Identify which functions have complex roots by selecting the function names on the provided coordinate plane,

Find the unit rate. Enter your answer as a mixed number.

lesya [120]

The unit rate is the change of a quantity over one unit of another thus the unit rate of fertilizer covers is (30/8) foot per hour.

<h3>What is the rate of change?</h3>

The rate of change is the change of a quantity over 1 unit of another quantity.

Most of the time the rate of change is the change with respect to time.

For example the speed 3meter/second.

As per the given,

A fertilizer covers 5/8 square feet in 1/6 hour.

5/8 square feet ⇒ 1/6 hour

Multiply both sides by 6

6 × 5/8 square feet ⇒ 6/6 hour

30/8 square feet ⇒ 1 hour.

Hence "The unit rate is the change of a quantity over one unit of another thus the unit rate of fertilizer covers is (30/8) foot per hour".

To learn more about the rate of change here,

brainly.com/question/12786410

#SPJ1

5 0

1 year ago

What is the value of x? Round your answer to the nearest tenth if necessary.

Nastasia [14]

X is equal to 1.2, I think.

6 0

3 years ago

Read 2 more answers

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