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3 years ago

Add fraction 6 4/5 + 3 7/10 how do i add this fraction

2 answers:

7 0

The answer is 10 1/2.

First you need to convert 4/5 and 7/10 so the denominator is the same. 4/5=8/10, 7/10 stays the same.

Now we add= 6 8/10 +3 7/10 =9 15/10.
9 15/10=10 5/10=10 1/2.

Hope this helps and hope you have a great day and please make brainiest

Eduardwww [97]3 years ago

6 0

Answer:

The answer would be 9 75/50 which would equal 9 3/2.

Step-by-step explanation: First you would multiply 4/5 by 10 and 7/10 by 5 because the common denominator is 50. Next, you would add it together. Finally, you would reduce that to 9 3/2 I am pretty sure.

You might be interested in

Please ΔEFU ~ ΔVWU. Find the length of UF.

koban [17]

<h3>Answer: UF = 11</h3>

=======================================================

Explanation:

In EFU, we see that EU are the first and last letters. In VWU, we see that VU are the first and last letters. This order is important to be able to pair EU with VU

So EU/VU is one ratio

The other ratio is UF and UW, which are the last two letters of EFU and VWU respectively. So the other ratio is UF/UW

Set the ratios equal to each other and solve for x

EU/VU = UF/UW

7/35 = x/55

7*55 = 35*x ... cross multiply

385 = 35x

35x = 385

x = 385/35 .... divide both sides by 35

x = 11

UF = 11

------------

Or you could note how UV = 35 is exactly 5 times larger compared to EU = 7

That must mean WU = 55 must also be 5 times larger compared to UF = x

So,

WU = 5*UF

55 = 5x

5x = 55

x = 55/5

x = 11

UF = 11

This is similar to the previous section because UV/EU = 35/7 = 5 is the scale factor.

4 0

4 years ago

Pls answe quick and correctly

andriy [413]

Answer:

9890

Step-by-step explanation

7×43×30+5×4×43=9890

4 0

2 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

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

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

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

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

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

Will brainlist please helping

wlad13 [49]

tried to do it on text but It was pretty hard

4 0

3 years ago

What is the coefficient in the expression 4x + 2 <br>O2<br>O4<br>Ox<br>O+

Amiraneli [1.4K]

A coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression (e.g. 4 in 4x).

The 2 and + are out since they don't have variables(and the + is just a +).

So 4 and x are left. x is a variable, not a coefficient, so the last(and correct) answer is 4.

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hope it helps

8 0

3 years ago