Since 100 cents = $1

Then to convert $20 to cents, all you have to do is multiply.

$20 × 100 cents

<em><u>=2000 cents. </u></em>

More examples to try:

Convert $0.10 dollars to cents.

$10 × 100 cents = 1000 cents.

Here is a guild for next time:

100 cents = $1
200 cents = $2
300 cents = $3
400 cents = $4
___ cents = $__

So.....now you can fill in the blanks by yourself. Tell me the answer if possible! :D

~Hope this helped!~

8 0

2 years ago

Find the midpoint of CD with endpoints C (6,-9) and D (0,15)

Nookie1986 [14]

Answer:

1.5

Step-by-step explanation:

6 0

2 years ago

Compute P(x) using the binomial probability formula. Then determine whether the normal distribution can be used to estimate thi

stich3 [128]

Answer:

The answers to the questions are;

A) P(X) where x = 12 is equal to 3.55 ×10⁻²

B) P(x) using the normal distribution is given by the probability distribution function and is equal to 3.453 ×10⁻²

C) The calculated probabilities of P(x=12) using the binomial probability formula and the probability distribution function differ by 9.7×10⁻⁴

D) The normal distribution be used to approximate this probability because a. because np(1-p) %u2265 ⇒ np(1-p) ≥ 10

E) c. the value of fi represents the expected proportion of observation less than or equal to the ith data value.

Step-by-step explanation:

To solve the question, we note that

n = 58

p = 0.3

x = 12

Here we have n·p = 17.4 and n·q = 40.6 both ≥ 5 that is the normal distribution can be used to estimate the probability

A) We are to use the binomial probability formula to find P(X)

Where x = 12, we have P(12) = ₅₈C₁₂×0.3¹²×0.7⁴⁶

= 891794789340×0.000000531441×7.49×10⁻⁸ = 3.55 ×10⁻²

B) Using the standard distribution table we have

the z score for x = 12 given as $z = \frac{x-\mu}{\sigma}$ = -1.55

From the normal distribution table, we have the probability that value is below 12 = .06057 while the normal probability distribution function which gives the probability of a number being 12 is given by

$\frac{1}{\sigma\sqrt{2 \pi } } e^{\frac{-(x-\mu)^2}{2\sigma^2} }$

where:

σ = Standard deviation = $\sqrt{npq} = \sqrt{np(1-p)} = \sqrt{58*0.3*(1-0.3)} = 3.48999$

μ = Sample mean = n·p = 58×0.3 =17.4

Therefore the probability density function is $\frac{1}{3.49\sqrt{2 \pi } } e^{\frac{-(12-17.4)^2}{2*3.49^2} } = 3.453*10^{-2}$

C) The probabilities differ by 3.55 ×10⁻² - 3.453 ×10⁻² = 9.7×10⁻⁴

D) The normal distribution be used to approximate this probability because n·p and n·p·(n-p) ≥ 5

E) f $_i$ represents the area under the curve towards left of the ith data observed in a normally distributed population.

Therefore, the value of fi represents the expected proportion of observation less than or equal to the ith data value

5 0

2 years ago