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Over [174]
3 years ago
13

Answer these two questions Please

Mathematics
1 answer:
alekssr [168]3 years ago
7 0

Answer:

2.

a) likely

b) unlikely

c) equally likely to occur or not occur

d) certain

e) unlikely

3.

In order of least to most likely to occur

A C B D

P(A) 0

P(B) 4/9

P(C) 1/9

P(D) 5/9

I hope this was helpful :-)

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What is the solution to the equation below?<br> log7+log(x-4)= 1
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Answer: x=38/7

Step-by-step explanation:

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3 years ago
What expression is equivalent to 6+3(n-4)-8+2n
lorasvet [3.4K]

Answer:

Step-by-step explanation:

Comment

Begin by removing the brackets.

6+3(n-4)-8+2n

6 + 3n - 12 - 8 + 2n                 Collect like terms

3n + 2n + 6 - 12 - 8                 Combine

5n - 14

Answer

5n - 14

4 0
2 years ago
Find the p-value: An independent random sample is selected from an approximately normal population with an unknown standard devi
vladimir1956 [14]

Answer:

(a) <em>p</em>-value = 0.043. Null hypothesis is rejected.

(b) <em>p</em>-value = 0.001. Null hypothesis is rejected.

(c) <em>p</em>-value = 0.444. Null hypothesis is not rejected.

(d) <em>p</em>-value = 0.022. Null hypothesis is rejected.

Step-by-step explanation:

To test for the significance of the population mean from a Normal population with unknown population standard deviation a <em>t</em>-test for single mean is used.

The significance level for the test is <em>α</em> = 0.05.

The decision rule is:

If the <em>p - </em>value is less than the significance level then the null hypothesis will be rejected. And if the <em>p</em>-value is more than the value of <em>α</em> then the null hypothesis will not be rejected.

(a)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> > <em>μ₀</em>

The sample size is, <em>n</em> = 11.

The test statistic value is, <em>t</em> = 1.91 ≈ 1.90.

The degrees of freedom is, (<em>n</em> - 1) = 11 - 1 = 10.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of 1.90 and degrees of freedom 10 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₁₀ > 1.91) = 0.043.

The <em>p</em>-value = 0.043 < <em>α</em> = 0.05.

The null hypothesis is rejected at 5% level of significance.

(b)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> < <em>μ₀</em>

The sample size is, <em>n</em> = 17.

The test statistic value is, <em>t</em> = -3.45 ≈ 3.50.

The degrees of freedom is, (<em>n</em> - 1) = 17 - 1 = 16.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of -3.50 and degrees of freedom 16 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₁₆ < -3.50) = P (t₁₆ > 3.50) = 0.001.

The <em>p</em>-value = 0.001 < <em>α</em> = 0.05.

The null hypothesis is rejected at 5% level of significance.

(c)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> ≠ <em>μ₀</em>

The sample size is, <em>n</em> = 7.

The test statistic value is, <em>t</em> = 0.83 ≈ 0.82.

The degrees of freedom is, (<em>n</em> - 1) = 7 - 1 = 6.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of 0.82 and degrees of freedom 6 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₆ < -0.82) + P (t₆ > 0.82) = 2 P (t₆ > 0.82) = 0.444.

The <em>p</em>-value = 0.444 > <em>α</em> = 0.05.

The null hypothesis is not rejected at 5% level of significance.

(d)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> > <em>μ₀</em>

The sample size is, <em>n</em> = 28.

The test statistic value is, <em>t</em> = 2.13 ≈ 2.12.

The degrees of freedom is, (<em>n</em> - 1) = 28 - 1 = 27.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of 0.82 and degrees of freedom 6 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₂₇ > 2.12) = 0.022.

The <em>p</em>-value = 0.444 > <em>α</em> = 0.05.

The null hypothesis is rejected at 5% level of significance.

5 0
3 years ago
Factor the polynomial. 7x2 + 68xy - 20y2 A) (7x - 2y)(x + 10y) B) (7x - 10y)(x - 2y) C) (7x + 10y)(x + 2y) D) (7x + 2y)(x - 10y)
Otrada [13]
For this case we have the following polynomial:
 7x2 + 68xy - 20y2
 Factoring we have:
 (7x-2y) (x + 10y)
 We verify the factorization:
 7x2 + 70xy - 2xy - 20y2
 Rewriting we have:
 7x2 + 68xy - 20y2
 Therefore, the factorization is correct.
 Answer:
 
A) (7x - 2y) (x + 10y)
5 0
3 years ago
Second​ dose, 122 of 676 subjects in the experimental group​ (group 1) experienced fever as a side effect. After the second​ dos
Oksanka [162]

Answer:

Step-by-step explanation:

Download docx
5 0
3 years ago
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