What is te value of x?

To test the belief that sons are taller than their fathers, a student randomly selects 13 fathers who have adult male children.

KATRIN_1 [288]

Answer:

1) B. The differences are normally distributed or the sample size is large

C. The sample size mus be large

E. The sampling method results in an independent sample

2) The null hypothesis H₀: $\bar x_1$ = $\bar x_2$

The alternative hypothesis Hₐ: $\bar x_1$ < $\bar x_2$

Test statistic, t = -0.00693

p- value = 0.498

Do not reject Upper H₀ because, the P-value is greater than the level of significance. There is sufficient evidence to conclude that sons are the same height as their fathers at 0.10 level of significance

Step-by-step explanation:

1) B. The differences are normally distributed or the sample size is large

C. The sample size mus be large

E. The sampling method results in an independent sample

2) The null hypothesis H₀: $\bar x_1$ = $\bar x_2$

The alternative hypothesis Hₐ: $\bar x_1$ < $\bar x_2$

The test statistic for t test is;

$t=\dfrac{(\bar{x}_1-\bar{x}_2)}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}-\dfrac{s _{2}^{2}}{n_{2}}}}$

The mean

Height of Father, h₁, Height of Son h₂

72.4, 77.5

70.6, 74.1

73.1, 75.6

69.9, 71.7

69.4, 70.5

69.4, 69.9

68.1, 68.2

68.9, 68.2

70.5, 69.3

69.4, 67.7

69.5, 67

67.2, 63.7

70.4, 65.5

$\bar x_1$ = 69.6

s₁ = 1.58

$\bar x_2$ = 69.9

s₂ = 3.97

n₁ = 13

n₂ = 13

$t=\dfrac{(69.908-69.915)}{\sqrt{\dfrac{3.97^{2}}{13}-\dfrac{1.58^{2} }{13}}}$

(We reversed the values in the square root of the denominator therefore, the sign reversal)

t = -0.00693

p- value = 0.498 by graphing calculator function

P-value > α Therefore, we do not reject the null hypothesis

Do not reject Upper H₀ because, the P-value is greater than the level of significance. There is sufficient evidence to conclude that sons are the same height as their fathers at 0.10 lvel of significance

8 0

4 years ago

Can anyone solve this? will give brainliest if correct.

miv72 [106K]

Answer:

See below

Step-by-step explanation:

<u>The missing reasons are:</u>

1. k. Given
2. j. Definition of parallelogram
3. d. Definition of linear pair
4. b. Linear pair postulate
5. e/m. Definition of supplementary
6. g. Same side interior angles theorem
7. e/m. Definition of supplementary
8. a/c. Substitution property of congruence
9. i. Subtraction property of congruence
10. f. Alternate interior angles theorem
11. l. Alternate exterior angles theorem
12. h. Angle congruence postulate

8 0

2 years ago

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PLEASE HELP WILL MARK BRAINLIEST

Romashka [77]

Answer:

A.The mean would increase.

Step-by-step explanation:

Outliers are numerical values in a data set that are very different from the other values. These values are either too large or too small compared to the others.

Presence of outliers effect the measures of central tendency.

The measures of central tendency are mean, median and mode.

The mean of a data set is a a single numerical value that describes the data set. The median is a numerical values that is the mid-value of the data set. The mode of a data set is the value with the highest frequency.

Effect of outliers on mean, median and mode:

Mean: If the outlier is a very large value then the mean of the data increases and if it is a small value then the mean decreases.
Median: The presence of outliers in a data set has a very mild effect on the median of the data.
Mode: The presence of outliers does not have any effect on the mode.

The mean of the test scores without the outlier is:

$\bar{x}=\frac{Total of the observations-Outlier value}{n-1} \\=\frac{(86*16)-72}{15} \\=\frac{1304}{15}\\ =86.9333$

*Here <em>n</em> is the number of observations.

So, with the outlier the mean is 86 and without the outlier the mean is 86.9333.

The mean increased.

Since the median cannot be computed without the actual data, no conclusion can be drawn about the median.

Conclusion:

After removing the outlier value of 72 the mean of the test scores increased from 86 to 86.9333.

Thus, the the truer statement will be that when the outlier is removed the mean of the data set increases.

4 0

3 years ago

carly mowed 2 more lawns than Roco each week for 6 weeks. carly mowed a total of 18 lawns. how many did Rocco mow each week?

mylen [45]

Answer:

Rocco mowed 1 lawn a week.

Step-by-step explanation:

If Carly mowed 18 lawns in six weeks that’s 3 lawns a week and 3-2=1.

Thus Rocco mowed 1 lawn a week.

4 0

3 years ago

−2x=x^2−6

Iteru [2.4K]

Step-by-step explanation:

Example 1

Solve the equation x3 − 3x2 – 2x + 4 = 0

We put the numbers that are factors of 4 into the equation to see if any of them are correct.

f(1) = 13 − 3×12 – 2×1 + 4 = 0 1 is a solution

f(−1) = (−1)3 − 3×(−1)2 – 2×(−1) + 4 = 2

f(2) = 23 − 3×22 – 2×2 + 4 = −4

f(−2) = (−2)3 − 3×(−2)2 – 2×(−2) + 4 = −12

f(4) = 43 − 3×42 – 2×4 + 4 = 12

f(−4) = (−4)3 − 3×(−4)2 – 2×(−4) + 4 = −100

The only integer solution is x = 1. When we have found one solution we don’t really need to test any other numbers because we can now solve the equation by dividing by (x − 1) and trying to solve the quadratic we get from the division.

Now we can factorise our expression as follows:

x3 − 3x2 – 2x + 4 = (x − 1)(x2 − 2x − 4) = 0

It now remains for us to solve the quadratic equation.

x2 − 2x − 4 = 0

We use the formula for quadratics with a = 1, b = −2 and c = −4.

We have now found all three solutions of the equation x3 − 3x2 – 2x + 4 = 0. They are: eftirfarandi:

x = 1

x = 1 + Ö5

x = 1 − Ö5

Example 2

We can easily use the same method to solve a fourth degree equation or equations of a still higher degree. Solve the equation f(x) = x4 − x3 − 5x2 + 3x + 2 = 0.

First we find the integer factors of the constant term, 2. The integer factors of 2 are ±1 and ±2.

f(1) = 14 − 13 − 5×12 + 3×1 + 2 = 0 1 is a solution

f(−1) = (−1)4 − (−1)3 − 5×(−1)2 + 3×(−1) + 2 = −4

f(2) = 24 − 23 − 5×22 + 3×2 + 2 = −4

f(−2) = (−2)4 − (−2)3 − 5×(−2)2 + 3×(−2) + 2 = 0 we have found a second solution.

The two solutions we have found 1 and −2 mean that we can divide by x − 1 and x + 2 and there will be no remainder. We’ll do this in two steps.

First divide by x + 2

Now divide the resulting cubic factor by x − 1.

We have now factorised

f(x) = x4 − x3 − 5x2 + 3x + 2 into

f(x) = (x + 2)(x − 1)(x2 − 2x − 1) and it only remains to solve the quadratic equation

x2 − 2x − 1 = 0. We use the formula with a = 1, b = −2 and c = −1.

Now we have found a total of four solutions. They are:

x = 1

x = −2

x = 1 +

x = 1 −

Sometimes we can solve a third degree equation by bracketing the terms two by two and finding a factor that they have in common.

6 0

3 years ago

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