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2 years ago

10

Write an equation of a line, in slope-intercept form, that is perpendicular to the line y=2x+5 and passes through the point (8,-

3)

1 answer:

Eva8 [605]2 years ago

7 0

Answer:

22

Step-by-step explanation:

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Any help, please! <br> I dont understand

frutty [35]

The answer would be 40% :)

3 0

2 years ago

a local hamburger shop sold a combined total of 613 hamburgers and cheeseburger on Friday. there were 63 more cheeseburger sold

kondaur [170]

Answer:

275 hamburgers was sold on Friday

3 0

1 year ago

If you could help I would really appreciate it, but if not that’s fine. Thank you.

dolphi86 [110]

Answer:

3

Step-by-step explanation:

j(2) means that x equals 2...

the first inequality in the function says that x is greater than or equal to 2

the second inequality states that x is greater than 2(not equal to) and is less than or equal to 5

the third one states that x is greater than 5

x applies to the first inequality since we know that x can be equal to 2

Now, to the number before the if

there is a 3 before the if in the first inequality

This means that if x is greater than or equal to 2 the y or the function would equal 3

therefore, j(2) would equal 3

I hope this helps!!!

6 0

2 years ago

1........... 5x (3x +2)

babymother [125]

You do 3832083082x 32u393829832983x to get this 5,418

7 0

2 years ago

Archaeologists can determine the diets of ancient civilizations by measuring the ratio of carbon-13 to carbon-12 in bones found

Vsevolod [243]

Answer:

Step-by-step explanation:

The null and the alternative hypothesis are:

H₀=μ₀=μ₁=μ₂=μ₃=μ₄

H₁=two or more μ are different X

Let $X_{ij}$ denote the jth observation of the ith sample. The mean and the variance of the ith sample is given by

$Xbar_{i}$ = $\frac{1}{J_{i}} summation(X_{ij})$ where J=1 to $J_{i}$ and $J_{i}$ is ith sample size

$s_{i} ^{2} =\frac{1}{J_{i}-1 }summation (X_{ij} - Xbar_{i})^{2}$

The sample sizes are J₁ =12 J₂=10 J₃=18 J₄=9

the total number in all samples combined is 49

finding Xbar₁ and s₁

Xbar₁= 1÷12(17.2+....+13.4)

Xbar₁= 17.0500

s₁²= 1÷(12-1) [(17.0500-17.2)+...+(17.0500-13.4)]

s₁²=5.1336

Similarly find the means and variances of other samples

$\left[\begin{array}{ccc}i&Mean&variance\\1&17.0500&5.1336\\2&15.4500&13.2317\\3&16.4972&6.9460\\4&15.4444&7.4428\end{array}\right]$

the sample grand mean denoted by Xbar is the average of all sampled items taken together:

Xbar= $\frac{1}{49}$ (17.2+.....+16.7)

Xbar=16.2255

Compute the treatment sum of squares SSTr, the sum of squares SSE and the total sum of squares SST

SSTr = ∑ $J_{i} (Xbar_{i} -Xbar)^{2}$ from i=1 to 49

SSTr= 20.9910

$SSE= \sum\limits^I_{i=1}\sum\limits^{J_i}_{j=1}(Xbar_{ij}-Xbar_i)^{2}$

$SSE=\sum\limits^I_{i=1}(J_i-1)s_i^{2}$

SSE=(12-1)(5.1336)+...(9-1)(7.4428)

SSE=353.1796

SST=SSTr+SSE

SST=374.1706

Find the treatment mean square MSTr and the error mean square MSE:

MSTr= SSTr/(I-1)

MSTr=6.9970

MSE=SSE/(N-I)

MSE=7.8484

$F=\frac{MSTr}{MSE}$

F=0.89

The degrees of freedom are I-1=3 for the numerator and N-I=45 for the denominator. Under H₀, F has an $F_{3,45}$ distribution. To find the P value we consult the F table.

P>0.100

The complete ANOVA table is below

$\left[\begin{array}{cccccc}source&DF&SS&MS&F&P\\Agegroup&3&20.9910&6.9970&0.89&0.453\\Error&45&353.1796&7.8484\\Total&48&374.1706\end{array}\right]$

(b) The P-value is large. A follower of 5% rule would not reject H₀. Therefore, we cannot conclude the concentration ratios differ among the age groups

4 0

2 years ago