The table below represents a linear function f(x) and the equation represents a function g(x):

I need help finding the area of 32,33,and 34

denis23 [38]

Area is 28 for number 33.

7*8=56
56/2=28

5 0

3 years ago

What is Ten percent of a number added to that number?

rodikova [14]

Let x be this number 10% of the number x is (10%)(x) → 0.1x
Then 10% of a number x, added to that number is (1.1x):

x + 0.1x = x(1+0.1) = x(1.1) = 1.1x

8 0

3 years ago

House

algol [13]

I can answer this for u

8 0

3 years ago

Read 2 more answers

(AP Calc) the graph of the function f shown above has three line segments....

I am Lyosha [343]

Answer:

B

Step-by-step explanation:

The average rate of change of f(x) in the closed interval [ a, b ] is

$\frac{f(b)-f(a)}{b-a}$

Here [ a, b ] = [ - 1, 6 ] and from the graph

f(b) = f(6) = 0

f(a) = f(- 1) = 0, thus

average rate of change = $\frac{0-0}{6+1}$ = $\frac{0}{7}$ = 0

7 0

3 years ago

A random sample of 49 measurements from one population had a sample mean of 15, with sample standard deviation 5. An independent

kotykmax [81]

Answer:

Comparing the p value with the significance level $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly different.

Step-by-step explanation:

$\bar X_{1}=15$ represent the mean for sample 1

$\bar X_{2}=18$ represent the mean for sample 2

$s_{1}=5$ represent the sample standard deviation for 1

$s_{f}=6$ represent the sample standard deviation for 2

$n_{1}=49$ sample size for the group 2

$n_{2}=64$ sample size for the group 2

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the difference in the population means, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}=0$

Alternative hypothesis: $\mu_{1} - \mu_{2}\neq 0$

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=49+64-2=111$

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$t=\frac{(15-18)-0}{\sqrt{\frac{5^2}{49}+\frac{6^2}{64}}}}=-2.897$

P value

Since is a bilateral test the p value would be:

$p_v =2*P(t_{111}$

Comparing the p value with the significance level $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly different.

7 0

3 years ago