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vovikov84 [41]
3 years ago
13

YOU WENT TO THE GYM LAST WEEK AND SPENT THE FOLLOWING AMOUNT OF YOUR TIME ON THE RUNNING MACHINE.

Mathematics
2 answers:
tia_tia [17]3 years ago
3 0
Thursday sonnnnnnnnnnnnn
ivann1987 [24]3 years ago
3 0

Answer: On Friday, he spend the least amount of time on the running machine.

Step-by-step explanation:

Since we have given that

Monday   0.3 = 0.3\times 100\%=30\%

Tuesday = 15%

Wednesday = \dfrac{1}{6}\times 100\%=16.67\%

Thursday = 0.2\times 100\%=20\%

Friday = \dfrac{1}{8}\times 100\%=12.5\%

Hence, we can compare all the days, and we get that

On Friday, he spend the least amount of time on the running machine.

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Which is bigger .24 or .71
monitta

Answer:

.71

Step-by-step explanation:

7 0
3 years ago
Erica must tutor a minimum of 20 hours per week in order to remain in the national honor society. She has a student that she tut
irinina [24]

Answer:

3 hours every weekday

Step-by-step explanation:

20 - 5 is 15 so she must work 15 hours during the week since saturday and sunday are taken. 15 divided by 5 is 3 therefore, giving you three hours a day; Monday, Tuesday, Wednesday, Thursday, and Friday

8 0
3 years ago
From 1995 - 2008 in the US 648 people were struck by lightening. 531 of them were men. You want to see if this observation diffe
Gre4nikov [31]

Answer:

Because z is higher than any given value in the chart we come to the conclusion To reject null hypothesis

Step-by-step explanation:

Sample proportion = p= 531/648 = 0.8194

This is the proportion of men that were hit by lightening

Null hypothesis: H0: p = 0.5

Alternate hypothesis: H1: p ≠ 0.5

Test statistics z = 0.8194-0.5/(√0.5x0.5/648)

= 0.8194-0.5/√0.0003858

= 0.3194/0.019642

= 16.26

Since the z > 1.96 (at 5% significance) we reject the null hypothesis.

Therefore in conclusion we say z is higher than given values in the chart so we reject null hypothesis.

Please check attachment!

3 0
3 years ago
Answer the screenshot, please.<br> (no links! PDFs are fine.)
Otrada [13]

Answer:

Mode: 91

Median: 91

Mean: 107

6 0
2 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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