=)

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3 years ago

Read 2 more answers

A segment has endpoints (-9,-20) and (14,12). What is the midpoint of this segment

Darya [45]

((-9+14)/2, (-20+12)/2)
(5/2,-8/2)
(5/2,-4)

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3 years ago

Stream Black Cult

riadik2000 [5.3K]

Sorry, but I cannot understand the first part of this question

Nor the second part it's not finished.

<h2>please change this so that i could help</h2><h2>:D</h2><h2 /><h2>From:</h2><h2>Kenny</h2>

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2 years ago

Please help me this is due in less than 45 minutes!

Natasha_Volkova [10]

Answer:

1.) It's 20th century painting

2.) 0.5 probability

Step-by-step explanation:

If the universal = 60

We need to first get the value of X. That is,

x (x - 2) + x + 2x + 8 + 10 = 60

First open the bracket

x^2 - 2x + x + 2x + 8 + 10 = 60

x^2 + x + 18 = 60

x^2 + x - 42 = 0

Factorise the above equation

x^2 + 7x - 6x - 42 = 0

x (x + 7) -6(x + 7) = 0

x = 6 or - 7

Since x can't be negative, so we will ignore -7

The value for T = 6(6 - 2) = 6×4 = 24

The value for B = 2(6) + 8 = 12 + 8 = 20

If a painting is chosen from random,

If it's from 20th century, the probability will be 34/60 = 0.567

If it's from British painting, the probability will be 30/60 = 0.5

We can therefore conclude that it's from 20th century painting since it has higher value of probability.

The the probability of choosing a British painting will be 30/60 = 0.5

5 0

3 years ago

What is the approximate area of the figure?

melisa1 [442]

Option D:

The approximate area of the figure is 109.6 square feet.

Solution:

The figure is splitted into three shapes.

One is rectangle and the other two is semi-circles.

Diameter of the semi-circle = 5 feet

Radius of the semi-circle = 5 ÷ 2 = 2.5 feet

Area of the semi-circle = $\frac{1}{2}\pi r^2$

$$=\frac{1}{2}\times 3.14\times (2.5)^2$

Area of the semi-circle = 9.8 square feet

Area of 2 semi-circles = 2 × 9.8

Area of 2 semi-circles = 19.6 square feet

Length of the rectangle = 18 feet

Width of the rectangle = 5 feet

Area of the rectangle = length × width

= 18 × 5

Area of the rectangle = 90 square feet

Area of the figure = Area of 2 semi-circles + Area of the rectangle

= 19.6 square feet + 90 square feet

= 109.6 square feet

The approximate area of the figure is 109.6 square feet.

Hence Option D is the correct answer.

4 0

3 years ago