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kvv77 [185]
3 years ago
14

Juan ran the 200 meters in 26.47 seconds, Scott ran it in 26.82 seconds, Kyrah ran it in 26.9 seconds, and Beth ran it in 30.07

seconds. Who ran the 200 meters the fastest?
Mathematics
2 answers:
Natalka [10]3 years ago
7 0

Answer:

Juan did.

Step-by-step explanation:

the two fastest speeds are 26.47 and 26.9 seconds. In order to be able to properly work with that, you need to add a 0 to 26.9 so that it turns into 26.47 and 26.90 seconds. If you were yo subtract those, you would have:

26.90-26.47= 0.43

Therefore, 26.47 seconds is faster than 26.9 seconds.

harina [27]3 years ago
7 0

Answer: Kyrah ran the fastest

Step-by-step explanation:

The other answers are wrong because remember these people were being timed so the bigger there number was the longer they took to make it, the smaller there number was the fastest they ran.

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N is an integer.<br> Write the values of n such that -15 &lt; 3 ≤ 6<br><br> Help please ❤️❤️
Alex17521 [72]

Answer:

-4,-3,-2,-1,0,1

Step-by-step explanation:

First, doublepound and simplify it.

-15<3n

3n<6

Solve:

-5<n

n<2

Compound:

-5<n<2.

So the values are -4,-3,-2,-1,0,1

Hope this helps plz hit the crown :D

8 0
3 years ago
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A company that manufacturers deep freezers has developed a new formula of refrigerant, the substance used to cause cooling in re
bonufazy [111]

Answer:

response variable- time required for water to turn to ice

control group- second freezer

explanatory variable- new formula

Step-by-step explanation:

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3 years ago
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A researcher tests five individuals who have seen paid political ads about a particular issue. These individuals take a multiple
devlian [24]

Answer:

t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606    

The degrees of freedom are given by:

df=n-1=5-1=4  

The p value wuld be given by:

p_v =2*P(t_{(4)}>2.606)=0.060  

For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40

The distribution with the critical values are in the figure attached

Step-by-step explanation:

Information given

48, 41, 40, 51, and 50

The sample mean and deviation can be calculated with these 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=46 represent the mean height for the sample  

s=5.148 represent the sample standard deviation

n=5 sample size  

\mu_o =40 represent the value that we want to test

\alpha=0.05 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

Hypothesis to test

We want to test if the true mean for this case is equal to 40, the system of hypothesis would be:  

Null hypothesis:\mu = 40  

Alternative hypothesis:\mu \neq 40  

The statistic is given by:

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

Replacing we got:

t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606    

The degrees of freedom are given by:

df=n-1=5-1=4  

The p value wuld be given by:

p_v =2*P(t_{(4)}>2.606)=0.060  

For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40

The distribution with the critical values are in the figure attached

6 0
3 years ago
Which of the following applies the law of cosines correctly and could be solved to find m∠E? ANSWERS: A) cos E = 312 + 392 – 2(3
defon

This question is incomplete because the options were not properly written.

Complete Question

Which of the following applies the law of cosines correctly and could be solved to find m∠E? ANSWERS:

A) cos E = 31²+ 39² – 2(31)(39)

C) 56² = 39² – 2(39) ⋅ cos E

D) 56² = 31² + 39² – 2(31)(39) ⋅ cos E

Answer:

D) 56² = 31² + 39² – 2(31)(39) ⋅ cos E

Step-by-step explanation:

From the above diagram, we see are told to apply the law of cosines to solve for m∠E i.e Angle E

The formula for the Law of Cosines is given as:

c² = a² + b² − 2ab cos(C)

Because we have sides d , e and f and we are the look for m∠E the law of cosines would be:

e² = d² + f² - 2df cos (E)

e = 56

d = 39

f = 31

56² = 39² + 31² - (2 × 39 × 31) × cos E

Therefore, from the above calculation and step by step calculation, the option that applies the law of cosines correctly and could be solved to find m∠E

Is option D: 56² = 31² + 39² – 2(31)(39) ⋅ cos E

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