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Archy [21]
3 years ago
11

What is 42-27x5 what is the answer

Mathematics
1 answer:
bulgar [2K]3 years ago
8 0

Answer:

-363

Step-by-step explanation:

42-27*5

42-405

Answer: -363

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7 0
3 years ago
Need help number the answers bye order plz
iragen [17]
OK so here it goes:
1a. 10³
1b. 10²
1c. 10(tiny 6)
1d. 10³
1e. 10(tiny6)
1f. 10(tiny 5)

2a. 400
2b. 640,000
2c. 5,400
2d. 5,301,000
2e. is it multipcation or addition
2f. 607,200
2g. 0.948
2h. 0.0094

3a) 0.02, 0.2, 2, 20, 200, 2000
3b) 3,400,000 ; 34,000; 340, 3.4, 0.034
3c) 85,700; 8,570; 857, 85.7, 8.57, 0.857
3d) 444, 4440, 44,400; 440,000; 4,400,000; 44,000,000
3e) 0.95, 9.5, 950, 95,000; 950,000; 9,500,000

Hope this helps.
6 0
3 years ago
Find an equation of the line that passes through the point (4, 3) and is perpendicular to the line 2x + 9y − 6 = 0.
vesna_86 [32]
Perpendicular has the reciprocal slope and u have to put the point given into point slope.

y-3=9/2(x-4)

(The slope of the original equation was -2/9)
3 0
3 years ago
Naval intelligence reports that 4 enemy vessels in a fleet of 17 are carrying nuclear weapons. If 9 vessels are randomly targete
icang [17]

Answer:

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

Step-by-step explanation:

The vessels are destroyed without replacement, 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 17 means that N = 17

4 are carrying nucleas weapons, which means that k = 4

9 are destroyed, which means that n = 9

What is the probability that more than 1 vessel transporting nuclear weapons was destroyed?

This is:

P(X > 1) = 1 - P(X \leq 1)

In which

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

So

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

P(X = 0) = h(0,17,9,4) = \frac{C_{4,0}*C_{13,9}}{C_{17,9}} = 0.0294

P(X = 1) = h(1,17,9,4) = \frac{C_{4,1}*C_{13,8}}{C_{17,9}} = 0.2118

Then

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0294 + 0.2118 = 0.2412

P(X > 1) = 1 - P(X \leq 1) = 1 - 0.2412 = 0.7588

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

8 0
3 years ago
Multiple 3/4 times 16/9
nataly862011 [7]
3/4 times 16/9 is equal to 1 1/3.
7 0
3 years ago
Read 2 more answers
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