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2 years ago

9

The opening balance of the May billing cycle for Marco's credit card was $3659. If he makes a new purchase of $100 on the 20th o

f May and doesn't make any payments, what is his average daily balance?

2 answers:

Sonbull [250]2 years ago

7 0

it is actually $3,694.48

VladimirAG [237]2 years ago

4 0

Answer:

Apex answer is 3,694.48

Step-by-step explanation:

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The diagram represents a difference of squares. A 2-column table with 2 rows. The labels for the columns and rows are blank. Fir

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I think its fisrts one

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2 years ago

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Suppose that 20% of the adult women in the United States dye or highlight their hair. We would like to know the probability that

Rasek [7]

Answer:

71.08% probability that pˆ takes a value between 0.17 and 0.23.

Step-by-step explanation:

We use the binomial approxiation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$p = 0.2, n = 200$ . So

$\mu = E(X) = np = 200*0.2 = 40$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.2*0.8} = 5.66$

In other words, find probability that pˆ takes a value between 0.17 and 0.23.

This probability is the pvalue of Z when X = 200*0.23 = 46 subtracted by the pvalue of Z when X = 200*0.17 = 34. So

X = 46

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{46 - 40}{5.66}$

$Z = 1.06$

$Z = 1.06$ has a pvalue of 0.8554

X = 34

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{34 - 40}{5.66}$

$Z = -1.06$

$Z = -1.06$ has a pvalue of 0.1446

0.8554 - 0.1446 = 0.7108

71.08% probability that pˆ takes a value between 0.17 and 0.23.

6 0

2 years ago

On Friday there were 8 snowboarders for every 3 skiers at Slovak ski resort. On Saturday the ratio of the number of snowboarders

atroni [7]

Answer:

There are 46 more skiers than snowboarder

Step-by-step explanation:

Given

Ratio of Snowboarders to Skiers

On Friday:

$Ratio = 8 : 3$

On Saturday:

$Ratio = 4 : 7$

Population = 168

Required

Determine the difference in the number of skiers and snowboarders on Saturday

On Saturday, we have

$Ratio = 4 : 7$

Calculate Total

$Total = 4 + 7$

$Total = 11$

Calculate the number of skiers

$Skiers = \frac{7}{11} * 168$

$Skiers = \frac{1176}{11}$

$Skiers = 107$ (Approximated)

Calculate the number of snowboarders

$Snowboarders = \frac{4}{11} * 168$

$Snowboarders = \frac{672}{11}$

$Snowboarders = 61$ (Approximated)

Calculate the difference

$Difference = 107 - 61$

$Difference = 46$

<em>Hence, there are 46 more skiers than snowboarder</em>

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2 years ago

Four trucks were used to make deliveries. The drivers recorded the number of miles driven and the amount of gasoline used. Truck

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Answer: Truck C

Step-by-step explanation: 14:1

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2 years ago

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6.8 Use the Normal approximation. Suppose we toss a fair coin 100 times. Use the Normal approximation to find the probability th

Maru [420]

Answer:

(a) The probability that proportion of heads is between 0.30 and 0.70 is 1.

(b) The probability that proportion of heads is between 0.40 and 0.65 is 0.9759.

Step-by-step explanation:

Let <em>X</em> = number of heads.

The probability that a head occurs in a toss of a coin is, <em>p</em> = 0.50.

The coin was tossed <em>n</em> = 100 times.

A random toss's result is independent of the other tosses.

The random variable <em>X</em> follows a Binomial distribution with parameters n = 100 and <em>p</em> = 0.50.

But the sample selected is too large and the probability of success is 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of $\hat p$ <em> </em>(sample proportion of <em>X</em>) if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

Check the conditions as follows:

$np=100\times 0.50=50>10\\n(1-p)=100\times (1-0.50)=50>10$

Thus, a Normal approximation to binomial can be applied.

So, $\hat p\sim N(p,\ \frac{p(1-p)}{n})$

$\mu_{p}=p=0.50\\\sigma_{p}=\sqrt{\frac{p(1-p)}{n}}=0.05$

(a)

Compute the probability that proportion of heads is between 0.30 and 0.70 as follows:

$P(0.30$

$=P(-4$

Thus, the probability that proportion of heads is between 0.30 and 0.70 is 1.

(b)

Compute the probability that proportion of heads is between 0.40 and 0.65 as follows:

$P(0.40$

$=P(-2$

Thus, the probability that proportion of heads is between 0.40 and 0.65 is 0.9759.

6 0

2 years ago