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Sedaia [141]
2 years ago
14

Which value is equivalent to 30 + 9 divided by (6 - 3) A) 13 B)30 C(32 D)33

Mathematics
2 answers:
KonstantinChe [14]2 years ago
6 0

Answer:

the answer is D 33

Step-by-step explanation:

TEA [102]2 years ago
4 0

Answer:

13

Step-by-step explanation:

30 plus 9 is 39

6 minus 3 is 3

39 divided by 3 is 13

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Step-by-step explanation:

y = |x - 2| - 1

Given equation is in the form of y=|x-h|+k

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3 years ago
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A recipe submitted to a magazine by one of its subscribers’ states that the mean baking time for a cheesecake is 55 minutes. A t
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Answer:

t=\frac{59.4-55}{\frac{3.239}{\sqrt{10}}}=4.296    

The degrees of freedom are given by:

df=n-1=10-1=9  

The p value for this case is given by:

p_v =P(t_{(9)}>4.296)=0.001  

And for this case the p value is lower than the significance level so we have enough evidence to reject the null hypothesis and then we can conclude that true mean is higher than 55.

Step-by-step explanation:

Information given

We have the following data: 54 55 58 59 59 60 61 61 62 65

The sample mean and deviation can be calculated with the following formulas:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X-i -\bar x)^2}{n-1}}

\bar X=59.4 represent the sample mean

s=3.239 represent the sample standard deviation

n=10 sample size  

\alpha=0.05 represent the significance level

t would represent the statistic  

p_v represent the p value

System of hypothesis

We want to test if the true mean is higher than 55, the system of hypothesis would be:  

Null hypothesis:\mu \leq 55  

Alternative hypothesis:\mu > 55  

Replacing the info given we got:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

And replacing the info given we got:

t=\frac{59.4-55}{\frac{3.239}{\sqrt{10}}}=4.296    

The degrees of freedom are given by:

df=n-1=10-1=9  

The p value for this case is given by:

p_v =P(t_{(9)}>4.296)=0.001  

And for this case the p value is lower than the significance level so we have enough evidence to reject the null hypothesis and then we can conclude that true mean is higher than 55

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