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andrew-mc [135]
3 years ago
6

Convert 2 7/8 into a decimal

Mathematics
2 answers:
rodikova [14]3 years ago
6 0

Answer:

the answer is 2.875 because if you make the fraction improper it becomes 23/8 and then you just divide it

AlekseyPX3 years ago
5 0

Answer:

2.875

Step-by-step explanation:

Brainliest Please!

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Use an algebraic approach to solve the problem.
qaws [65]

Answer:

Her first exam score was a 78, her second exam score was a 89, and his third exam score was a 94.

Step-by-step explanation:

Use the formula for the mean: sum of elements / number of elements

Let x represent her first exam score.

Her second exam score can be represented by x + 11, since it was 11 points better than her first.

Her third exam score can be represented by (x + 11) + 5, since it was 5 points better than her second.

Plug in all of these expressions into the mean formula. Plug in 87 as the mean, and plug in 3 as the number of elements (since there are 3 scores):

mean = sum of elements / number of elements

87 =  ( (x) + (x + 11) + (x + 11) + 5 ) / 3

Add like terms and solve for x:

87 = (3x + 27) / 3

261 = 3x + 27

234 = 3x

78 = x

So, her first score was a 78.

Find her second score by adding 11 to this:

78 + 11 = 89

Find her third score by adding 5 to the second score:

89 + 5 = 94

Her first exam score was a 78, her second exam score was a 89, and his third exam score was a 94.

6 0
3 years ago
Find the length of the missing side. The triangle is not drawn to scale
goblinko [34]

Answer:

MOnkey mode

Step-by-step explanation:

cause why not

5 0
3 years ago
Read 2 more answers
Let C(x) be the statement "x has a cat," let D(x) be the statement "x has a dog," and let F(x) be the statement "x has a ferret.
jek_recluse [69]

Answer:

\mathbf{a)} \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\\\mathbf{b)} \left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee \; F(x)\\\mathbf{c)} \left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)\\\mathbf{d)} \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)\\\mathbf{e)} \left((\exists x\in X)C(x) \right) \wedge  \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)

Step-by-step explanation:

Let X be a set of all students in your class. The set X is the domain. Denote

                                        C(x) -  ' \text{$x $ has a cat}'\\D(x) -  ' \text{$x$ has a dog}'\\F(x) -  ' \text{$x$ has a ferret}'

\mathbf{a)}

Consider the statement '<em>A student in your class has a cat, a dog, and a ferret</em>'. This means that \exists x \in X so that all three statements C(x), D(x) and F(x) are true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

                         \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)

\mathbf{b)}

Consider the statement '<em>All students in your class have a cat, a dog, or a ferret.' </em>This means that \forall x \in X at least one of the statements C(x), D(x) and F(x) is true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

                        \left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee F(x)

\mathbf{c)}

Consider the statement '<em>Some student in your class has a cat and a ferret, but not a dog.' </em>This means that \exists x \in X so that the statements C(x), F(x) are true and the negation of the statement D(x) . We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

                      \left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)

\mathbf{d)}

Consider the statement '<em>No student in your class has a cat, a dog, and a ferret..' </em>This means that \forall x \in X none of  the statements C(x), D(x) and F(x) are true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as a negation of the statement in the part a), as follows

\neg \left( \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\right) \iff \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)

\mathbf{e)}

Consider the statement '<em> For each of the three animals, cats, dogs, and ferrets, there is a student in your class who has this animal as a pet.' </em>

This means that for each of the statements C, F and D there is an element from the domain X so that each statement holds true.

We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

           \left((\exists x\in X)C(x) \right) \wedge  \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)

5 0
4 years ago
NEED HELP!!! 100PTS!!!!
lutik1710 [3]

answer:

x=2=3=y=10 i got the person above give them brainlesss plz

Step-by-step explanation:

3 0
2 years ago
Subtract 4/5-1/6 and write it in a simplest form
svetlana [45]
\frac{4}{5}-\frac{1}{6}=\ \ \ \ | making\ common\ denominator\\\\&#10;\frac{24}{30}-\frac{5}{30}=\frac{24-5}{30}=\frac{19}{30}\\\\Solution \ is\ \frac{19}{30}.
3 0
3 years ago
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