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2 years ago

Is 6/3 > 8/3?

Mathematics

1 answer:

Novay_Z [31]2 years ago

7 0

No 8/3 is greater because if you turn them into mixed numbers 6/3 would be 2 and 8/3 would be 2 1/2

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I need help with this someone please help!

Dafna1 [17]

Answer:

Step-by-step explanation:

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3 years ago

Your cell phone company charges $20 per month for 500 minutes plus an additional

Alex_Xolod [135]

Answer:

$21

Step-by-step explanation:

$20 for the 500 minutes

We still have to pay for 10 minutes.

10 cents per minute.

10centsx10 minutes.

100 cents, also known as 1 dollar.

$20 plus $1 means the answer is $21.

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-AxekickNebulite

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2 years ago

What is ¼(2.5 + c) (8) = 11

artcher [175]

The answer is this c= 3

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2 years ago

Read 2 more answers

Which of the following is the correct factorization of -y²+y+6?

SpyIntel [72]

Answer:

Step-by-step explanation:

I'm just going to factor it for you.

Factor −y^2+y+6

−y^2+y+6

=(−y−2)(y−3)

Answer:

(−y−2)(y−3)

6 0

2 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