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Arte-miy333 [17]
3 years ago
14

a pizza costs $11 and $1.50 per topping write a equation for cost in dollars and the nunber of toppings ​

Mathematics
2 answers:
romanna [79]3 years ago
8 0

Answer:

y= 11 + 1.50x. (y is the price, x is number of toppings)

Step-by-step explanation:

olga55 [171]3 years ago
5 0
Y= 11+1.50x because y is the total, x is your independent variable, and 11 is ur baseline
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Step-by-step explanation:

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$104 For 8 Hours

Step-by-step explanation:

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Evgen [1.6K]

The idea of using 4 because it's a factor is the wrong choice. If the equation were 5x = 88 and you divide by 4, you aren't getting a single 1x alone. So the "divide because it's a factor" choice is wrong.




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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

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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)

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SST=SSTr+SSE

SST=374.1706

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

MSTr= SSTr/(I-1)

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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

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