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3 years ago

Y is greater than or equal to 3and less than 7

Mathematics

1 answer:

In-s [12.5K]3 years ago

5 0

Answer:

7 > y ≥ 3

Step-by-step explanation:

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When pointing A(-6. - 12) is reflected over the y-axis to get point B, what are the

Fudgin [204]

Answer:

Step-by-step explanation:

under a reflection in the y- axis

a point (x, y ) → (- x, y ) , then

A (- 6, - 12 ) → B (6, - 12 )

4 0

2 years ago

Please help me on this

Marizza181 [45]

Answer: m=2.8

Step-by-step explanation:

You can reverse the equation and instead solve...

4-1.2 which equals 2.8.

m=2.4

8 0

3 years ago

Read 2 more answers

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

4 years ago

I have some answers the these problems but I just wanna make sure there right.. PLEASE Help??

yanalaym [24]

7x+12+63=180
8x+12=180
Subtract 12 from both sides
8x= 168
=c=21

10x-40=50
Add 40 to both sides
10x=90
X=9

4x+72+2x-12=180
6x +60 =180
Subtract 60 from both sides
6x=120
X=20

3 0

3 years ago

In a beach town, 13% of the residents own boats. A random sample of 100 residents was selected. What is the probability that les

mariarad [96]

Answer:

Approximately $0.038$ (or equivalently $3.8\%$ ,) assuming that whether each resident owns boats is independent from one another.

Step-by-step explanation:

Assume that whether each resident of this town owns boats is independent from one another. It would be possible to model whether each of the $n = 100$ selected residents owns boats as a Bernoulli random variable: for $k = \lbrace 1,\, \dots,\, 100\rbrace$ , $X_k \sim \text{Bernoulli}(\underbrace{0.13}_{p})$ .

$X_k = 0$ means that the $k$ th resident in this sample does not own boats. On the other hand, $X_k = 1$ means that this resident owns boats. Therefore, the sum $(X_1 + \cdots + X_{100})$ would represent the number of residents in this sample that own boats.

Each of these $100$ random variables are all independent from one another. The mean of each $X_k$ would be $\mu = 0.13$ , whereas the variance of each $X_k\!$ would be $\sigma = p\, (1 - p) = 0.13 \times (1 - 0.13) = 0.1131$ .

The sample size of $100$ is a rather large number. Besides, all these samples share the same probability distribution. Apply the Central Limit Theorem. By this theorem, the sum $(X_1 + \cdots + X_{100})$ would approximately follow a normal distribution with:

mean $n\, \mu = 100 \times 0.13 = 13$ , and
variance $\sigma\, \sqrt{n} = p\, (1 - p)\, \sqrt{n} = 0.1131 \times 10 = 1.131$ .

$11\%$ of that sample of $100$ residents would correspond to $11\% \times 100 = 11$ residents. Calculate the $z$ -score corresponding to a sum of $11$ :

$\begin{aligned}z &= \frac{11 - 13}{1.131} \approx -1.77 \end{aligned}$ .

The question is (equivalently) asking for $P( (X_1 + \cdots + X_{100}) < 11)$ . That is equal to $P(Z < -1.77)$ . However, some $z$ -tables list only probabilities like $P(Z > z)$ . Hence, convert $P(Z < -1.77)\!$ to that form: