What is five eighths divided by twenty four twenty fifths

Find the constant of variation for the relation and use it to write an equation for the statement. Then solve the equation

Olenka [21]

Answer:

y = $\frac{49}{3}$

Step-by-step explanation:

Given that y varies directly as x then the equation relating them is

y = kx ← k is the constant of variation

To find k use the condition y = 7 when x = 3

k = $\frac{y}{x}$ = $\frac{7}{3}$ , thus

y = $\frac{7}{3}$ x ← equation of variation

When x = 7, then

y = $\frac{7}{3}$ × 7 = $\frac{49}{3}$

8 0

2 years ago

Please help!❤️ 9/4y-2=25 Please solve and show work!!

STatiana [176]

9/4y-2=25
9/4y-2+2=27+2
9/4y=27
9/4y*4y=27*4y
9=4y*27,
1/3=4y, (9/27 simpltfies to 1/3, 1/3 divided by 4: 1/3*1/4=1/12)
1/12=y

Answer:y=1/12

8 0

3 years ago

Read 2 more answers

Simplify the following expression.( Fraction form) 7-5/6 * 7-7/6

Ipatiy [6.2K]

Answer:

1295/36

Step-by-step explanation:

the statement tell us:

(7-(5/6))*(7-(7/6))

we have:

(7-(5/6))=((6*7)-5)/6=(42-5)/6=37/6

and we have:

(7-(7/6))=((6*7)-7)/6=(42-7)/6=35/6

we need multiply both terms:

(37/6)*(35/6)=(37*35)/(6*6)

finally we have

1295/36

5 0

3 years ago

You ran for 25 minutes and burned 220 calories. How many calories did you burn per minute?

Bezzdna [24]

Answer:

8.8, you have to do long devision

Step-by-step explanation:

Hope this helps!

3 0

2 years ago

Read 2 more answers

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.

4 0

3 years ago