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LenaWriter [7]
3 years ago
7

You spin the spinner below once: what is the P(getting a number less than 5)?

Mathematics
2 answers:
Alex_Xolod [135]3 years ago
8 0

Answer:

3/4

Step-by-step explanation:

wariber [46]3 years ago
6 0

Answer: it would be 3/4 hope this helps you!!!

Step-by-step explanation:

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True or False?
saveliy_v [14]

Answer:

Step-by-step explanation:

false

false false

3 0
2 years ago
All the factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48.<br><br><br> True or False
Gekata [30.6K]
Factors of 48 are 1,2,3,4,6,8,12,16,24,48.

True.
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Since 1990, the number of gallons of whole milk each person in the United States drinks has decreased 4.5% each year. In 1990, e
MariettaO [177]

Answer:

11.2 gallons

Step-by-step explanation:

by multiplying 16 by 0.965 (as it is a 4.5% decrease per year) and then multiplying the answer by 0.965 and so on until done a total of 10 times gets you the answer 11.2 gallons. there is a quicker way to do it but I cannot remember sorry

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Find sin q if q is an angle in standard position and the point with coordinates (4, 3) lies on the terminal side of the angle.
jolli1 [7]
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4 0
2 years ago
A heavy rope, 50 ft long, weighs 0.6 lb/ft and hangs over the edge of a building 120 ft high. Approximate the required work by a
Anastasy [175]

Answer:

Exercise (a)

The work done in pulling the rope to the top of the building is 750 lb·ft

Exercise (b)

The work done in pulling half the rope to the top of the building is 562.5 lb·ft

Step-by-step explanation:

Exercise (a)

The given parameters of the rope are;

The length of the rope = 50 ft.

The weight of the rope = 0.6 lb/ft.

The height of the building = 120 ft.

We have;

The work done in pulling a piece of the upper portion, ΔW₁ is given as follows;

ΔW₁ = 0.6Δx·x

The work done for the second half, ΔW₂, is given as follows;

ΔW₂ = 0.6Δx·x + 25×0.6 × 25 =  0.6Δx·x + 375

The total work done, W = W₁ + W₂ = 0.6Δx·x + 0.6Δx·x + 375

∴ We have;

W = 2 \times \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= 2 \times \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 750

The work done in pulling the rope to the top of the building, W = 750 lb·ft

Exercise (b)

The work done in pulling half the rope is given by W₂ as follows;

W_2 =  \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 562.5

The work done in pulling half the rope, W₂ = 562.5 lb·ft

6 0
2 years ago
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