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3 years ago

Convert 5 feet to inches. There are 12 inches in 1 foot.

Mathematics

2 answers:

cupoosta [38]3 years ago

7 0

Answer:D

Step-by-step explanation:

12=60

gregori [183]3 years ago

3 0

The answer is D, you have to do 12 x 5 your welcome

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This table shows a value that represents an exponential function what is the average rate of change for this function for the in

aliya0001 [1]

Answer:

Option A is correct, i.e. 12.

Step-by-step explanation:

Given is the table of x&y relationship which represents an exponential function.

The average rate of change for a function can be found using the following formula:-

F_average = { f(b) - f(a) } / (b-a)

Given a = 3 and b = 5.

From the table, f(3) = 8 and f(5) = 32.

So, F_average = (32-8)/(5-3)

F_average = 24/2 = 12.

Hence, option A is correct, i.e. 12.

4 0

2 years ago

The entry fee for a charity event is $15 per pair of people. The fee can be

Genrish500 [490]

Answer:

660

Step-by-step explanation:

First, to find out how much money they have collected so far, you need to add 7100 + 1300 which is 8400. 8400= $8.40. So now you know that they have collected 8.40, you need to subtract 1500-840 which = 660

Hope this helps!

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2 years ago

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Identify the range of the function shown in the graph.

dezoksy [38]

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2 years ago

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klemol [59]

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Step-by-step explanation:

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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