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Alenkasestr [34]

3 years ago

9

find the slope and y-intercept for a line passing through the following two points,then create a slope intercept equation (-2,2)

and (-5,-4)

1 answer:

slava [35]3 years ago

4 0

Answer:

Slope: 2

y intercept: 6

Equation: y = 2x+6

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B. Convert quarts to gallons.

jeyben [28]

Answer:

There are 4 quarts in one gallon. 36 divided by 4 is 9 gallons.

Step-by-step explanation:

9 gallons.

5 0

2 years ago

A test has 20 questions What percent I get if answered 17 questions correct.

bazaltina [42]

Number of answer right divided by total number of questions:

17/20 = 85%

4 0

2 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

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

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

(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 <u>Two-sample t test statistics</u> 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, <u>test statistics</u> = $\frac{(24.51-22.69)-(0)}{4.16 \sqrt{\frac{1}{35}+\frac{1}{40} } }$ ~ $t_7_3$

= 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 <u>we fail to reject our null hypothesis</u>.

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 = <u>0.0662</u>

4 0

2 years ago

3 regions are defined in the figure find the volume generated by rotating the given region about the specific line

anastassius [24]

The volume generated by rotating the given region $R_{3}$ about OC is $\frac{4}{g} \pi$

<h3>Washer method</h3>

Because the given region ( $R_{3}$ ) has a look like a washer, we will apply the washer method to find the volume generated by rotating the given region about the specific line.

solution

We first find the value of x and y

$y=2(x)^{\frac{1}{4} }$

$x=(\frac{y}{2} )^{4}$

$y=2x$

$x=\frac{y}{2}$

$\int\limits^a_b {\pi } \, (R_{o^{2} } - R_{i^{2} } ) dy$

$R_{o} = x = \frac{y}{2}$

$R_{i} = x= (\frac{y}{2}) ^{4}$

$a=0, b=2$

$v= \int\limits^2_o {\pi } \, [(\frac{y}{2})^{2} - ((\frac{y}{2}) ^{4} )^{2} ) dy$

$v= \pi \int\limits^2_o= [\frac{y^{2} }{4} - \frac{y^{8} }{2^{8} }} ] dy$

$v= \pi [\int\limits^2_o {\frac{y^{2} }{4} } \, dy - \int\limits^2_o {\frac{y}{2^{8} } ^{8} } \, dy ]$

$v=\pi [\frac{1}{4} \frac{y^{3} }{3} \int\limits^2_0 - \frac{1}{2^{8} } \frac{y^{g} }{g} \int\limits^2_o\\v= \pi [\frac{1}{12} (2^{3} -0)-\frac{1}{2^{8}*9 } (2^{g} -0)]\\v= \pi [\frac{2}{3} -\frac{2}{g} ]\\v= \frac{4}{g} \pi$

A similar question about finding the volume generated by a given region is answered here: brainly.com/question/3455095

6 0

1 year ago

Mari used a thermometer to record temperatures of -3.4 Celsius and 1.6 Celsius . Which temperature in degrees is lower

madam [21]

I think is would be -3.4

3 0

2 years ago