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3 years ago

Which is greater, 8.0 or 0.8? Explain .

Mathematics

1 answer:

VashaNatasha [74]3 years ago

6 0

8.0 is greater than 0.8 because 8.0 is a unit but 0.8 is lesser than uint

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Pls help...mhdusbdhsj

Karo-lina-s [1.5K]

Answer:

270

Step-by-step explanation:

-6 multiply 5 gives you -30

-3 multiply -3 gives you -9

-30 multiply 9 gives you

-270

4 0

2 years ago

There are 30boys in a sporting club 20of them play hockey and 15 play volley ball each boy plays atleast one of the true game.il

Fed [463]

5 boys play both hockey and volleyball (30-20-15=5)

7 0

2 years ago

A triangle has a hypotenuse of length 25 and one leg of length 24. What is the tangent of the larger acute angle of the triangle

cestrela7 [59]

Answer:

$\sin( \theta) = \frac{24}{25} \\ \theta = { \sin }^{ - 1} ( \frac{24}{25} ) \\ \theta = 73.7 \degree \\ \\ \tan( \theta) = \tan(73.7 \degree) \\ = 3.4$

8 0

2 years ago

Read 2 more answers

In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the

earnstyle [38]

Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

A = video components need repair within 1 year

B = electronic components need repair within 1 year

C = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events A, B and C are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

$=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278$

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

$=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277$

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

7 0

2 years ago

Mrs. White's students are making school spirit packs. They have 24 bumper stickers and 18 window clings. Every pack must have th

ExtremeBDS [4]

Answer:

6 packet.

Step-by-step explanation:

Given:

Mrs. White's students are making school spirit packs.

They have 24 bumper stickers and 18 window clings.

Every pack must have the same contents and there should be no leftover items.

Question asked:

What is the greatest number of packs they can make?

Solution:

To find the greatest number of packs could be made by 24 bumper stickers and 18 window clings by using same contents in each pack with no left over items, we will simply use Highest Common Factor.

Factor of 24 - 2, 3, 4, 6, 8, 12, 24

Factor of 18 - 2, 3, 6, 9, 18

Highest Common Factor is 6.

Thus, they can make 6 packet at most.

3 0

2 years ago