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3 years ago

Please answer question 4

Mathematics

1 answer:

olasank [31]3 years ago

3 0

#4)
cross multiply
2(8x+10) = 4 * 5x
16x + 20 = 20x
4x = 20
x = 5

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For which sample size (n) and sample proportion (p) can a normal curve be

Nonamiya [84]

Answer:

Option c.

Step-by-step explanation:

Using the normal curve to approximate a sampling distribution:

For a sample size n and a proportion n, the normal curve can be used if:

$np \geq 10$ and $n(1-p) \geq 10$

Option a:

$np = 35*0.8 = 28 > 10$

$n(1-p) = 35*0.2 = 7 < 10$

So option a cannot be used.

Option b:

$np = 65*0.9 = 58.5 > 10$

$n(1-p) = 65*0.1 = 6.5 < 10$

So option b cannot be used.

Option c:

$np = 65*0.8 = 52 > 10$

$n(1-p) = 65*0.2 = 13 > 10$

So option c can be used, and is the answer

Option d:

$np = 35*0.9 = 31.5 > 10$

$n(1-p) = 35*0.1 = 3.5 < 10$

So option d cannot be used.

The answer is given by option c.

7 0

3 years ago

Population Growth Using 20th-century U.S. census data,

asambeis [7]

The population of Ohio will be 10 million in the year 1970

The population growth of Ohio is represented by the function:

$P(t)=\frac{12.79}{1+2.402e^{-}0.0309t}$

P is the population in millions

t is the number of years since 1900

To find the year that the population of Ohio will be 10 million, substitute P = 10 into the function P(t) and solve for t

$10=\frac{12.79}{1+2.402e^{-}0.0309t}\\\\10(1+2.402e^{-}0.0309t)=12.79\\\\1+2.402e^{-}0.0309t=\frac{12.79}{10} \\\\1+2.402e^{-}0.0309t=1.279\\\\2.402e^{-}0.0309t=1.279-1\\\\2.402e^{-}0.0309t = 0.279\\\\e^{-}0.0309t =\frac{0.279}{2.402} \\\\e^{-}0.0309t =0.116\\\\-0.0309t=ln0.116\\\\-0.0309t=-2.154\\\\t=\frac{-2.154}{-0.0309} \\\\t=69.7$