Which situation can be modeled by an exponential growth function?

Ralph owns a circular farm as shown. He is planning to fence a triangular pigpen ABC in the farm. Point M is the midpoint of the

Flauer [41]

Answer:

$Total = \$942$

Step-by-step explanation:

Given

$A = (0,0)$

$C = (3,4)$

See attachment for farm

Required

The cost of fencing the farm

First, calculate the radius of the farm.

The radius is represented with AC.

So, we have:

$AC = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$ ---- distance formula

So, we have:

$AC = \sqrt{(0 - 3)^2 + (0 - 4)^2}$

$AC = \sqrt{(- 3)^2 + ( - 4)^2}$

$AC = \sqrt{9+16}$

$AC = \sqrt{25}$

$AC = 5$

Hence:

$r = 5$ --- radius

Calculate the circumference of the circle

$C = 2\pi r$

$C = 2 * 3.14 * 5$

$C = 31.4yd$

If 1 yard costs $30, 31.4 yards will cost:

$Total = 31.4 * 30$

$Total = \$942$

5 0

2 years ago

Which products result in a difference of squares or a perfect square trinomial? Check all that apply.

GrogVix [38]

All you need to do is convert all equations into standard form by using the distributive property on each equation.

6 0

3 years ago

Read 2 more answers

What is a congruent triangle

sweet-ann [11.9K]

When two triangles are congruent they will have exactly the same three sides and exactly the same three angles. The equal sides and angles may not be in the same position (if there is a turn or a flip), but they are there

4 0

3 years ago

I’LL MARK AS BRAINLIEST!!! Plz no links

Rufina [12.5K]

Answer:

6 people

Step-by-step explanation:

75% of 8 is 6

multiply 8 by the decimal .75

4 0

3 years ago

A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. Assume that the distr

Marianna [84]

Answer:

(a) Decision rule for 0.01 significance level is that we will reject our null hypothesis if the test statistics does not lie between t = -2.651 and t = 2.651.

(b) The value of t test statistics is 1.890.

(c) We conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) P-value of the test statistics is 0.0662.

Step-by-step explanation:

We are given that a recent study focused on the number of times men and women who live alone buy take-out dinner in a month.

Also, following information is given below;

Statistic : Men Women

The sample mean : 24.51 22.69

Sample standard deviation : 4.48 3.86

Sample size : 35 40

Let $\mu_1$ = mean number of times men order take-out dinners in a month.

$\mu_2$ = mean number of times women order take-out dinners in a month

(a) So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2$ = 0 {means that there is no difference in the mean number of times men and women order take-out dinners in a month}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2\neq$ 0 {means that there is difference in the mean number of times men and women order take-out dinners in a month}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_-_n_2_-_2$

where, $\bar X_1$ = sample mean for men = 24.51

$\bar X_2$ = sample mean for women = 22.69

$s_1$ = sample standard deviation for men = 4.48

$s_2$ = sample standard deviation for women = 3.86

$n_1$ = sample of men = 35

$n_2$ = sample of women = 40

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(35-1)\times 4.48^{2}+(40-1)\times 3.86^{2} }{35+40-2} }$ = 4.16

So, test statistics = $\frac{(24.51-22.69)-(0)}{4.16 \sqrt{\frac{1}{35}+\frac{1}{40} } }$ ~ $t_7_3$

= 1.890

(b) The value of t test statistics is 1.890.

(c) Now, at 0.01 significance level the t table gives critical values of -2.651 and 2.651 at 73 degree of freedom for two-tailed test.

Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) Now, the P-value of the test statistics is given by;

P-value = P( $t_7_3$ > 1.89) = 0.0331

So, P-value for two tailed test is = 2 $\times$ 0.0331 = 0.0662

4 0

3 years ago