How do I solve this??

Given the number, two-fifths, generate its equivalent forms as a fraction, a decimal, and a percent. In addition, give a real wo

vichka [17]

The number in the three different forms is as follows.

2/5 (fraction), 0.2 (decimal), and 20% (percent).

You could use percentages for the results of a poll. For example, if you held a poll for which flavour of ice cream is your favorite. Let's say you surveyed 100 kids and 50 kids voted chocolate, 30 vanilla, and 20 strawberry. Then you could say 20% voted for strawberry.

You could use the decimal form for money. For example, let's say you want to buy something that costs $10.20 and you only have 10 dollars. You could say that you would need 20 more cents, or $0.20.

As for the fraction, you could use this for measurements. For example, let's say you have a piece of fabric. You want to cut it up into 5 equal pieces, but only use two of those pieces. You could say that you would only use 2/5 of the fabric.

I hope I helped you out! If anything is wrong, please let me know about it! :)

6 0

3 years ago

Samantha's notebook has 14 pages more than three times as much as Lisa's notebook.

Bumek [7]

X - Lisa's notebook

$3x+14=260\\
3x=246\\x=82$

7 0

2 years ago

Answer correctly please

velikii [3]

Answer:

The answer would be Image A.

8 0

2 years ago

The cargo of the truck weighs at most 2,000 pounds.

Vadim26 [7]

$w \geqslant 2000$

which is, W is less than or equal to 2000 lbs.

6 0

2 years ago

Suppose a baker claims that the average bread height is more than 15cm. Several of this customers do not believe him. To persuad

Deffense [45]

Answer:

a) $17-2.26\frac{1.9}{\sqrt{10}}=15.64$

$17+2.26\frac{1.9}{\sqrt{10}}=18.36$

So on this case the 95% confidence interval would be given by (15.64;18.36)

b) 1. n=15, conf =95% $\bar X= 35$ s=2.7

> round(qt(p=1-0.025,df=15-1),2)

[1] 2.14

> round(qt(p=0.025,df=15-1),2)

[1] -2.14

2. n=37, conf =99% $\bar X= 82$ s=5.9

> round(qt(p=1-0.005,df=37-1),2)

[1] 2.72

> round(qt(p=0.005,df=37-1),2)

[1] -2.72

3. n=1009, conf =90% $\bar X= 0.9$ s=0.04

> round(qt(p=1-0.05,df=1009-1),2)

[1] 1.65

> round(qt(p=0.05,df=1009-1),2)

[1] -1.65

Step-by-step explanation:

Part a: What is the lower bound to this confidence interval? 2 cm (round to 2 decimal places) What is the upper bound to this confidence interval? cm (round to 2 decimal places)

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

We have the following data:

$\bar x= 17$ represent the sample mean

$s = 1.9$ represent the sample deviation

n =10 represent the sample size

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=10-1=9$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that $t_{\alpha/2}=2.26$

Now we have everything in order to replace into formula (1):

$17-2.26\frac{1.9}{\sqrt{10}}=15.64$

$17+2.26\frac{1.9}{\sqrt{10}}=18.36$

So on this case the 95% confidence interval would be given by (15.64;18.36)

Part b

1. n=15, conf =95% $\bar X= 35$ s=2.7

> round(qt(p=1-0.025,df=15-1),2)

[1] 2.14

> round(qt(p=0.025,df=15-1),2)

[1] -2.14

2. n=37, conf =99% $\bar X= 82$ s=5.9

> round(qt(p=1-0.005,df=37-1),2)

[1] 2.72

> round(qt(p=0.005,df=37-1),2)

[1] -2.72

3. n=1009, conf =90% $\bar X= 0.9$ s=0.04

> round(qt(p=1-0.05,df=1009-1),2)

[1] 1.65

> round(qt(p=0.05,df=1009-1),2)

[1] -1.65

5 0

2 years ago