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3 years ago

Shop only had 2 types of stamps left using

Mathematics

1 answer:

saveliy_v [14]3 years ago

7 0

Answer:

{3¢, 28¢} or {4¢, 19¢} or {7¢, 10¢}

Step-by-step explanation:

54 = 1×54 = 2×27 = 3×18 = 6×9

Possible values of the stamps are 1 more than the values of a pair of factors. Of course, a 2¢ and 55¢ stamp will not permit paying 54¢ in postage, so that combination won't work. However, other pairs that will work are ...

3¢ and 28¢
4¢ and 19¢
7¢ and 10¢

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Plz help me

KiRa [710]

Answer:

$\left(-4,-2\right)$ , $\left(4,-2\right)$ , $\left(4,4\right)$ , and $\left(-4,4\r the perimeter and the area of the secret chamber

Step-by-step explanation:

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3 years ago

34% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem, we have that:

10 student are sampled, so $n = 10$

34% of college students say they use credit cards because of the rewards program, so $\pi = 0.34$

(a) exactly two

This is P(X = 2).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157$

$P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162$

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573$

$P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320$

$P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82$

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0

3 years ago

The definition of parallel lines requires the undefined terms line and plane, while the definition of perpendicular lines requir

Tomtit [17]

Point, line, and plane are the undefined expression that relinquish the starting location for geometry. When we define words, we ordinarily use simpler words, and these simpler words are in turn defined using yet simpler words. This procedure must eventually abort; at some stage, the definition must use a word whose meaning is accepted as intuitively clear. Because that meaning is accepted without definition, we refer to these words as undefined terms. These terms will be used in defining other terms. Although these expressions are not formally defined, a brief intuitive dialogue is needed. A point is the most fundamental object in geometry. It is represented by a dot and named by a capital letter. A point constitute position only. A line (straight line) can be thought of as a connected set of infinitely many points. It extends infinitely far in two opposite directions. A line has boundless length, zero width, and zero height. Any two points on the line name it. The symbol ↔ written on top of two letters is used to denote that line. <span>A plane may be contemplating as an infinite set of points creating a connected flat surface extending infinitely far in all directions. It is usually represented in drawings by a four‐sided figure. A single capital letter is used to designate a plane.</span>

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3 years ago

Read 2 more answers

Simplify the given expression. Assume that no variable equal 0.

marissa [1.9K]

Answer:

The answer I think is correct is B

Step-by-step explanation:

I hope that's corrected?

5 0

2 years ago

According to the United States Health and Human Services, the mean height for Americans is 1.757 meters for men and 1.618 meters