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fredd [130]
3 years ago
6

Simplify. 1/4y+1 1/ 2+2−1 3/4y−12

Mathematics
2 answers:
Kamila [148]3 years ago
6 0
Y=-1.5
   

hope this helps
:)


zhannawk [14.2K]3 years ago
3 0

\frac{1}{4} y+ 1 \frac{1}{2} + 2 - 1 \frac{3}{4}  y-12

It is easy to calculate when we convert all the mixed fractions to proper fractions

\frac{1}{4} y+  \frac{3}{2} + 2 -  \frac{7}{4}  y-12

lets add the similar terms

(\frac{1}{4} -\frac{7}{4} )y + \frac{3}{2} +(2-12)

\frac{-3}{2} y + \frac{3}{2} -10

\frac{-3}{2} y - \frac{17}{2}

-1.5y-8.5

As we cannot simplify this further, this will be our simplified answer

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Answer:(x+10)(x+9)

Step-by-step explanation:

x^2+19x+90

x^2+10x+9x+90

x(x+10)+9(x+10)

(x+10)(x+9)

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Two angles are supplementary. The first angle is 3x degrees. The second angle is (2x+25) degrees. Determine the measure of the s
Alex17521 [72]
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5 0
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We have n = 100 many random variables Xi ’s, where the Xi ’s are independent and identically distributed Bernoulli random variab
777dan777 [17]

Answer:

(a) The distribution of X=\sum\limits^{n}_{i=1}{X_{i}} is a Binomial distribution.

(b) The sampling distribution of the sample mean will be approximately normal.

(c) The value of P(\bar X>0.50) is 0.50.

Step-by-step explanation:

It is provided that random variables X_{i} are independent and identically distributed Bernoulli random variables with <em>p</em> = 0.50.

The random sample selected is of size, <em>n</em> = 100.

(a)

Theorem:

Let X_{1},\ X_{2},\ X_{3},...\ X_{n} be independent Bernoulli random variables, each with parameter <em>p</em>, then the sum of of thee random variables, X=X_{1}+X_{2}+X_{3}...+X_{n} is a Binomial random variable with parameter <em>n</em> and <em>p</em>.

Thus, the distribution of X=\sum\limits^{n}_{i=1}{X_{i}} is a Binomial distribution.

(b)

According to the Central Limit Theorem if we have an unknown population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.  

The sample size is large, i.e. <em>n</em> = 100 > 30.

So, the sampling distribution of the sample mean will be approximately normal.

The mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu=p=0.50

And the standard deviation of the distribution of sample mean is given by,

\sigma_{\bar x}=\sqrt{\frac{\sigma^{2}}{n}}=\sqrt{\frac{p(1-p)}{n}}=0.05

(c)

Compute the value of P(\bar X>0.50) as follows:

P(\bar X>0.50)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{0.50-0.50}{0.05})\\

                    =P(Z>0)\\=1-P(Z

*Use a <em>z</em>-table.

Thus, the value of P(\bar X>0.50) is 0.50.

8 0
3 years ago
Help answer pleaseeee
tankabanditka [31]

Answer:

5%  students were absent on Monday

Step-by-step explanation:

(12/240) *100=5%

3 0
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