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3 years ago

Simplify 4^3 x 4^5 a. 16x15 b. 16^8 c. 4^15 d. 4^8

Mathematics

1 answer:

Mrac [35]3 years ago

3 0

Answer:

4^8

Step-by-step explanation:

4^3 x 4^5 = 4^3+5 = 4^8 = 65536

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Julli [10]

Answer:

Step-by-step explanation:

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3 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