Solution. 6) -162-5-X 8 9 10 11 12 14 15 16 17 18

Match the information on the left with the appropriate equation on the right.

Slav-nsk [51]

The equation of the line passing the points (-2,3),(-3,4) is y = -x + 1 or x+y = 1

<h3>Equation of a line</h3>

A line is the shortest distance between two points. The equation of a line in slope-intercept form is expressed as;

y = mx + b

m is the slope

b is the y-intercept

Given that m=1, b=-3, the required equation will be y = x - 3 or x - y = 3

For the parameter m=1, (-1,2)

y - 2 =1(x+1)

y - 2 = x+1

y =x + 3

Hence the equation of the line for the parameters m=1, (-1,2) is y = x + 3

For the coordinates (-2,3),(-3,4)

Slope = 4-3/-3+2
Slope = -1

For the intercept;

4 =-1(-3) + b

b = 4 - 3

b = 1

Hence the equation of the line passing the points (-2,3),(-3,4) is y = -x + 1 or x+y = 1

Learn more on equation of a line here: brainly.com/question/18831322

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5 0

2 years ago

3.5x-81=360 Please help this is due tomorrow

postnew [5]

Answer:

Step-by-step explanation:

so what your gonna do is you are going to add 81 to 360 and then sum it up. after you are going to divide it by 3.5.

This might be wrong but there is always a point where you have to try :)

3 0

2 years ago

Read 2 more answers

A driver descended to a depth of -65 feet in 5 minutes. If she descended at the same rate each minute, what’s the rate?

Sidana [21]

This is an easy one.

65/5= 13.

Therefore, she descended 13 feet per minute.

7 0

2 years ago

Solve this pls -5x - 10 = 10

Vanyuwa [196]

Step-by-step explanation:

-5x = 10 + 10

-5x = 20

X = 20/-5

X = -4

check :

-5 x -4 =20

20 =20

5 0

1 year 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

2 years ago