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3 years ago

14

Four boxes were mailed and weighted at the post office and the first one is 8.25pounds the second one weights 8.200 and the he t

hird box ways 8.225 which one is the heaviest?

2 answers:

hoa [83]3 years ago

5 0

Answer:

The first one would be the heaviest

Step-by-step explanation:

becuase if you compare all three of them vertically you wil see that the tenth in the first one is larger than the other two

dezoksy [38]3 years ago

3 0

Answer:The third box! Hope this helped

Step-by-step explanation:8.250

8.200 8.225 Five is the greater number because at the end of all of these decimals are smaller than the five because you had a zero to line them up and in the end and give is greater:)

You might be interested in

Can someone help me with question 2

Sunny_sXe [5.5K]

You just have to add I think

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4 years ago

In a random sample of 150 customers of a high-speed Internetprovider, 63 said that their service had been interrupted one ormore

erastovalidia [21]

Answer:

a) The 95% confidence interval would be given by (0.341;0.499)

b) The 99% confidence interval would be given by (0.316;0.524)

c) n=335

d)n=649

Step-by-step explanation:

1) Notation and definitions

$X_{IS}=63$ number of high speed internet users that had been interrupted one or more times in the past month.

$n=150$ random sample taken

$\hat p_{IS}=\frac{63}{150}=0.42$ estimated proportion of high speed internet users that had been interrupted one or more times in the past month.

$p_{IS}$ true population proportion of high speed internet users that had been interrupted one or more times in the past month.

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".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

1) Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96$

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

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.42 - 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.341$

$0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.499$

The 95% confidence interval would be given by (0.341;0.499)

2) Part b

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$t_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58$

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

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.42 - 2.58\sqrt{\frac{0.42(1-0.42)}{150}}=0.316$

$0.42 + 2.58\sqrt{\frac{0.42(1-0.42)}{150}}=0.524$

The 99% confidence interval would be given by (0.316;0.524)

3) Part c

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.42(1-0.42)}{(\frac{0.05}{1.96})^2}=374.32$

And rounded up we have that n=335

4) Part d

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.42(1-0.42)}{(\frac{0.05}{2.58})^2}=648.599$

And rounded up we have that n=649

5 0

3 years ago

Ellie is numbering her 150 pages scrapbook. how many times will she write the digit 4?

Oksi-84 [34.3K]

Answer:

33

Step-by-step explanation:

4,14,24,34,40,41,42,43,44,45,46,47,48,49,54,64,74,84,94,

104,114,124,134,140,141,142,143,144,145,146,147,148,149

3 0

2 years ago

Do the ratios 9/4 and 11/6 make a proportion please explain. Thank You!

svetoff [14.1K]

No they're not proportionate because 9 divided by 4 = 2.25 and 11 divided by 6 is 1.83 so they're not the same in value

3 0

3 years ago

A new car is purchased for 22300 dollars. The value of the car depreciates at 10.25% per year. What will the value of the car be

Roman55 [17]

Answer:

$11654 approx

Step-by-step explanation:

Given data

Principal amount = 22300 dollars

Rate= 10.25%

time= 6 years

The expression for the decrease is given as

A=P(1-r)^t-----------Note the negative sign is because of the decrease

substitute

A= 22300(1-0.1025)^6

A=22300(0.8975)^6

A=22300*0.5226

A=$11653.98

Hence the new value is $11654 approx

6 0

3 years ago