Kx + 3x = 4 solve for x.

The owner of a sandwich shop researches the cost of buying different numbers of pounds of sandwich meat. The owner creates a sca

IgorC [24]

Answer:

what r the statementzsssssssssssssssssssssssssssssssssss

Step-by-step explanation:

3 0

2 years ago

Conscientiousness is a tendency to show self-discipline, act dutifully, and aim for achievement. The trait shows a preference fo

MAVERICK [17]

Answer:

a

The Null hypothesis represented as

$H_0: \mu_1 - \mu_2 = 0$

The Alternative hypothesis represented as

$H_a: \mu_1 - \mu_2 < 0$

b

p-value = P(Z < -3.37 ) = 0.000376

c

There is insufficient evidence to conclude that graduate students score higher, on average, on the HPI than undergraduate students

Step-by-step explanation:

From the question we are told that

The population size is $n= 650$

The sample size for graduates is $n_1 = 300$

The sample mean for graduates is $\= x _1 = 148$

The sample standard deviation for graduates is $\sigma_1 = 16$

The sample size for under-graduates is $n _2 = 350$

The sample mean for under-graduates is $\= x _2 = 153$

The sample standard deviation for graduates is $\sigma_2 = 21$

The Null hypothesis represented as

$H_0: \mu_1 - \mu_2 = 0$

The Alternative hypothesis represented as

$H_a: \mu_1 - \mu_2 < 0$

Where $\mu_1 \ and \ \mu_2$ are the population mean

Now the test statistic is mathematically represented as

$t = \frac{(\= x_1 - \= x_2 ) }{ \sqrt{ \frac{ (n_1 - 1 )\sigma_1 ^2 + (n_2 - 1)\sigma_2^2}{n_1 +n_2 -2} } * \sqrt{ \frac{1}{n_1} + \frac{1}{n_2} } }$

substituting values

$t = \frac{(148 - 153 ) }{ \sqrt{ \frac{ (300- 1 )16 ^2 + (350 - 1) 21^2}{300 +350 -2} } * \sqrt{ \frac{1}{300} + \frac{1}{350} } }$

$t = -3.37$

The p-values is mathematically evaluated as

p-value = P(Z < -3.37 ) = 0.000376

The above answer is gotten using a p-value calculator at (0.05) level of significance

Looking the p-value we see that it is less than the level of significance (0.05) so Null hypothesis is rejected

Hence there is insufficient evidence to conclude that graduate students score higher, on average, on the HPI than undergraduate students

6 0

2 years ago

Simplify: –3(y + 2)2 – 5 + 6y What is the simplified product in standard form?

Lina20 [59]

-3(y+2)^2-5+6y
-3(y^2+4y+4)-5+6y
-3y^2-12y-12-5+6y
-3y^2-6y-17

8 0

2 years ago

Anna set up a lemonade stand on her block over the summer. She recorded each day’s high temperature and the number of cups of le

Alexxx [7]

Answer:

Residual = -2

The negative residual value indicates that the data point lies below the regression line.

Step-by-step explanation:

We are given a linear regression model that relates daily high temperature, in degrees Fahrenheit and number of lemonade cups sold.

$y = -34+ \frac{3}{5} x$

Where y is the number of cups sold and x is the daily temperature in Fahrenheit.

Residual value:

A residual value basically shows the position of a data point with respect to the regression line.

A residual value of 0 is desired which means that the regression line best fits the data.

The Residual value is calculated by

Residual = Observed value - Predicted value

The predicted value of number of lemonade cups is obtained as

$y = -34+ \frac{3}{5} (95)\\\\y = -34+ 3 (19)\\\\y = -34+ 57\\\\y = 23$

So the predicted value of number of lemonade cups is 23 and the observed value is 21 so the residual value is

Residual = Observed value - Predicted value

Residual = 21 - 23

Residual = -2

The negative residual value indicates that the data point lies below the regression line.

7 0

2 years ago

"Immediately after a ban on using hand-held cell phones while driving was implemented, compliance with the law was measured. A r

sergiy2304 [10]

Answer:

(a) Null Hypothesis, $H_0$ : $p_1-p_2=0$ or $p_1= p_2$

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq 0$ or $p_1\neq p_2$

(b) We conclude that there is a statistical difference in these two proportions measured initially and then one year later.

Step-by-step explanation:

We are given that a random sample of 1,250 drivers found that 98.9% were in compliance. A year after the implementation, compliance was again measured to see if compliance was the same (or not) as previously measured.

A different random sample of 1,100 drivers found 96.9% compliance."

Let $p_1$ = proportion of drivers that were in compliance initially

$p_2$ = proportion of drivers that were in compliance one year later

(a) Null Hypothesis, $H_0$ : $p_1-p_2=0$ or $p_1= p_2$ {means that there is not any statistical difference in these two proportions measured initially and then one year later}

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq 0$ or $p_1\neq p_2$ {means that there is a statistical difference in these two proportions measured initially and then one year later}

The test statistics that will be used here is Two-sample z proportion statistics;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{ \frac{\hat p_1(1-\hat p_1)}{n_1} + \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of drivers in compliance initially = 98.9%

$\hat p_2$ = sample proportion of drivers in compliance one year later = 96.9%

$n_1$ = sample of drivers initially = 1,250

$n_2$ = sample of drivers one year later = 1,100

(b) So, the test statistics = $\frac{(0.989-0.969)-(0)}{\sqrt{ \frac{0.989(1-0.989)}{1,250} + \frac{0.969(1-0.969)}{1,100}} }$

= 3.33

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

P-value = P(Z > 3.33) = 1 - P(Z $\leq$ 3.33)

= 1 - 0.99957 = 0.00043

Since in the question we are not given with the level of significance so we assume it to be 5%. Now at 5% significance level, the z table gives critical values between -1.96 and 1.96 for two-tailed test.

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

Therefore, we conclude that there is a statistical difference in these two proportions measured initially and then one year later.

7 0

2 years ago