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Galina-37 [17]
3 years ago
15

If M=5x^2+7x-4 and N=-3x^2-4x+5, then M-N equals

Mathematics
1 answer:
Phantasy [73]3 years ago
4 0

Answer:

8x² + 11x - 9

Step-by-step explanation:

M - N

= 5x² + 7x - 4 - 1(- 3x² - 4x + 5) ← distribute parenthesis by - 1

= 5x² + 7x - 4 + 3x² + 4x - 5 ← collect like terms

= 8x² + 11x - 9

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How many three-digit multiples of 5 have three different digits?
aalyn [17]

There are 180 numbers which are three-digit multiples of 5.

According to the statement

we have to find the three digit numbers which are multiples of 5.

we know that three digit numbers are start from 100 to 999 and in 1 line of counting means 100 to 110 there are 2 numbers are multiples of 5.

Here we use Multiplication method to find the number of digit which are a multiple of 5.

It means there are two numbers which are multiples of 5 in the one line of counting and there are 90 lines of  counting from 100 to 999.

So, it means the answer will become

2*90 = 180.

So, There are 180 numbers which are three-digit multiples of 5.

Learn more about NUMBERS here brainly.com/question/1094036

#SPJ1

8 0
2 years ago
Write yhr equation in standard form of the line with m=3/4 and b=6
ozzi

Answer:

y = 3/4 x + 6

Step-by-step explanation:

y = mx + b

y = 3/4 x + 6

3 0
3 years ago
Show your work, type the answer 6y + 5(6 + 4у + Зу)​
Andreas93 [3]
6y + 5*(6 + 7y)
6y + 30 + 35y
41y + 30
6 0
3 years ago
Which shows 2.6x10^4 in standard notation
timofeeve [1]
Hi. The answer is 26000
7 0
3 years ago
The score on an exam from a certain MAT 112 class, X, is normally distributed with μ=78.1 and σ=10.8.
salantis [7]

a) X

b) 0.1539

c) 0.1539

d) 0.6922

Step-by-step explanation:

a)

In this problem, the score on the exam is normally distributed with the following parameters:

\mu=78.1 (mean)

\sigma = 10.8 (standard deviation)

We call X the name of the variable (the score obtained in the exam).

Therefore, the event "a student obtains a score less than 67.1) means that the variable X has a value less than 67.1. Mathematically, this means that we are asking for:

X

And the probability for this to occur can be written as:

p(X

b)

To find the probability of X to be less than 67.1, we have to calculate the area under the standardized normal distribution (so, with mean 0 and standard deviation 1) between z=-\infty and z=Z, where Z is the z-score corresponding to X = 67.1 on the s tandardized normal distribution.

The z-score corresponding to 67.1 is:

Z=\frac{67.1-\mu}{\sigma}=\frac{67.1-78.1}{10.8}=-1.02

Therefore, the probability that X < 67.1 is equal to the probability that z < -1.02 on the standardized normal distribution:

p(X

And by looking at the z-score tables, we find that this probability is:

p(z

And so,

p(X

c)

Here we want to find the probability that a randomly chosen score is greater than 89.1, so

p(X>89.1)

First of all, we have to calculate the z-score corresponding to this value of X, which is:

Z=\frac{89.1-\mu}{\sigma}=\frac{89.1-78.1}{10.8}=1.02

Then we notice that the z-score tables give only the area on the left of the values on the left of the mean (0), so we have to use the following symmetry property:

p(z>1.02) =p(z

Because the normal distribution is symmetric.

But from part b) we know that

p(z

Therefore:

p(X>89.1)=p(z>1.02)=0.1539

d)

Here we want to find the probability that the randomly chosen score is between 67.1 and 89.1, which can be written as

p(67.1

Or also as

p(67.1

Since the overall probability under the whole distribution must be 1.

From part b) and c) we know that:

p(X

p(X>89.1)=0.1539

Therefore, here we find immediately than:

p(67.1

7 0
3 years ago
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