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Papessa [141]
3 years ago
12

What is the purpose of sampling?

Mathematics
1 answer:
Kryger [21]3 years ago
6 0

Answer:

A. to make inferences about a sampling technique

Step-by-step explanation:


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Lines c and d are parallel lines cut by transversal p.
lilavasa [31]

Answer:

Option (B).

Step-by-step explanation:

Here there are two parallel lines c and d cuts by a transversal p.

The angles are formed as shown in diagram.

Here,

\angle 1 = \angle 7 (alternate)\\\\\angle 2 = \angle 6 (corresponding)\\\\\angle 3 = \angle 5 (alternate)\\\\\angle 5 = \angle 7 (alternate)

So, the option (B) is correct.

5 0
3 years ago
I really need help if anyone can help that would be great
Oliga [24]
Factor the denominator:

(3x-1)(x+2)

every time x=-2 (because of (x+2), it would cause a warp in the graph, and not have an input. that is the vertical asymptote.

since you have factor 3x-1 on top and bottom, in which the zero is 1/3, there would be no solution at x=1/3. it would be a hole.

the answer here is C.
5 0
4 years ago
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uranmaximum [27]
It's the second one for sure
6 0
3 years ago
Solve for n <br> 4 + 3n = –6 + 4n
Nikolay [14]

Answer: 10

Step-by-step explanation:

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
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