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3 years ago

Find the least common multiple of 3x and 3(x-2).

Mathematics

1 answer:

larisa [96]3 years ago

7 0

Answer:

the least common multiple is 3 x ( x − 2 )

Step-by-step explanation:

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Just answer no need to explain.

Bogdan [553]

Shifting to left mean adding the number from each occurrence of x from the given function.

So, shifting x³ , 3 units to left will result in:

f(x) = (x + 3)³

Shifting down means subtracting the number from the function value.

So, shifting 2 units down will result in:

f(x) = (x + 3)³ - 2

Therefore, option A is the correct graph

8 0

3 years ago

Read 2 more answers

Area of sector................................

sveticcg [70]

Answer:

A ≈ 25.1 cm²

Step-by-step explanation:

the area (A) of the sector is calculated as

A = area of circle × fraction of circle

= πr² × $\frac{\frac{4\pi }{9} }{2\pi }$

= π × 6² × $\frac{\frac{4\pi }{9} }{2\pi }$

= 36π × $\frac{\frac{4\pi }{9} }{2\pi }$

= 18 × $\frac{4\pi }{9}$

= 2 × 4π

= 8π

≈ 25.1 cm² ( to the nearest tenth )

3 0

2 years ago

Their 15,000 kids in the multipurpose room.there is only 16 rows with 5 seats in each.how many kids will not have a seat to sit

sp2606 [1]

Answer: 14,920 kids

Step-by-step explanation:

15,000-(16×5)=x

15,000-80=x

14,920=x

8 0

3 years ago

Read 2 more answers

Tan^-1[cos(pi)] find the exact value without calculator

nata0808 [166]

Answer:

- $\frac{\pi }{4}$

Step-by-step explanation:

Evaluate cosπ then inverse tan, that is

cosπ = - 1

$tan^{-1}$ (- 1) = - $\frac{\pi }{4}$

4 0

4 years ago

A random sample of 500 registered voters in Phoenix is asked if they favor the use of oxygenated fuels year-round to reduce air

Stells [14]

Answer:

a) 0.0853

b) 0.0000

Step-by-step explanation:

Parameters given stated that;

H₀ : p = 0.6

H₁ : p = 0.6, this explains the acceptance region as;

p° ≤ $\frac{315}{500}$ =0.63 and the region region as p°>0.63 (where p° is known as the sample proportion)

a).

the probability of type I error if exactly 60% is calculated as :

∝ = P (Reject H₀ | H₀ is true)

= P (p°>0.63 | p=0.6)

where p° is represented as pI in the subsequent calculated steps below

= P $[\frac{p°-p}{\sqrt{\frac{p(1-p)}{n}}} >\frac{0.63-p}{\sqrt{\frac{p(1-p)}{n}}} |p=0.6]$

= P $[\frac{p°-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} >\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} ]$

= P $[Z>\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500} } } ]$

= P [Z > 1.37]

= 1 - P [Z ≤ 1.37]

= 1 - Ф (1.37)

= 1 - 0.914657 ( from Cumulative Standard Normal Distribution Table)

≅ 0.0853

The probability of Type II error β is stated as:

β = P (Accept H₀ | H₁ is true)

= P [p° ≤ 0.63 | p = 0.75]

where p° is represented as pI in the subsequent calculated steps below

= P $[\frac{p°-p} \sqrt{\frac{p(1-p)}{n} } }\leq \frac{0.63-p}{\sqrt{\frac{p(1-p)}{n} } } | p=0.75]$

= P $[\frac{p°-0.6} \sqrt{\frac{0.75(1-0.75)}{500} } }\leq \frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]$

= P $[Z\leq\frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]$

= P [Z ≤ -6.20]

= Ф (-6.20)

≅ 0.0000 (from Cumulative Standard Normal Distribution Table).

6 0

3 years ago