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3 years ago

Estimate the sum9/50 + 7/15A.0B.1/2C.1

1 answer:

vladimir1956 [14]3 years ago

3 0

9/50 is close to 10/50 or 1/5 or 0.2
7/15 is close to 7/14 or 1/2 or 0.5
0.2 + 0.5 = 0.7
0.7 is closer to 0.5 than 1
The answer would be B

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Can anyone help me with 24,27,30 and 33 please

mamaluj [8]

24) 9/19

27 is a bit tricky for me.. sorry ):

30) On the 7th it will be 1/120
33) C

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3 years ago

The scale on a map is 1 cm: 7 km. If two cities are 15 cm apart on the map, what is the actual distance between the cities?

Lostsunrise [7]

Answer: C. 105 km

Step-by-step explanation:

By finding out that 1 cm is equal to 7 km, and that the two cities are 15 cm apart you can find it out by doing 7 x 15.

Enjoy. :)

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3 years ago

Read 2 more answers

A rancher has 300 feet of fencing to enclose a pasture bordered on one side by a river. the river side of the pasture needs no f

Mashcka [7]

I think your best answer is a pasture that is 75 feet by 150 feet, which will give you 11,250 sq ft of pasture

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3 years ago

Let X be a set of size 20 and A CX be of size 10. (a) How many sets B are there that satisfy A Ç B Ç X? (b) How many sets B are

Svetlanka [38]

Answer:

(a) Number of sets B given that

A⊆B⊆C: 2¹⁰. (That is: A is a subset of B, B is a subset of C. B might be equal to C)
A⊂B⊂C: 2¹⁰ - 2. (That is: A is a proper subset of B, B is a proper subset of C. B≠C)

(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.

Step-by-step explanation:

Let $x_1, x_2, \cdots, x_{20}$ denote the 20 elements of set X.

Let $x_1, x_2, \cdots, x_{10}$ denote elements of set X that are also part of set A.

For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain $x_1, x_2, \cdots, x_{10}$ .

For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from $x_1, x_2, \cdots, x_{20}$ .

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for set B.

In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves $2^{10} -2 = 1024 - 2 = 1022$ possibilities.

Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.

Start by considering the case when set A and set B are indeed disjoint.

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for a set B that is disjoint with set A.

There are 20 elements in X so that's $2^{20} = 1048576$ possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only $2^{20} - 2^{10}$ possibilities left.

5 0

3 years ago

Four friends bought 2 pizzas and 4 large soft drinks. They shared the cost equally. Each pizza cost

valina [46]

Answer:

$7.85

Step-by-step explanation:

2 pizzas at 11.99=23.98

4 soft drinks at 1.85=$7.40

23.98+7.40=31.38

31.38/4=7.845

Each friend pays $7.845, or $7.85 .........

7 0

2 years ago