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3 years ago

If y equals 6 , what is the value of x given 8x plus 5y equals 62

Mathematics

2 answers:

OverLord2011 [107]3 years ago

6 0

8x+5(6)=62
8x+30=62
8x=32
x=4

Brilliant_brown [7]3 years ago

5 0

Answer: 8x+5(6)=62

8x+30=62

8x=32

x=4

Step-by-step explanation:

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3 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

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Answer:

(a + c) + 7

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2 years ago

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Answer:

66.3

Step-by-step explanation:

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3 years ago

The following table shows the weights, in hundreds of pounds, for six selected cars. Also shown is the corresponding fuel eff

Reika [66]

Answer:

Step-by-step explanation:

Given is a table showing the weights, in hundreds of pounds, for six selected cars. Also shown is the corresponding fuel efficiency, in miles per gallon (mpg), for the car in city driving.

Weight Fuel eff. x^2 xy y^2

X Y

28 20 784 560 400

3 22 9 66 484

35 19 1225 665 361

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30 23 900 690 529

29 21 841 609 441

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Covariance -553.8611111

r -0.341120235

Correlaton coefficient =cov (xy)/S_x S_y

Covariance (x,y) = E(xy)-E(x)E(y)

The correlation coefficient between the weight of a car and the fuel efficiency is -0.341

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3 years ago