Which statement describes the changes that were made to the graph of f to make the graph of g

Evaluate the expression (a^2+b)^2 - (a^2-b)^2, if a=4 and b=1.

Sergio039 [100]

(a^2+b)^2 - (a^2-b)^2, if a=4 b=1.

4 0

3 years ago

Can someone help me with this question I’ll give you a brainslist alg 1

bogdanovich [222]

Answer:

Its the first one 2^(x-1) -3

3 0

3 years ago

Write an exponential function to model the situation. Find each amount after the specified time. A population of 120,000 grows 1

Savatey [412]

Exponential growth function is given by f(t) = A(1 + k%)^t; where A is the initial value, k is the growth rate and t is the time period.

Therefore, the required function is
f(t) = 120,000(1 + 0.012)^t = 120,000(1.012)^t

After 15 years, the new population is 120,000(1.012)^15 = 120,000 x 1.1959 = 143,512

8 0

3 years ago

Evaluate the following expression <br> 6 - 2 x 1

torisob [31]

6 - 2 x 1

= 6 - 2

= 4

We have to use the PEDMAS method in solving all kinds of calculation expression.

P = Parentheses (or brackets)
E = Exponents (or Index)
D = Division
M = Multiplication
A = Addition
S = Subtraction

3 0

3 years ago

Rothamsted Experimental Station (England) has studied wheat production since 1852. Each year many small plots of equal size but

podryga [215]

Answer:

(a) 0.653

(b) 0.0198

(c) Yes, after testing we conclude that the population variance of annual wheat production for the first plot is larger than that for the second plot.

Step-by-step explanation:

(a) We are give the sample of one year annual production of wheat (in pounds) ;

4.46, 4.21, 4.40, 4.81, 2.81, 2.90, 4.93, 3.54, 4.16, 4.48, 3.26, 4.74, 4.97, 4.02, 4.91, 2.59

For calculating sample variance, firstly we will calculate mean of the above data;

Mean of above data, $X_1bar$ = Sum of all values ÷ n (no. of values)

= $\frac{4.46 + 4.21+4.40+.......+4.91+2.59}{16}$ = 4.074

Sample Variance, $s_1^{2}$ = $\frac{\sum (X-X_1bar)^{2} }{n-1}$ = $\frac{(4.46-4.074)^{2}+(4.21-4.074)^{2}+.........+(4.91-4.074)^{2}+(2.59-4.074)^{2} }{16-1}$ = 0.653

(b) Another sample for annual wheat production (in pounds);

3.89, 3.81, 3.95, 4.07, 4.01, 3.73, 4.02, 3.78, 3.72, 3.96, 3.62, 3.76, 4.02, 3.73, 3.94, 4.03

Mean of above data, $X_2bar$ = Sum of all values ÷ n (no. of values)

= $\frac{3.89+3.81+3.95.......+3.94+4.03}{16}$ = 3.88

Sample Variance, $s_2^{2}$ = $\frac{\sum (X-X_2bar)^{2} }{n-1}$ = $\frac{(3.89-3.88)^{2}+(3.81-3.88)^{2}+.........+(3.94-3.88)^{2}+(4.03-3.88)^{2} }{16-1}$ = 0.0198

(c) Now, we have to test the claim that population variance of annual wheat production for the first plot is larger than that for the second plot i.e.;

Null Hypothesis, $H_0$ : $\sigma_1^{2} = \sigma_2^{2}$ or $H_0$ : $\frac{\sigma_1^{2} }{\sigma_2^{2} } = 1$

Alternate Hypothesis, $H_0$ : $\sigma_1^{2} > \sigma_2^{2}$ or $H_0$ : $\frac{\sigma_1^{2} }{\sigma_2^{2} } > 1$

The test statistics used here is;

$\frac{s_1^{2} }{s_2^{2} }* \frac{\sigma_1^{2} }{\sigma_2^{2}}$ ~ $F_n__1-1_,n_2-1$ where, $n_1 = 16$ and $n_2 =16$

Test Statistics = $\frac{0.653 }{ 0.0198}* 1$ ~ $F_1_5_,_1_5$

= 32.98

Since, we are not provided with any significance level so we assume it to be 5% and at this level, the F table gives critical value of 2.4282.

<em>Since our test statistics is higher than the critical value and it falls in the rejection region so we have sufficient evidence to reject null hypothesis and conclude that the population variance of annual wheat production for the first plot is larger than that for the second plot.</em>

8 0

3 years ago