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sergejj [24]
3 years ago
5

A partial variation has an equation of the form y=mx+b, where m represents the

Mathematics
1 answer:
Nastasia [14]3 years ago
6 0

Answer:

false

Step-by-step explanation:

b represents y value

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4/25, 13%, 0.28, 7%, 21/100, 0.15 least to greatest
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7%, 13%, 0.15, 4/25, 21/100, 0.28
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In one basketball season, Susan threw the ball and it went in the basket 56 times. If 7/8 of the baskets Susan threw were worth
Ket [755]

Answer:

119

Step-by-step explanation:

Divide the number of baskets (56) by the denominator:

56 ÷ 8 = 7

Multiply by 7 to get the amount of 2 point baskets and multiply by 1 to get the amount of 3 point baskets. (they both add up to 56 baskets)

7 × 7 = 49

7 × 1 = 7

multiply the 49, 2 point baskets by 2 and the 7, 3 point baskets by 3 and add them together to get the total points.

49 × 2 = 98

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98 + 21 = 119

6 0
3 years ago
What is 83 + 119 + 17 =
Fittoniya [83]

Answer:

219

Step-by-step explanation:

83 + 119 = 202

202 + 17 = 219

3 0
3 years ago
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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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