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2 years ago

A box contains ten cards labeled Q, R, S, T, U, V, W, X, Y, and Z. One card will be randomly chosen.

Mathematics

2 answers:

lina2011 [118]2 years ago

4 0

Answer:

2/5

Step-by-step explanation:

Picking 4 cards out of 10, and as a fraction, it would be 2/5

inessss [21]2 years ago

4 0

Answer:

2/5 chance

Step-by-step explanation:

Well there 10 letters and only 4 letters are between Q and t (Q,R,S,T) so 4/10 simplified is 2/5.

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A distributor buys packs of trading cards from a wholesaler for $0.80 each. The distributor marks up the cards by 100% before se

Llana [10]

Answer:

$1.60

$2.88

Step-by-step explanation:

To find the selling price of the distributor and the retailer, we first need to find how much the distributor sells each pack of cards.

To find the selling price we use the formula.

Selling Price = Cost + Markup

Cost = $0.80

Markup rate = 100% or 1

Selling Price = 0.80 + (0.80*1)

Selling Price = $1.60

So the distributor sells each pack of cards at $1.60 to the retailer.

Now to find the selling of the retailer, we need to use the selling price of the distributor.

Cost = $1.60

Markup rate = 80% or 0.80

Selling price = 1.60 + (1.60 * 0.80)

Selling price = $2.88

So the retailer sells each pack of cards at $2.88 to the customers.

8 0

2 years ago

Read 2 more answers

A website randomly selects among 10 products to discount each day. The color printer of interest to you is discounted today. Det

Varvara68 [4.7K]

Answer:

a) P = 0.039

b) The expected number of days is 10 days.

Step-by-step explanation:

The most appropiate distribution to use in this case is the geometric distribution, in order to calculate the probability of a success after k failure trials.

The probability of success, as each of the 10 products are assumed to have fair probabilities, is:

$p=1/10=0.1$

Then, the probability that our product is not selected any given day is:

$q=1-p=1-0.1=0.9$

a) The probability that exactly this product is selected exactly 10 days from now is the probability that is not selected (probbility q) for the next 9 days and selected (probability p) at the 10th day:

$P=q^9p^1=0.9^9\cdot0.1=0.3874\cdot0.1=0.039$

b) The expected number of days is calculated as:

$E(X)=\dfrac{1}{p}=\dfrac{1}{0.1}=10$

6 0

2 years ago

What triangle is 3,7,9

Marta_Voda [28]

9^ ? 3^2 + 7^2
81 ? 9 + 49
81 > 58

c^ > a^2 + b^2
answer is obtuse triangle

5 0

2 years ago

Maria studied the traffic trends in India. She found that the number of cars on the roads increases by 10% each year. If there w

Nitella [24]

Year 2 is 88 million and in year 3 it is 96.8 million. 96.8 million-88 million it makes 8.8 million. So the answer is 8.8 million people.

8 0

2 years ago

Read 2 more answers

Employees in the marketing department of a large regional restaurant chain are researching the amount of money that households i

kap26 [50]

Answer:

The z-score for the group "25 to 34" is 0.37 and the z-score for the group "45 to 54" is 0.25.

Step-by-step explanation:

The data provided is as follows:

25 to 34 45 to 54

1329 2268

1906 1965

2426 1149

1826 1591

1239 1682

1514 1851

1937 1367

1454 2158

Compute the mean and standard deviation for the group "25 to 34" as follows:

$\bar x=\frac{1}{n}\sum x=\frac{1}{8}\times [1329+1906+...+1454]=\frac{13631}{8}=1703.875\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{8-1}\times 1086710.875}=394.01$

Compute the z-score for the group "25 to 34" as follows:

$z=\frac{x-\bar x}{s}=\frac{1851-1703.875}{394.01}=0.3734\approx 0.37$

Compute the mean and standard deviation for the group "45 to 54" as follows:

$\bar x=\frac{1}{n}\sum x=\frac{1}{8}\times [2268+1965+...+2158]=\frac{14031}{8}=1753.875\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{8-1}\times 1028888.875}=383.39$

Compute the z-score for the group "45 to 54" as follows:

$z=\frac{x-\bar x}{s}=\frac{1851-1753.875}{383.39}=0.25333\approx 0.25$

Thus, the z-score for the group "25 to 34" is 0.37 and the z-score for the group "45 to 54" is 0.25.

8 0

2 years ago