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Viefleur [7K]
3 years ago
15

There are more answers please help

Mathematics
1 answer:
Dominik [7]3 years ago
3 0

No you type the numbers press = and poof

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A cylindrical container of five rubber balls has a height of 30 centimeters and a diameter of 6 centimeters. Each ball in the co
Artyom0805 [142]

Answer:

There is 170 cm³ free space in the container.

Step-by-step explanation:

Cylindrical container: height = 18 cm ; diameter = 6 cm.

3 balls each have a radius of 3 cm.

Volume of a cylinder = π r² h

V = 3.14 * (3cm)² * 18 cm

V = 508.68 cm³

Volume of rubber ball = 4/3  π r³

V = 4/3 * 3.14 * (3cm)³

V = 113.04 cm³

113.04 cm³  * 3 balls = 339.12 cm³

508.68 cm³ - 339.12 cm³ = 169.56 cm³ or 170 cm³

3 0
3 years ago
Which is bigger 2/3 1/8 or 3/4
sergeinik [125]

Answer:

3/4

Step-by-step explanation:

2/3= 0.6

1/8= 0.125

3/4= 0.75

<u>I convert the fractions into </u><u>decimals the decimal with the highest number in the tenths place is the biggest </u>

<u>Hope This Helps!</u>

6 0
4 years ago
Read 2 more answers
If a seed is planted, it has a 80% chance of growing into a healthy plant.
AlladinOne [14]

Answer:

0.336

Step-by-step explanation:

Use binomial probability:

P = nCr p^r q^(n-r)

where n is the number of trials,

r is the number of successes,

p is the probability of success,

and q is the probability of failure (1-p).

Here, n = 8, r = 7, p = 0.8, and q = 0.2.

P = ₈C₇ (0.8)⁷ (0.2)⁸⁻⁷

P = 0.336

3 0
3 years ago
What is the interquartile range (IQR) of the following data set 17 16 21 15 25 22 18 23 17
Pepsi [2]

IQR = 6

First locate the median Q_{2} at the centre of the data arranged in ascending order. Then locate the lower and upper quartiles Q_{1} and Q_{3} located at the centre of the data to the left and right of the median.

Note that if any of the above are not whole values then they are the average of the values either side of the centre.

rearrange data in ascending order

15 16 ↓17 17 18 21 22 ↓23 25

                     ↑

Q_{2} = 18

Q_{1} = \frac{16+17}{2} = 16.5

Q_{3} = \frac{22+23}{2} = 22.5

IQR = Q_{3} - Q_{1} = 22.5 - 16.5 = 6




4 0
3 years ago
Q11.
Kamila [148]

Answer:

Step-by-step explanation:

The area of a trapezium is given as;

Area = \frac{1}{2}(a + b)h

where: a is the length of the upper side, b is the length of its base and h is its height.

This implies that; a = x - 4, b = x + 5, h = 2x and area = 351

Thus,

351 =  \frac{1}{2}((x - 4) + (x + 50)) 2x

      = (x - 4 + x + 5) x

531 = (2x + 1) x

      = 2x^{2} + x

So that;

2x^{2} + x - 531 = 0

This gives a quadratic equation.

5 0
3 years ago
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