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weqwewe [10]
3 years ago
11

18+27 write the sum of the numbers as the product of their GCF and another sum

Mathematics
1 answer:
Eva8 [605]3 years ago
3 0
Ghhfbdnxndnxnxnxxnxnxnbxbzbzbbxbx
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Three six-sided fair dice are rolled. The six sides of each die are numbered 1; 2; : : : ; 6. Let A be the event that the first
valkas [14]

Answer:

1.) [A, B, C]

2.) [A', B', C']

3.) [A', B, C] or [A, B', C] or [A, B, C'] or [ A', B', C] or [A', B, C'] or [A, B', C'] or [A', B', C']

4.) A', B', C] or [A', B, C'] or [A, B', C'] or [A', B, C] or [A, B', C] or [A, B, C'] or [A, B, C]

Step-by-step explanation:

If A is the event that the first die shows an even number, then A' is the event that first die DOES NOT show an even number.

If B is the event that the Second die shows and even number, then B' is the event that second die DOES NOT show an even number.

If C is the event that the third die shows an even number, then C' is the event that third die DOES NOT show an even number.

Hence, our possible events are A, A', B, B', C, C'.

For question 1, in the event that all three dice show and even number, the expression of the event = [A, B, C]

For question two, in the event that no die shows an even number, the expression of the event = [A', B', C']

For question 3,In the event that at least one die shows an odd number, the expression of the event = [A', B, C] or [A, B', C] or [A, B, C'] or [ A', B', C] or [A', B, C'] or [A, B', C'] or [A', B', C']

For question 4,in the event that at most two dice show odd numbers, the expression of the event = [A', B', C] or [A', B, C'] or [A, B', C'] or [A', B, C] or [A, B', C] or [A, B, C'] or [A, B, C]

6 0
2 years ago
A circle has a diameter of 20 ft. What is the diameter? *
fenix001 [56]

Answer:

20ft

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
A jewelry company prints a hidden logo watermark. The watermark is a chord that is 0.8 cm from the center of a circular ring tha
Sergeeva-Olga [200]

Answer:

length of chord is 6cm

Step-by-step explanation:

Here, we are to calculate the length of the chord.

It should be understood that the chord has a length of 0.8cm from the center of the circle of radius 3cm, thereby forming two right-angled triangles with the radius 3cm being the hypotenuse of each and 0.8cm being the height of each.

Now, the chord is divided into 2 by this height dropping from the center of the circle. To calculate the first half, we use Pythagoras’ theorem with 3cm being hypotenuse and 0.8cm being the other side.

mathematically;

3^2 = 0.8^2 + l^2

9 = 0.64 + l^2

l^2 = 9-0.64

l^2 = 8.36

l = √(8.36)

l = 2.89 approximately

The length of the chord would be 2l = 2 * 2.89 = 5.78 cm which is 6cm to the nearest length

5 0
2 years ago
Find the measure of each acute angle in a right triangle where the measure of one acute angle is 8 times the measure of the othe
Arlecino [84]

Answer:

10° and 80°

Step-by-step explanation:

All triangles have internal angles that always add up to 180°.

8 0
2 years ago
A company is making a new label for one of their containers. The container is a cylinder that is 9 inches tall and 5 inches in d
nydimaria [60]

Answer:

180.55 in².

Step-by-step explanation:

Data obtained from the question include the following:

Height (h) = 9 in.

Diameter (d) = 5 in

Pi (π) = 3.14

Area of the label =..?

Next, we shall determine the radius.

Diameter (d) = 5 in

Radius (r) =.. ?

Radius (r) = Diameter (d) /2

r = d/2

r = 5/2

r = 2.5 in.

Next, we shall determine the area of the label that needs to be printed to go around the new container by calculating the surface area of the cylinder.

This is illustrated below:

Height (h) = 9 in.

Pi (π) = 3.14

Radius (r) = 2.5 in.

Surface Area (SA) =.?

SA = 2πrh + 2πr²

SA = (2×3.14×2.5×9) + (2×3.14×2.5²)

SA = 141.3 + 39.25

SA = 180.55 in²

The surface area of the cylinder is 180.55 in².

Therefore, the area of the label that needs to be printed to go around the new container is 180.55 in².

5 0
3 years ago
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