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ankoles [38]
3 years ago
15

Find (3) if f(x)=x2 + 2x2-x-1

Mathematics
1 answer:
Olenka [21]3 years ago
7 0

Answer:

The correct answer is:

f(3)=23

Step-by-step explanation:

By evaluating first, and then the mathematical calculation, we have:

f(3)= 3^{2} + 2 ×3^{2} - 3 -1 

= 9 +18 -3-1  

= 23

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Write fraction in "Simpilist Form". If already , then write "Simpilist Form". The Fraction is: 3 over 100 - 3/100​
dezoksy [38]

Answer:

see explanation

Step-by-step explanation:

A fraction is in simplest form when no other factor but 1 will divide into the numerator/ denominator

(1)

\frac{3}{100} ← is in simplest form

(2)

\frac{2}{5} ← is in simplest form

(3)

\frac{10}{12} ( divide numerator/ denominator by 2 )

= \frac{5}{6} ← in simplest form

8 0
3 years ago
I need help PLEASE!!!!!!!!
PIT_PIT [208]

Answer:

Step-by-step explanation:

<h2>the letter in the problem is the missing number</h2>
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3 years ago
A(5−x)=bx−8 what is x
nata0808 [166]

5a+8 /a+b  is the answer.



 


6 0
3 years ago
Read 2 more answers
You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confide
salantis [7]

Answer:

With​ 90% confidence, it can be said that the population mean price lies in the first interval. With​ 95% confidence, it can be said that the population mean price lies in the second interval. The​ 95% confidence interval is wider than the​ 90%.

Step-by-step explanation:

We are given that a random sample of 60 home theater systems has a mean price of​$131.00. Assume the population standard deviation is​$18.80.

  • Firstly, the pivotal quantity for 90% confidence interval for the  population mean is given by;

                            P.Q. = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean price = $131

            \sigma = population standard deviation = $18.80

            n = sample of home theater = 60

            \mu = population mean

<em>Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.</em>

<u>So, 90% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90  {As the critical value of z at 5% level

                                                   of significance are -1.645 & 1.645}  

P(-1.645 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.645) = 0.90

P( -1.645 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

P( \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ]

                                                  = [131-1.645 \times {\frac{18.8}{\sqrt{60} } } , 131+1.645 \times {\frac{18.8}{\sqrt{60} } } ]

                                                  = [127.01 , 134.99]

Therefore, 90% confidence interval for the population mean is [127.01 , 134.99].

  • Now, the pivotal quantity for 95% confidence interval for the  population mean is given by;

                            P.Q. = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean price = $131

            \sigma = population standard deviation = $18.80

            n = sample of home theater = 60

            \mu = population mean

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.</em>

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                   of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ]

                                                  = [131-1.96 \times {\frac{18.8}{\sqrt{60} } } , 131+1.96 \times {\frac{18.8}{\sqrt{60} } } ]

                                                  = [126.24 , 135.76]

Therefore, 95% confidence interval for the population mean is [126.24 , 135.76].

Now, with​ 90% confidence, it can be said that the population mean price lies in the first interval. With​ 95% confidence, it can be said that the population mean price lies in the second interval. The ​95% confidence interval is wider than the​ 90%.

7 0
3 years ago
Geometry question Need help with #10
Lena [83]

Answer:

x = 6

Step-by-step explanation:

Line segment SU is a bisector of angle <TSV which means it divided the angle into two equal parts:

The measure of <TSV = 62 in this case the measure of <VSU = 62/2 = 31

we can write the following equation to find the value of x

5x + 1 = 31 subtract 1 from both sides

5x = 30 divide both sides by 5

x = 6

6 0
1 year ago
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