since it goes down a certain way and it goes down then B will go the same way as a evil ghost literally down to the phone the top right to the left like see if I just a little bit shorter Jay will go all the way up to age and we will go to k

4 0

2 years ago

Hmmmmmmmmmmmmmmmmmmmmmmmmmmm

belka [17]

$\dfrac{6}{7}cd=\dfrac{6}{7}\cdot c\cdot d\\\\Factors:\ \dfrac{6}{7},\ c,\ d\\\\6x^2+7xy+3+9\\\\Constans\ (without\ x\ and\ y):\ 3,\ 9\\\\2x-4y+8=(2x)+(-4y)+(8)\\\\Terms:\ 2x,\ -4y,\ 8$

4 0

3 years ago

Some transportation experts claim that it is the variability of speeds, rather than the level of speeds, that is a critical fact

scZoUnD [109]

Answer:

Explained below.

Step-by-step explanation:

The claim made by an expert is that driving conditions are dangerous if the variance of speeds exceeds 75 (mph)².

(1)

The hypothesis for both the test can be defined as:

H₀: The variance of speeds does not exceeds 75 (mph)², i.e. σ² ≤ 75.

Hₐ: The variance of speeds exceeds 75 (mph)², i.e. σ² > 75.

(2)

A Chi-square test will be used to perform the test.

The significance level of the test is, α = 0.05.

The degrees of freedom of the test is,

df = n - 1 = 55 - 1 = 54

Compute the critical value as follows:

$\chi^{2}_{\alpha, (n-1)}=\chi^{2}_{0.05, 54}=72.153$

Decision rule:

If the test statistic value is more than the critical value then the null hypothesis will be rejected and vice-versa.

(3)

Compute the test statistic as follows:

$\chi^{2}=\frac{(n-1)\times s^{2}}{\sigma^{2}}$

$=\frac{(55-1)\times 94.7}{75}\\\\=68.184$

The test statistic value is, 68.184.

Decision:

$cal.\chi^{2}=68.184$

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

Conclusion:

The variance of speeds does not exceeds 75 (mph)². Thus, concluding that driving conditions are not dangerous on this highway.

7 0

3 years ago