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SVETLANKA909090 [29]
3 years ago
5

Is the product the same when multiplying 22×(-5) and multiplying (-5)×22

Mathematics
1 answer:
uranmaximum [27]3 years ago
4 0

Answer:

Yes they should be the same.

Step-by-step explanation:

The answers should always be the same as long as the negative stays with the 5 and the 22 remains positive.

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Point K is on line JL. given JL =4x, JK= 2x+3, and KL=x, determine the numerical length of kl​
balandron [24]

Hi! I'm happy to help!

Our total line is JL (4x), and it is split into two parts: JK, and KL. We have our values, and we know that JK+KL=JL, so we can substitute our values and solve for x:

4x=(2x+3)+(x)

4x=3x+3

To solve for x, we have to isolate it on one side of the equation.

First, let's subtract 3x from both sides so that we can isolate x:

4x=3x+3

-3x -3x

x=3

<u>So, our x=3, which means that KL=3.</u>

I hope this was helpful, keep learning! :D

5 0
3 years ago
What is 6 1/4 - 3 1/3
Anettt [7]
Hey There!

Remember, first find the gcf of 4,3 in this problem it is 12.

Steps to solve

1.6\frac{1}{4} - 3\frac{1}{3}

2.6\frac{1}{4} - 3 \frac{1}{3} = 2\frac{11}{12}

So, The answer is <span>2\frac{11}{12}</span>

Hope This Helps :)
6 0
3 years ago
Read 2 more answers
In a large statistics course, 74% of the students passed the first exam, 72% of the students pass the second exam, and 58% of th
11111nata11111 [884]

Answer:

Required probability is 0.784 .

Step-by-step explanation:

We are given that in a large statistics course, 74% of the students passed the first exam, 72% of the students pass the second exam, and 58% of the students passed both exams.

Let Probability that the students passed the first exam = P(F) = 0.74

     Probability that the students passed the second exam = P(S) = 0.72

     Probability that the students passed both exams = P(F \bigcap S) = 0.58

Now, if the student passed the first exam, probability that he passed the second exam is given by the conditional probability of P(S/F) ;

As we know that conditional probability, P(A/B) = \frac{P(A\bigcap B)}{P(B) }

Similarly, P(S/F) = \frac{P(S\bigcap F)}{P(F) } = \frac{P(F\bigcap S)}{P(F) }  {As P(F \bigcap S) is same as P(S \bigcap F) }

                          = \frac{0.58}{0.74} = 0.784

Therefore, probability that he passed the second exam is 0.784 .

5 0
3 years ago
Written as fractions, the decimal numbers 0.3 and 0.11 are 3/10 and 11/100, respectively
Vlada [557]
What exactly is the question here?
4 0
2 years ago
Read 2 more answers
Box of 15 gadgets is known to contain 5 defective gadgets if 4 gadgets are drawn at random what is the probability of finding no
baherus [9]

To solve this problem, we make use of the Binomial Probability equation which is mathematically expressed as:

P = [n! / r! (n – r)!] p^r * q^(n – r)

where,

n = the total number of gadgets = 4

r = number of samples = 1 and 2 (since not more than 2)

p = probability of success of getting a defective gadget

q = probability of failure = 1 – p

 

Calculating for p:

p = 5 / 15 = 0.33

So,

q = 1 – 0.33 = 0.67

 

Calculating for P when r = 1:

P (r = 1) = [4! / 1! 3!] 0.33^1 * 0.67^3

P (r = 1) = 0.3970

 

 

Calculating for P when r = 2:

P (r = 2) = [4! / 2! 2!] 0.33^2 * 0.67^2

P (r = 2) = 0.2933

 

Therefore the total probability of not getting more than 2 defective gadgets is:

P = 0.3970 + 0.2933

P = 0.6903

 

Hence there is a 0.6903 chance or 69.03% probability of not getting more than 2 defective gadgets.

5 0
3 years ago
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