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allochka39001 [22]
2 years ago
12

Mr. McLean manages playgrounds for the 8 elementary schools in the Franklin School District. This year, he wants to put new sand

boxes at each of the schools. He estimates that he will need 3 tons of sand for this project. How many 50-pound bags of sand should he order for each school?
Mathematics
2 answers:
Tems11 [23]2 years ago
8 0
He should order 120 bags for each school
eduard2 years ago
7 0
He will need 15 sandbags per school

3 tons = 6000

6000/8= 750lbs for each school
750/50=15 (50#) bags needed
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A rock climber is planning to climb a mountain cliff over two days. The climb begins at an altitude of 6,400 feet above sea leve
Furkat [3]

Answer:

D

Step-by-step explanation:

360h+6,400 >8,020, with a solution is h> 4.

3 0
1 year ago
one side of a right angled triangle is 10cm the other two are both of length x calculate x to 2 decimal place
shepuryov [24]

Answer: x = 7.07

Step-by-step explanation:

Using Pythagoras theorem , since it is a right angled triangle ,

side 1 = 10cm

side 2 = x cm

side 3 = x cm

Pythagoras theorem  states that the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares on the other two sides. That is

10^{2}=x^{2}  +x^{2}

100 = 2x^{2}

divide through by 2 , we have

50 = x^{2}

find the square root of both sides

x = \sqrt{50}

x = 7.07

8 0
2 years ago
Mr. Barrow scored his high school students' essays on a scale of 0 to 10. He recorded data about the scores in the table below.
Fynjy0 [20]

09 08 07 06 10  1

wrryetryytrwertry (It was too short)

7 0
1 year ago
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Cory took out a loan at a 13.5% APR, compounded monthly, to buy a car, and he is making monthly payments to pay off the loan. Wh
Irina-Kira [14]
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2 years ago
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After the 2008 elections, it is desired to estimate the proportion of Florida voters who now regret that they did not vote.
Sonja [21]

Answer:

n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681  

And rounded we got 1681

Step-by-step explanation:

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".  

p represent the real population proportion of interest

\hat p represent the estimated proportion for the sample

n is the sample size required (variable of interest)

z represent the critical value for the margin of error

Solution to the problem

The population proportion have the following distribution  

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

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 90% of confidence, our significance level would be given by \alpha=1-0.90=0.10 and \alpha/2 =0.05. And the critical value would be given by:  

z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64  

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 0.02 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)

We can assume that the estimates proportion is 0.5 since we don't have other info provided to assume a different value. And replacing into equation (b) the values from part a we got:  

n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681  

And rounded we got 1681

8 0
2 years ago
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