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3 years ago

Plz help this due today

Mathematics

1 answer:

Butoxors [25]3 years ago

5 0

I saw two of these and they are both from you so i’m gonna try to help.. the last question on the last page is 4. I’m so sorry. I don’t know anything else. I tried looking it up but nothing worked i’m sorry. I just wish i could help and do more..

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4 years ago

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I don't understand proportional relationship can u help?

Reptile [31]

Answer: I’ll explain it in simpler terms for you. A proportional relationship is one in which two quantities vary directly with each other. Ratios are proportional if they represent the same relationship. One way to see if two ratios are proportional is to write them as fractions and then reduce them. If the reduced fractions are the same, your ratios are proportional. An example of a proportional relationship is simple interest, tax, markups and markdowns, gratuities and commissions, fees, percent increase and decrease, percent error. Hope this helps! :D

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3 years ago

At the begging of the month a store had a balance of -554

makvit [3.9K]

What is the question? Do you mean "At the beginning of the month a store had the balance of $554"? If so, what is the question?

5 0

4 years ago

Use inductive reasoning to predict the most probable next number in each list.

PolarNik [594]

Answer:

Step-by-step explanation:

First differences increase by 4 each time. The given numbers are a quadratic sequence that can be described by 2n² -2n +9. The 7th term would be 93.

_____

First differences are ...

13-9 = 4, 21-13 = 8, 33-21 = 12, 49-33 = 16, 69-49 = 20

Second differences are ...

8-4 = 4, 12-8 = 4, 16-12 = 4, 20-16 = 4

When second differences are constant, the sequence can be represented by a second-degree polynomial function. The leading coefficient is half the value of the second differences.

3 0

3 years ago

In a regression analysis involving 30 observations, the following estimated regressionequation was obtained.y^ =17.6+3.8x 1 −2.3

USPshnik [31]

Answer:

(a) There is a significant relationship between y and $x_1, x_2, x_3, x_4$

(b) $SSE_{(x_1 ,x_2 ,x_3 ,x_4) }= 45$

(d) $x_1$ and $x_4$ are significant

Step-by-step explanation:

Given

$y = 17.6+3.8x_1 - 2.3x_2 +7.6x_3 +2.7x_4$ --- estimated regression equation

$n = 30$

$p = 4$ --- independent variables i.e. x1 to x4

$SSR = 1760$

$SST = 1805$

$\alpha = 0.05$

Solving (a): Test of significance

We have:

$H_o :$ There is no significant relationship between y and $x_1, x_2, x_3, x_4$

$H_a :$ There is a significant relationship between y and $x_1, x_2, x_3, x_4$

First, we calculate the t-score using:

$t = \frac{SSR}{p} \div \frac{SST - SSR}{n - p - 1}$

$t = \frac{1760}{4} \div \frac{1805- 1760}{30 - 4 - 1}$

$t = 440 \div \frac{45}{25}$

$t = 440 \div 1.8$

$t = 244.44$

Next, we calculate the p value from the t score

Where:

$df = n - p - 1$

$df = 30 -4 - 1=25$

The p value when $t = 244.44$ and $df = 25$ is:

$p =0$

So:

$p < \alpha$ i.e. $0 < 0.05$

Solving (b): $SSE(x_1 ,x_2 ,x_3 ,x_4)$

To calculate SSE, we use:

$SSE = SST - SSR$

Given that:

$SSR = 1760$ ----------- $(x_1 ,x_2 ,x_3 ,x_4)$

$SST = 1805$

So:

$SSE_{(x_1 ,x_2 ,x_3 ,x_4)} = 1805 - 1760$

$SSE_{(x_1 ,x_2 ,x_3 ,x_4) }= 45$

Solving (c): $SSE(x_2 ,x_3)$

To calculate SSE, we use:

$SSE = SST - SSR$

Given that:

$SSR = 1705$ ----------- $(x_2 ,x_3)$

$SST = 1805$

So:

$SSE_{(x_2,x_3)} = 1805 - 1705$

$SSE_{(x_2,x_3)} = 100$

Solving (d): F test of significance

The null and alternate hypothesis are:

We have:

$H_o :$ $x_1$ and $x_4$ are not significant

$H_a :$ $x_1$ and $x_4$ are significant

For this model:

$y =11.1 -3.6x_2+8.1x_3$

$SSE_{(x_2,x_3)} = 100$

$SST = 1805$

$SSR_{(x_2 ,x_3)} = 1705$

$SSE_{(x_1 ,x_2 ,x_3 ,x_4) }= 45$

$p_{(x_2,x_3)} = 2$

$\alpha = 0.05$

Calculate the t-score

$t = \frac{SSE_{(x_2,x_3)}-SSE_{(x_1,x_2,x_3,x_4)}}{p_{(x_2,x_3)}} \div \frac{SSE_{(x_1,x_2,x_3,x_4)}}{n - p - 1}$

$t = \frac{100-45}{2} \div \frac{45}{30 - 4 - 1}$

$t = \frac{55}{2} \div \frac{45}{25}$

$t = 27.5 \div 1.8$

$t = 15.28$

Next, we calculate the p value from the t score

Where:

$df = n - p - 1$

$df = 30 -4 - 1=25$

The p value when $t = 15.28$ and $df = 25$ is:

$p =0$

So:

$p < \alpha$ i.e. $0 < 0.05$

<em>Hence, we reject the null hypothesis</em>

7 0

3 years ago