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Gelneren [198K]
3 years ago
11

Enter the decimal equivalent of 9/4.

Mathematics
2 answers:
Ludmilka [50]3 years ago
5 0

Answer:

2.25

Step-by-step explanation:

9 divided by 4 is 2.25

Finger [1]3 years ago
4 0

Answer:

2.25

Step-by-step explanation:

\frac{9}{4} = 2\frac{1}{4} = 2.25.

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frutty [35]

Answer:

-39 2/3

Step-by-step explanation:

t - 122= 2t - 3

subtract 't' from each side to get:

-122 = 3t - 3

add 3 to each side to get:

-119 = 3t

divide each side by 3 to get:

t = -39 2/3

5 0
3 years ago
THIS IS PROBABLY MY LAST QUESTION!!!!!!!!
Lady bird [3.3K]

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2.c 3.a 4.b

Step-by-step explanation:

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What the expression (4f-3+2g)-(4g+2)
galben [10]
For this question you can say: 
4f -3 +2g - 4g -2
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8 0
3 years ago
Please help I’ll mark you as brainliest if correct!
Pavlova-9 [17]

Answer:

3 ways

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6 0
2 years ago
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The manager of a music store has kept records of
Marysya12 [62]

Answer:

(a) P(x\le 3) = 0.75

(b) P(x\le 3) = 0.75

<em>(b) is the same as (a)</em>

(c) P(x \ge 5) = 0.10

(d) P(x = 1\ or\ 2) = 0.55

(e) P(x > 2) = 0.45

Step-by-step explanation:

Given

\begin{array}{ccccccc}{CDs} & {1} & {2} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.30} & {0.25} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}

Solving (a): Probability of 3 or fewer CDs

Here, we consider:

\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}

This probability is calculated as:

P(x\le 3) = P(1) + P(2) + P(3)

This gives:

P(x\le 3) = 0.30 + 0.25 + 0.20

P(x\le 3) = 0.75

Solving (b): Probability of at most 3 CDs

Here, we consider:

\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}

This probability is calculated as:

P(x\le 3) = P(1) + P(2) + P(3)

This gives:

P(x\le 3) = 0.30 + 0.25 + 0.20

P(x\le 3) = 0.75

<em>(b) is the same as (a)</em>

<em />

Solving (c): Probability of 5 or more CDs

Here, we consider:

\begin{array}{ccc}{CDs} & {5} & {6\ or\ more}\ \\ {Prob} & {0.05} & {0.05}\ \ \end{array}

This probability is calculated as:

P(x \ge 5) = P(5) + P(6\ or\ more)

This gives:

P(x\ge 5) = 0.05 + 0.05

P(x \ge 5) = 0.10

Solving (d): Probability of 1 or 2 CDs

Here, we consider:

\begin{array}{ccc}{CDs} & {1} & {2} \ \\ {Prob} & {0.30} & {0.25} \ \ \end{array}

This probability is calculated as:

P(x = 1\ or\ 2) = P(1) + P(2)

This gives:

P(x = 1\ or\ 2) = 0.30 + 0.25

P(x = 1\ or\ 2) = 0.55

Solving (e): Probability of more than 2 CDs

Here, we consider:

\begin{array}{ccccc}{CDs} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}

This probability is calculated as:

P(x > 2) = P(3) + P(4) + P(5) + P(6\ or\ more)

This gives:

P(x > 2) = 0.20+ 0.15 + 0.05 + 0.05

P(x > 2) = 0.45

3 0
3 years ago
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