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3 years ago

The table lists the top four times for the 50-yard freestyle at a swim meet. Who had the fastest time?

Mathematics

1 answer:

Advocard [28]3 years ago

4 0

Answer:

Step-by-step explanation:

Nobody likes a short, one-word-answer. Let’s make sure other people understand everything.

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Can someone give me the answers plz:)

Alborosie

Remark

There is no short way to do this problem and no obvious way to get the answer other that to solve each part.

Solve

$\dfrac{x + 1.6}{2} =\text{x + 0.1}$ Multiply by 2

x + 1.6 = 2(x + 0.1) Remove the brackets

x + 1.6 = 2x + 0.1*2

x + 1.6 = 2x + 0.2 Subtract x from both sides

1.6 = x + 0.2 Subtract 0.2 from both sides

1.6 - 0.2 = x

1.4 = x

Circle A

Subtract 2x from both sides.

3x - 2x = 1.4

Circle B

Remove the brackets.

4x + 6 = 2x - 6 Add 6 to both sides

4x + 12 = 2x Subtract 4x from both sides.

12 = -2x Divide by - 2

12/-2 = x

x = - 6 Don't circle C

I'm going to be very scant in my solution of this. You can fill in the steps.

3x = 4.2

x = 4.2/3

x = 1.4

Circle D

5 0

3 years ago

Read 2 more answers

What is 1000 less than 1416

lara31 [8.8K]

1,416 - 1,000 = 416

6 0

2 years ago

Read 2 more answers

Explain how the value of the 7 in 327,902 will change if you move it to the tens place

rjkz [21]

It will change from having a value of 7,000 to having a value of 70

6 0

3 years ago

Completează tabelul de mai jos:

KiRa [710]

I’ll what is this language

4 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)

7 0

3 years ago