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3 years ago

Need help on steps (3/7-2/3)+(4/5)(2/3)

1 answer:

6 0

$\bf \left( \cfrac{3}{7}-\cfrac{2}{3} \right)+\left( \cfrac{4}{5} \right)\left( \cfrac{2}{3} \right)\implies \stackrel{LCD~21}{\left( \cfrac{3}{7}-\cfrac{2}{3} \right)}+\left( \cfrac{4}{5} \cdot \cfrac{2}{3} \right)$

$\bf {\left( \cfrac{3\cdot 3-7\cdot 2}{21} \right)}+\left( \cfrac{4\cdot 2}{5\cdot 3} \right)\implies \left( \cfrac{9-14}{21} \right)+\left( \cfrac{8}{15} \right)
\\\\\\
\stackrel{LCD~105}{\cfrac{-5}{21}+\cfrac{8}{15}}\implies \cfrac{-5\cdot 5+8\cdot 7}{105}\implies \cfrac{-25+56}{105}\implies \boxed{\cfrac{31}{105}}$

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Someone pls help 0xy + 32y

Thepotemich [5.8K]

Answer:

32y

Step-by-step explanation:

0xy=0

If you change x or y on any number, you will get 0: 5*0=0, 100*0=0

So, 0+32y = 32 y

7 0

3 years ago

Read 2 more answers

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

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

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

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

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

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

Please answer correctly !!!!! Will mark Brianliest !!!!!!!!!!!!!!

Sedbober [7]

Answer:

24/25

Step-by-step explanation:

Please see the attached image.

The cosine of an angle is the adjacent side relative to that angle over the hypotenuse.

Answer: 24/25

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E%7B2%7D-5x%2B6%7D%7B2x%5E%7B2%7D-7x%2B6%20%7D" id="TexFormula1" title="\frac{x^{

il63 [147K]

Answer:

$\frac{x-3}{2x-3}$ . hole or removable discontinuity at x=2

Step-by-step explanation:

Well generally if you want the simplest form, you factor each the denominator and numerator and then see if you can cancel any of the factors out (because they're in the denominator and numerator)

So let's start by factoring the first equation:

$x^2-5x+6$

Now let's find what ac is (it's just c since a=1...)

$AC= 6$

List factors of -6

$\pm1, \pm2, \pm3, \pm6$ .

Now we have to look for two numbers that add up to -5. It's a bit obvious here since there isn't many factors, but it's -2 and -3, and they're both negative since 6 is positive, and -5 is negative...

So using these two factors we get

$(x-2)(x-3)$

Ok now let's factor the second equation:

$2x^2-7x+6$

Multiply a and c

$AC = 12$

List factors of 12:

$\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$ .

Factors that add up to -7 and multiply to 12:

$-3\ and\ -4$

Rewrite equation:

$2x^2-4x-3x+6$

Group terms:

$(2x^2-4x)+(-3x+6)$

Factor out GCF:

$2x(x-2)-3(x-2)$

Rewrite:

$(2x-3)(x-2)$

Now let's write out the equation using these factors:

$\frac{(x-2)(x-3)}{(2x-3)(x-2)}$ .

Here we can factor out the x-2 and the simplified form is:

$\frac{x-3}{2x-3}$

So we can "technically" define f(2) using the most simplified form, but it's removable discontinuity, so it has a hole as x=2. since it makes (x-2) equal to 0 (2-2) = 0.

8 0

2 years ago

Which of the following tables matches the graph below?

navik [9.2K]

Right answer is C

We can solve it with the formula

y=mx+b

b is the point where graphic starts

Table C shows us 0 by x, and -4 by y

So the graph starts at (0, -4 )

3 0

2 years ago