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4 years ago

The expression 32−48 factored using the GCF is .

Mathematics

1 answer:

Ulleksa [173]4 years ago

3 0

Answer:

Step-by-step explanation:

The largest factor that will divide evenly into both 32 and 48 is 16.

Thus, 32 - 48 = 16(2 - 3), or 16 (-1), or just -16.

Are you sure there wasn't more to this problem?

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3 years ago

The number of small lawns Isabella can rake depends on the amount of time she spends raking. Isabella can rake 3 lawns in 3.75 h

ycow [4]

Answer:

1) 4 2)7

Step-by-step explanation:

1) If we divide 3.75/3=1.25 which is the amount of time she takes per yard so if we add another yard to what she has already done we find she only could do one more yard in 5 hours 3.75+1.25=5.00 So she can only rake 4 yards in 5 hours

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4 years ago

Naval intelligence reports that 99 enemy vessels in a fleet of 1818 are carrying nuclear weapons. If 99 vessels are randomly tar

oee [108]

Answer:

0.001687 = 0.1687% probability that no more than 1 vessel transporting nuclear weapons was destroyed.

Step-by-step explanation:

The vessels are destroyed and then not replaced, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

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

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

In this question:

Fleet of 18 means that $N = 18$

9 are carrying nuclear weapons, which means that $k = 9$

9 are destroyed, which means that $n = 9$

What is the probability that no more than 1 vessel transporting nuclear weapons was destroyed?

This is:

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

In which

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,18,9,9) = \frac{C_{9,0}*C_{9,9}}{C_{18,9}} = 0.000021$

$P(X = 1) = h(1,18,9,9) = \frac{C_{9,1}*C_{9,8}}{C_{18,9}} = 0.001666$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.000021 + 0.001666 = 0.001687$

0.001687 = 0.1687% probability that no more than 1 vessel transporting nuclear weapons was destroyed.

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3 years ago

You are given 3 to 1 odds against tossing three tails with three coins, meaning you win $3 if you succeed and you lose $1 if

valentina_108 [34]

Step-by-step explanation:

To find the E(X) expected value, you come up with the different probabilities for each outcome
your set of outcomes after 3 tosses would be = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} where H is heads and T is tails
Each element has a probability of 1/8 so let x represent number of tails
$Probability(x)=P(x)\\P(0)=\frac{1}{8} \\P(1)=\frac{3}{8} \\P(2)=\frac{3}{8} \\P(3)=\frac{1}{8}$
The E(x)=Summation (x times P(x))
$E(X)=(0)(0.125)+(1)(0.375)+(2)(0.375)+(3)(0.125)=1.5$
Now which probability is 1.5 tails? None, so it is either 2 tails or 1 tails
So you can expect to lose money in 1 game
But as you play more games the probability of getting 3 tails becomes more and more likely, so you can expect to win in a 100 games

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3 years ago