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Sati [7]
3 years ago
14

Please help!! Brainliest and Points will be given!! thank you!!

Mathematics
2 answers:
Reil [10]3 years ago
4 0
Answer:
712

Step-by-step:
First we need to find the total number of small packets:

Anne buys 600(34%)=204 small packets
Ben buys 400(3/10)=120 small packets
Chas buys 210 small packets
Total number of small packets is 204+120+210=534

Now, we know that the number of small packets to the number of medium packets is 3/4, if we let x be the number of medium packets we can write:

3/4=534/x
x=712
:)
Dmitrij [34]3 years ago
3 0

Answer:

12:4

Step-by-step explanation:

im sorry if its wrong because i tried.. SORRYYY

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Explain how you would find 45 x 7/9 mentally​
Artemon [7]

45*7/9

cross out 45 and 9

divide by 9

45/9=5

9/9=1

5*1*7

=587

=35

answer:

35

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60

Step-by-step explanation:

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Andrej [43]
Y-12=28
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6 0
3 years ago
A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro
-Dominant- [34]

Answer:

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

Step-by-step explanation:

Data given and notation  

\bar X=1.6 represent the sample mean

s=0.46 represent the sample deviation

n=32 sample size  

\mu_o =1.35 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:  

Null hypothesis:\mu =1.35  

Alternative hypothesis:\mu \neq 1.35  

The statistic is given by:

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

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

P-value  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

5 0
3 years ago
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