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yanalaym [24]
3 years ago
12

For f(x)=3x+1 and g(x)=(x^2)-6, find (f+g)(x).

Mathematics
2 answers:
Anon25 [30]3 years ago
3 0

f(x)=3x+1;\ g(x)=x^2-6\\\\(f+g)(x)=(3x+1)+(x^266)=3x+1+x^2-6=x^2+3x+(1-6)\\\\=\boxed{x^2+3x-5}

marishachu [46]3 years ago
3 0

Answer:   (f+g)(x)=x^2+3x-5.

Step-by-step explanation:  We are given the following two functions :

f(x)=3x+1,\\\\g(x)=x^2-6.

We are to find the value of (f + g)(x).

We know that

for any two functions p(x) and q(x),

(p+q)(x)=p(x)+q(x).

Therefore, we get

(f+g)(x)\\\\=f(x)+g(x)\\\\=(3x+1)+(x^2-6)\\\\=3x+1+x^2-6\\\\=x^2+3x-5.

Thus, (f+g)(x)=x^2+3x-5.

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Is 1.7 rational or irrational
Tpy6a [65]

Answer:

rational

Step-by-step explanation:

5 0
3 years ago
The coordinate plane has the following points A=(2, 1), B=(5, 1), C=(7,2), and 1 point
dusya [7]

Answer:

<em>d = 3</em>

Step-by-step explanation:

Coordinate Plane

The image provided shows the four points given:

A=(2, 1), B=(5, 1), C=(7,2), D=(4,2).

It can be clearly seen the length of CD is just the difference of their x-coordinates:

CD = 7 - 4 = 3

We can also use the formula of the distance.

Given two points C(x,y) and D(w,z), the distance between them is:

d=\sqrt{(z-y)^2+(w-x)^2}

d=\sqrt{(2-2)^2+(7-4)^2}

d=\sqrt{0^2+3^2}

d=\sqrt{9}

d = 3

3 0
3 years ago
As part of the Pew Internet and American Life Project, researchers conducted two surveys in late 2009. The first survey asked a
REY [17]

Answer:

The 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social networking sites is (0.223, 0.297). This means that we are 95% sure that the true difference of the proportion is in this interval, between 0.223 and 0.297.

Step-by-step explanation:

Before building the confidence interval we need to understand the central limit theorem and the subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Sample of 800 teens. 73% said that they use social networking sites.

This means that:

p_T = 0.73, s_T = \sqrt{\frac{0.73*0.27}{800}} = 0.0157

Sample of 2253 adults. 47% said that they use social networking sites.

This means that:

p_A = 0.47,s_A = \sqrt{\frac{0.47*0.53}{2253}} = 0.0105

Distribution of the difference:

p = p_T - p_A = 0.73 - 0.47 = 0.26

s = \sqrt{s_T^2+s_A^2} = \sqrt{0.0157^2+0.0105^2} = 0.019

Confidence interval:

Is given by:

p \pm zs

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

Lower bound:

p - 1.96s = 0.26 - 1.96*0.019 = 0.223

Upper bound:

p + 1.96s = 0.26 + 1.96*0.019 = 0.297

The 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social networking sites is (0.223, 0.297). This means that we are 95% sure that the true difference of the proportion is in this interval, between 0.223 and 0.297.

3 0
2 years ago
Are these right?!?! Cannot fail
jolli1 [7]
The first one is correct but I don’t know about the second one
4 0
3 years ago
In a certain month a family spent $750 on food .In third month the family spent $729 on food .What was percentage decrease in th
seraphim [82]

Answer: 2.8%

Step-by-step explanation:

Give, In a certain month a family spent $750 on food

In third month the family spent $729 on food .

i.e.  Expenditure on food in a certain month = $750

Expenditure on food in third month = $729

Decrease in monthly spending on food = (Expenditure in a certain month )- (Expenditure in third month )

= $750  -  $729

= $21

Now , the percentage decrease in the month spent on food will be :

\dfrac{\text{Decrease in monthly spending on food }}{\text{Expenditure on food in a certain month}}\times100\\\\=\dfrac{21}{750}\times100\\\\=2.8\%

Hence, the percentage decrease in the month spent on food  is 2.8%.

3 0
3 years ago
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