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murzikaleks [220]
3 years ago
12

In the simple linear regression model, the ____________ accounts for the variability in the dependent variable that cannot be ex

plained by the linear relationship between the variables. a. error term b. constant term c. residual d. model parameter
Mathematics
1 answer:
Elanso [62]3 years ago
5 0

Answer:jfjsbdjfbsj

Step-by-step explanation:

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Find abc !!!!!!!! !!!!!
KonstantinChe [14]

Answer:

58 degreess

3x+11=4x-5

-3x      -3x

11=x-5

+5   +5

16=x

x=6

3(6)+11

18+11

29

29*2

58

Step-by-step explanation:

8 0
2 years ago
In most microcomputers the addresses of memory locations are specified in hexadecimal. These addresses are sequential numbers th
iren [92.7K]

Considering that the addresses of memory locations are specified in hexadecimal.

a) The number of memory locations in a memory address range ( 0000₁₆ to FFFF₁₆ )  = 65536 memory locations

b) The range of hex addresses in a microcomputer with 4096 memory locations is ;  4095

<u>applying the given data </u>:

a) first step : convert FFFF₁₆ to decimal           ( note F₁₆ = 15 decimal )

( F * 16^3 ) + ( F * 16^2 ) + ( F * 16^1 ) + ( F * 16^0 )

= ( 15 * 16^3 ) + ( 15 * 16^2 ) + ( 15 * 16^1 ) + ( 15 * 1 )

=  61440 + 3840 + 240 + 15 = 65535

∴ the memory locations from  0000₁₆ to FFFF₁₆ = from 0 to 65535 = 65536 locations

b) The range of hex addresses with a memory location of 4096

= 0000₁₆ to FFFF₁₆ =  0 to 4096

∴ the range = 4095

Hence we can conclude that the memory locations in ( a ) = 65536 while the range of hex addresses with a memory location of 4096 = 4095.

Learn more : brainly.com/question/18993173

6 0
2 years ago
Help pls and explain thank you :)
Solnce55 [7]
25 because Becky read 45 and Jason read 20

45-20 = 25
5 0
3 years ago
Read 2 more answers
Some researchers have conjectured that stem-pitting disease in peach-tree seedlings might be controlled with weed and soil treat
stiks02 [169]

Answer:

Part A

b. 14.6 ± 7.38

Part B

b. 3.43

Part C

a. P-value < 0.01

Part D

b. There is sufficient evidence to reject the null hypothesis

Step-by-step explanation:

Part A

The given data are;

The number of seedlings in the field = 20

The number of seedlings selected to receive herbicide A = 10

The number of seedlings selected to receive herbicide B = 10

The height in centimeters of seedlings treated with Herbicide A, \overline x _1 = 94.5 cm

The standard deviation, s₁ = 10 cm

The height in centimeters of seedlings treated with Herbicide B, \overline x _2 = 109.1 cm

The standard deviation, s₂ = 9 cm

The 90% confidence interval for μ₂ - μ₁, is given as follows;

\left (\bar{x}_{2}- \bar{x}_{1}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

The critical-t at 95% and n₁ + n₂ - 2 degrees of freedom is given as follows;

The degrees of freedom, df = n₁ + n₂ - 2 = 10 + 10 - 2 = 18

α = 100% - 90% = 10%

∴ For two tailed test, we have, α/2 = 10%/2 = 5% = 0.05

t_{(0.025, \, 18)} = 1.734

C.I. = \left (109.1- 94.5 \right )\pm 1.734 \times \sqrt{\dfrac{10^{2}}{10}+\dfrac{9^{2}}{10}}

C.I. ≈ 14.6 ± 7.37714603353

The 90% C.I. ≈ 14.6 ± 7.38

b. 14.6 ± 7.38

Part B

With the hypotheses are given as follows;

H₀; μ₂ - μ₁ = 0

Hₐ; μ₂ - μ₁ ≠ 0

The two sample t-statistic is given as follows;

t=\dfrac{(\bar{x}_{2}-\bar{x}_{1})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}+\dfrac{s _{2}^{2}}{n_{2}}}}

t-statistic=\dfrac{(109.1-94.5)}{\sqrt{\dfrac{10^{2} }{10}+\dfrac{9^{2}}{10}}} \approx 3.43173361147

The two sample t-statistic ≈ 3.43

b. 3.43

Part C

From the t-table, the p-value, we have, the p-value < 0.01

a. P-value < 0.01

Part D

Given that a significance level of 0.05 level is used and the p-value of 0.01 is less than the significance level, there is enough statistical evidence to reject the null hypothesis

b. There is sufficient evidence to reject the null hypothesis.

4 0
3 years ago
Jackie received 78% of the votes in an election. in a random sample of 40 eligible voters, 28 said they voted for jackie. which
Artyom0805 [142]
The answer to this question is C since <span>78% of the votes in an election. in a random sample of 40 eligible voters, 28 said they voted for jackie.</span>
6 0
3 years ago
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