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3 years ago

9

Taylor and Desmond ran laps after school to train for the football team.The ratio of the number of laps Taylor ran to the number

of laps Desmond ran was 2.3. If Taylor ran 4 laps, how many laps did Desmond run?

2 answers:

Fittoniya [83]3 years ago

4 0

Answer:

6 laps

Step-by-step explanation:

Since the ratio is 2:3, and 4 is twice the number of 2, then you'd have to times 3 by 2, in which you'd get 6.

klasskru [66]3 years ago

4 0

Answer:

6 laps

Step-by-step explanation:

I hope this helps!

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2 years ago

Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean. See Attached Excel for Da

alexgriva [62]

Answer:

The 99% confidence interval for the mean germination time is (12.3, 19.3).

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean: 18, 12, 20, 17, 14, 15, 13, 11, 21, 17. Assume that the population germination time is normally distributed. Find the 99% confidence interval for the mean germination time.</em>

<em />

We start calculating the sample mean M and standard deviation s:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{10}(18+12+20+17+14+15+13+11+21+17)\\\\\\M=\dfrac{158}{10}\\\\\\M=15.8\\\\\\$

$s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{9}((18-15.8)^2+(12-15.8)^2+(20-15.8)^2+. . . +(17-15.8)^2)}\\\\\\s=\sqrt{\dfrac{101.6}{9}}\\\\\\s=\sqrt{11.3}=3.4\\\\\\$

We have to calculate a 99% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=15.8.

The sample size is N=10.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{3.4}{\sqrt{10}}=\dfrac{3.4}{3.162}=1.075$

The degrees of freedom for this sample size are:

df=n-1=10-1=9

The t-value for a 99% confidence interval and 9 degrees of freedom is t=3.25.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=3.25 \cdot 1.075=3.49$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 15.8-3.49=12.3\\\\UL=M+t \cdot s_M = 15.8+3.49=19.3$

The 99% confidence interval for the mean germination time is (12.3, 19.3).

8 0

3 years ago