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3 years ago

Luis can swim 1 ½ meters per second. How many meters can he swim in a minute? Show your work and state the answer.

Mathematics

1 answer:

just olya [345]3 years ago

7 0

Luis can swim 90 meters per minute
1 1/2 x 60 = 90

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Nine less than four times a number equals fifteen

OverLord2011 [107]

4n-9=15. N=6 (n stands for number)

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4 years ago

Read 2 more answers

An urn contains 8 pink and 5 red balls. Six balls arerandomly drawn from the urn in succession, with replacement.That is, after

slega [8]

0.088

Step-by-step explanation:

The total number of balls in the urn is;

8 + 5 = 12

Because the pink balls are 8, then the probability of picking a pink ball from the urn is;

8/12

To get the probability that all 6 balls drawn from the urn are pink, we will use the AND probability rule of the mutually exclusive events which means we’ll multiply the probabilities of each of the six pink balls;

8/12 * 8/12 * 8/12 * 8/12 * 8/12 * 8/12

= 0.088

Learn More:

For more on probabilities check out;

brainly.com/question/11248705

brainly.com/question/1177537

#LearnWithBrainly

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4 years ago

The sum of a number y and -3 is -8 .

Allushta [10]

Answer:

-5 IDS THE AANSERRRRR thanks

Step-by-step explanation:

3 0

3 years ago

A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall

Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : $\mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7$

Standard deviation : $\sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}$

$=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}$

$=\dfrac{30.265}{10}=3.0265$

The confidence interval for the population mean (for sample size <30) is given by :-

$\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given significance level : $\alpha=1-0.95=0.05$

Critical value : $t_{n-1,\alpha/2}=t_{9,0.025}=2.262$

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

$4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)$

Hence, 95% confidence interval for the true mean= (2.54,6.86)

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4 years ago

Please help!

poizon [28]

The answer is b

$x^{2} - 2x + 1 + y ^{2} + 6y + 2$

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3 years ago