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vitfil [10]
3 years ago
12

Rosa is filling up the back of her dump truck with cedar mulch for a delivery to a landscaped shopping center. The mulch is bein

g pumped into the truck. After 30 minutes, the truck is 75% full and weighs 33 tons. Fifteen minutes later, Rosa's truck is fully loaded and weighs 42 tons. If Rosa's truck weighs 15 tons when it is empty, what was the rate in tons per minute that the mulch pump took to load the truck?
A. 0.6 tons per minute
B. 0.93 tons per minute
C. 1.07 tons per minute
D. 1.67 tons per minute
Mathematics
1 answer:
Illusion [34]3 years ago
3 0
In 30 minutes, the truck weighs 33 tons, which means the mulch is 18 tons (33 tons -15 tons from the truck).
In 45 minutes (15 minutes later), the truck weighs 42 tons, which means the mulch is 27 tons.

So from 30 to 45 minutes, 15 minutes elapsed. And in those 15 minutes, 9 tons of mulch was loaded into the truck (27 tons of mulch - 18 tons of mulch).
This can be expressed as \frac{9 tons}{15 minutes}=0.6tons/minute
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Robbin earned $80 over the weekend and Beth made $70. Which statement shows who made more money over the weekend?
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Robbin made more over the weekend

8 0
3 years ago
Read 2 more answers
A town's population went from 25,800 to 42,600 in 15 years . What is the percent in change
Grace [21]

Answer:

61%

Step-by-step explanation:

To solve this you have to find the change in the town's population.

Percentage wise we simply divide 25,800 by 42,600.

\frac{25800}{42600}= 0.6056

Now we round.

0.6056 The six rounds the 5 up, which will then round the 0 up.

Our change is 0.61.

Now to find the percentage we move the decimal point over two times to the right.

61% is the percentage of change.

<em>~CaityConcerto</em>

4 0
3 years ago
A random sample of 1014 adults in a certain large country was asked "Do you pretty much think televisions are a necessity or a l
arlik [135]

Answer:

a) \hat p=\frac{538}{1014}=0.531 estimated proportion of people indicated that televisions are a luxury they could do without

b) Verify that the requirements for constructing a confidence interval about p are satisfied. The sample is a simple random sample, the value of np(1-p)=1014*0.531*(1-0.531)=252.53 is, which is > 10, and the less than or equal to 5% of the population.

c) The 95% confidence interval would be given by (0.500;0.562)

B. We are 95% confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between 0.500 and 0.562

d) We can say that is possible that more than 60% of adult Americans believe that television is a luxury they could do without because we need to remember that the confidence interval is just a point of estimation and iits probable that the true proportion would be not contained on the interval. But we can say that is not that likely that the true proportion is greater than 60%

e) The 95% confidence interval would be given by (0.438;0.500)

Step-by-step explanation:

Notation and definitions

X=538 number of people indicated that televisions are a luxury they could do without

n=1014 random sample taken

\hat p=\frac{538}{1014}=0.531 estimated proportion of people indicated that televisions are a luxury they could do without

p true population proportion of people indicated that televisions are a luxury they could do without

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{p(1-p)}{n}})

a) \hat p=\frac{538}{1014}=0.531 estimated proportion of people indicated that televisions are a luxury they could do without.

b) Verify that the requirements for constructing a confidence interval about p are satisfied. The sample is a simple random sample, the value of np(1-p)=1014*0.531*(1-0.531)=252.53 is, which is > 10, and the less than or equal to 5% of the population.

c) 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:

z_{\alpha/2}=-1.96, z_{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:

0.531 - 1.96\sqrt{\frac{0.531(1-0.531)}{1014}}=0.500

0.531 + 1.96\sqrt{\frac{0.531(1-0.531)}{1014}}=0.562

The 95% confidence interval would be given by (0.500;0.562)

B. We are 95% confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between 0.500 and 0.562

d) We can say that is possible that more than 60% of adult Americans believe that television is a luxury they could do without because we need to remember that the confidence interval is just a point of estimation and iits probable that the true proportion would be not contained on the interval. But we can say that is not that likely that the true proportion is greater than 60%

e) The new propotion on this case would be

\hat p =1-0.531=0.469

0.469 - 1.96\sqrt{\frac{0.469(1-0.469)}{1014}}=0.438

0.469 + 1.96\sqrt{\frac{0.469(1-0.469)}{1014}}=0.500

The 95% confidence interval would be given by (0.438;0.500)

4 0
4 years ago
A high school coach needs to buy new athletic shorts for the 15 members of the basketball team. The coach must spend less than $
AfilCa [17]
If the budget is $200 and he have 15 members then we have divide the two. 200 / 15 = $13.33 per shorts. 15x =< $200. x represents 13.33. So the solution represents the coach may spend up to $13.33 per pair of shorts. If it was even 1 cent more than $13.33 than he wouldn't have enough.So he can spend up to $13.33 or less per pair of shorts.
5 0
4 years ago
Read 2 more answers
Will choose brainliest
Inga [223]

Answer:

1st one

Step-by-step explanation:

8 0
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