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Papessa [141]
3 years ago
9

Help. Me!!!!!!!!!!!!!!!!!!! Hi

Mathematics
1 answer:
bearhunter [10]3 years ago
6 0

Answer:

Terminating decimals: -1/5, 4/25

Repeating decimals: 7/9, 4/11, -1/6

Step-by-step explanation:

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Multiply. Check picture.
VashaNatasha [74]

The answer is 3x^4-13x^3-x^2-11x+6.

Solution:

Use algebraic identity: a^m\times a^n=a^{m+n}

For example: x^2\times x=x^{2+1}=x^3

Given expression (x^2-5x+2) and (3x^2+2x+3).

To multiply these equations.

(x^2-5x+2)\times(3x^2+2x+3)

             =x^2(3x^2+2x+3)-5x(3x^2+2x+3)+2(3x^2+2x+3)

             =(3x^4+2x^3+3x^2)+(-15x^3-10x^2-15x)+(6x^2+4x+6)

             =3x^4+2x^3+3x^2-15x^3-10x^2-15x+6x^2+4x+6

Combine like terms together.

             =3x^4+(2x^3-15x^3)+(3x^2-10x^2+6x^2)-15x+4x+6

             =3x^4-13x^3-x^2-11x+6

(x^2-5x+2)\times(3x^2+2x+3)=3x^4-13x^3-x^2-11x+6

Hence the answer is 3x^4-13x^3-x^2-11x+6.

3 0
3 years ago
Match the vocabulary to the part of expression below
Fantom [35]

Answer:

Gas up in my room I don’t know how much you have to do that yet but

Step-by-step explanation:

Abc news I didn’t see you yesterday I didn’t answer you I didn’t answer your phone I didn’t answer your question I don’t understand why

5 0
2 years ago
The manager of a computer retails store is concerned that his suppliers have been giving him laptop computers with lower than av
34kurt

Answer:

Probability that the 50 randomly selected laptops will have a mean replacement time of 3.1 years or less is 0.0092.

Yes. The probability of this data is unlikely to have occurred by chance alone.

Step-by-step explanation:

We are given that the replacement times for the model laptop of concern are normally distributed with a mean of 3.3 years and a standard deviation of 0.6 years.

He then randomly selects records on 50 laptops sold in the past and finds that the mean replacement time is 3.1 years.

<em>Let M = sample mean replacement time</em>

The z-score probability distribution for sample mean is given by;

            Z = \frac{ M-\mu}{\frac{\sigma}{\sqrt{n} } }} }  ~ N(0,1)

where, \mu = population mean replacement time = 3.3 years

            \sigma = standard deviation = 0.6 years

            n = sample of laptops = 50

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the 50 randomly selected laptops will have a mean replacement time of 3.1 years or less is given by = P(M \leq 3.1 years)

 P(M \leq 3.1 years) = P( \frac{ M-\mu}{\frac{\sigma}{\sqrt{n} } }} } \leq \frac{ 3.1-3.3}{\frac{0.6}{\sqrt{50} } }} } ) = P(Z \leq -2.357) = 1 - P(Z \leq 2.357)

                                                           = 1 - 0.99078 = <u>0.0092</u>  or  0.92%          

<em>So, in the z table the P(Z </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.357 in the z table which will lie between x = 2.35 and x = 2.36 which has an area of 0.99078.</em>

Hence, the required probability is 0.0092 or 0.92%.

Now, based on the result above; <u>Yes, the computer store has been given laptops of lower than average quality</u> because the probability of this data is unlikely to have occurred by chance alone as the probability of happening the given event is very low as 0.92%.

8 0
3 years ago
How do I complete the linear equation
kifflom [539]

Answer:

Follow along with the pattern, for example if it is going by 2's then you would write a 4

3 0
3 years ago
Can some explain this to me do i need to covert something or can someone explain it to me how in gonna get the answer i try to d
solniwko [45]
Cross multiply:

2.5 ft x 16.5 ft = 41.25 ft

now divide:

41.25 / 8.25 = 5

L = 5 feet 
7 0
3 years ago
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