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mel-nik [20]
3 years ago
9

What’s the value of [-1.1]

Mathematics
2 answers:
arlik [135]3 years ago
5 0
The value is 1.1. Because it is 1.1 spaces away. the negative symbol isn’t,t important in this case.
ELEN [110]3 years ago
5 0
Hello, there!


I think you mean the absolute value of -1.1

This would be 1.1 as absolute value is how many units away from 0 a number is. 

Even if you were asking the absolute value of 1.1, the answer would be still be 1.1

Absolute value is always positive, regardless of the number.


I hope I helped!

Let me know if you need anything else!

~ Zoe
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remainder = 14

2x^{4} - 3x^{3} - 6x² + 11x + 8

= 2x³ + x² - 4x + 3 + \frac{14}{x - 2}


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Gina says that for all real numbers a and b, ab=ba. is this statement true or false? If false, give a counterexample
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True

Step-by-step explanation:

The communative property states that the order in addition and multiplication dosen't matter.

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Cameron buys 2.45 pounds of apples and 1.65 pounds of pears. Apples and pears each cost c dollars per pound. If the total cost a
Leto [7]

Answer:

2.45c + 1.65c = 4.12 + 0.75

Step-by-step explanation:

To write an equation to find the value for c, we need to declare what c is first.

c = price of fruit

2.45c + 1.65c = 4.12 + 0.75

Now we multiplied c to 2.45 and 1.65 and added them together, because whatever the value of c is will give us the equivalence of the sum of 4.12 + 0.75.

Now to check if the equation is right, let's solve for c.

2.45c + 1.65c = 4.12 + 0.75

4.1c = 4.87

Now to get the value of c, we divide both sides of the equation by 4.1.

\dfrac{4.1c}{4.1}=\dfrac{4.87}{4.1}

c = 1.19

Now let's substitute the value of c in the equation to see if we got it right.

2.45(1.19) + 1.65(1.19) = 4.12 + 0.75

2.92 + 1.96 = 4.87

4.87 = 4.87

Therefore concluding that the value of c is 1.19.

7 0
3 years ago
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The standard error of the mean is the standard deviation which is 6.50
8 0
3 years ago
Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits.
DENIUS [597]

Answer:

The probability that a randomly chosen plate contains the number 2222 is 0.000028 approximately.

The probability that a randomly chosen plate contains the sub-string HI is 0.002548  approximately.

Step-by-step explanation:

Consider the provided information.

Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits.

Part (A)

Let A is the ways in which plates consist of three letters followed by three digits and B is the ways in which two letters followed by four digits.

Here repetition is allow. The number of alphabets are 26 and the number of distinct digits are 10.

The numbers of ways in which three letters followed by three digits can be chosen is: 26\times 26\times 26 \times 10 \times 10 \times10

26^3\times 10^3=17576000

The numbers of ways in which two letters followed by four digits can be chosen is: 26\times 26\times 10 \times 10 \times 10 \times10

26^2\times 10^4=6760000

Hence, the total number of ways are 17576000 + 6760000 = 24336000

Randomly chosen plate contains the number 2222 that means the first two letter can be any alphabets but the rest of the digit should be 2222.

Thus, the total number of ways that a randomly chosen plate contains the number 2222 number are: 26^2=676

The probability that a randomly chosen plate contains the number 2222 is:

\frac{676}{24336000} \approx 0.000028

Part (B)

The number of ways in which chosen plate contains the sub-string HI:

If three letters followed by three digits plate contains the sub-string HI, then the number of possible ways are:

26\times 1\times10^3+1\times 26\times10^3

If two letters followed by four digits plate contains the sub-string HI, then the number of possible ways are:

1\times 10^4

Thus, the total number of ways that a randomly chosen plate contains the sub-string HI are:

1\times 10^4+26\times 1\times10^3+1\times 26\times10^3

62000

From part (A) we know that the total number of ways to chose a number plate is 24336000.

The probability that a randomly chosen plate contains the sub-string HI is:

\frac{62000}{24336000} \approx 0.002548

7 0
3 years ago
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