Solve for x. a = bx A) (-b)\a B) b - a C) (-a) + b D) (-a)/b

A company want to find out if there is a linear relationship between indirect labor expense (ILE), in dollars, and direct labor

trapecia [35]

Answer:

y = 115.0977+12.2992X

91.59%

Step-by-step explanation:

Given :

DLH(X) ILE(Y)

20 361_0

25 400

22 376

23 384

20 361_0

19 360

24 427.2

28 458.4

26 450_8

29 475.2

DLH (X) : Direct Labor Hours

ILE (Y) : Indirect Labor Expense

The regression model obtained by fitting the data using technology is :

y = 115.0977+12.2992X

Where ;

Intercept = 115.0977 ;

Slope = 12.2992

The Coefficient of determination, R² gives the proportion of explained variance due to the regression line.

For the data given above, the Coefficient of determination R² obtained using the Coefficient of determination calculator is 0.9159 ; which means that (0.9159 * 100%) about 91.59% of the variation in indirect labor expense is explained by the regression line while 8.41% is due to other factors.

7 0

3 years ago

Given the g(x)=f(x) + k, identify a value of k that transforms f into g

Yuliya22 [10]

Answer:

a vertical shift of 6 units up

Step-by-step explanation:

We have the transformation:

g(x) = f(x) + k

This is what we call a vertical shift, this transformation moves the graph of f(x) k units, and the motion is upwards if k is positive, and downwards if k is negative.

Here we can see that the graph of f(x) intersects the y-axis at y = -2

While the graph of g(x) intersects the y-axis at y = 4.

Then the distance between these two points is:

4 - (-2) = 6

This means that the graph of g(x) is 6 units above the graph of f(x)

Then we have that k must be equal to 6, then the transformation is:

g(x) = f(x) + 6

This is a vertical shift of 6 units up.

8 0

3 years ago

Part D

posledela

Can’t help you then...

7 0

3 years ago

Read 2 more answers

A simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute

GREYUIT [131]

Answer:

(a) Null Hypothesis, $H_0$ : $\sigma$ = 10 beats per minute

Alternate Hypothesis, $H_A$ : $\sigma\neq$ 10 beats per minute

(b) The value of chi-square test statistics is 35.704.

(c) P-value = 0.4360.

(d) We conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.

Step-by-step explanation:

We are given that a simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute.

If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute.

Let $\sigma$ = population standard deviation for the pulse rates of men.

(a) So, Null Hypothesis, $H_0$ : $\sigma$ = 10 beats per minute {means that the pulse rates of men have a standard deviation equal to 10 beats per minute}

Alternate Hypothesis, $H_A$ : $\sigma\neq$ 10 beats per minute {means that the pulse rates of men have a standard deviation different from 10 beats per minute}

The test statistics that will be used here is One-sample chi-square test for standard deviation;

T.S. = $\frac{(n-1)\times s^{2} }{\sigma^{2} }$ ~ $\chi^{2}__n_-_1$

where, s = sample standard deviation = 10.1 beats per minute

n = sample of men = 36

So, the test statistics = $\frac{(36-1)\times 10.1^{2} }{10^{2} }$ ~ $\chi^{2}__3_5$

= 35.70 4

(b) The value of chi-square test statistics is 35.704.

(c) Also, the P-value of the test statistics is given by;

P-value = P( $\chi^{2}__3_5$ > 35.704) = 0.4360

(d) Since the P-value of our test statistics is more than the level of significance as 0.4360 > 0.10, so we have insufficient evidence to reject our null hypothesis as the test statistics will not fall in the rejection region.

Therefore, we conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.

3 0

4 years ago

82. A company spends x hundred dollars on an advertising

crimeas [40]

The company spends $649 on advertising

The sales function after advertising is given as:

$S(x) = 5 + 7\ln(x + 1)$