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2 years ago

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The number of wins in the American Baseball League, Central Division, for a recent year is given in the table. Find the mean num

ber of wins. Round your answer to one decimal place.
The mean number of wins for the five teams is ______wins?

1 answer:

Strike441 [17]2 years ago

3 0

Answer:

add sll number so you will get answer

dd

Step-by-step explanation:

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If you shift the exponential parent function, f(x) = 2^x , right 2 units, what is the equation of the new function?

aev [14]

Answer:

$g(x)=2^{x-2}$

Step-by-step explanation:

<u>Translations</u>

For $a > 0$

$f(x+a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units left}$

$f(x-a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units right}$

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$f(x)-a \implies f(x) \: \textsf{translated}\:a\:\textsf{units down}$

Parent function: $f(x)=2^x$

Translated 2 units right: $g(x)=f(x-2)=2^{x-2}$

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What is <br> 10x + 300 = 800<br> Pls

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Step-by-step explanation:

10x + 300 = 800

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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:

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

$P(x > 2) = 0.45$

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3 years ago

A computer code is based on the prime facttorization of 160. find the prime factorization of 160

igor_vitrenko [27]

Prime factorization of 160 = 2 x 2 x 2 x 2 x 2 x 5

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