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Margaret [11]
3 years ago
5

Lee has saved 65% of the money she needs to buy a pair of jeans that cost $52. How much money does lee have, and how much more m

oney does she need to buy the jeans?
Mathematics
1 answer:
schepotkina [342]3 years ago
4 0
Lee has $33.80 and needs $18.20 more to buy the jeans

youre welcome
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A rectangular block has square base of area 64 cm². If the thickness of the block is 5 cm, find its total surface area. ​
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Answer:

total surface area of cube is <u>6 * area of one side</u>

therefore area of one side is given 64cm^2

thickness is also given 5cm .

TSA is 64*5 = 320cm^2

hope it is helpful

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Which situation can be modeled by the equation 705 – m= 430?<br> help !!!!
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Real world example could be:
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A baseball team's total wins and losses for one season is 162. The number of wins is 6 less than 2 times the number of losses. W
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Answer:

x + y = 162 and x = 2y - 6

Step-by-step explanation:

Let the total wins be x

Let the total loss be y

If a baseball team's total wins and losses for one season is 162, this can be expressed as;

x + y = 162 ....... 1

If the number of wins is 6 less than 2 times the number of losses, this is expressed as;

2 times the number of losses = 2y

6 less than 2 times the number of losses = 2y - 6

Now, If the number of wins is 6 less than 2 times the number of losses this is expressed as;

x = 2y - 6 .... 2

Hence the required system of equation are;

x + y = 162 and x = 2y - 6

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3 years ago
In a sample of 42 water specimens taken from a construction site, 26 contained detectable levels of lead. Construct 90% confiden
hram777 [196]

Answer:

(0.4958, 0.7422)

Step-by-step explanation:

Let p be the true proportion of water specimens that contain detectable levels of lead. The point estimate for p is \hat{p}=26/42=0.6190. The estimated standard deviation is given by \sqrt{\hat{p}(1-\hat{p})/n}=\sqrt{(0.6190)(1-0.6190)/42}=0.0749. Because we have a large sample, the 90% confidence interval for p is given by 0.6190\pm z_{0.05}0.0749 where z_{0.05}=1.6448 is the value that satisfies that above this and under the standard normal density there is an area of 0.05. So, the confidence interval is 0.6190\pm (1.6448)(0.0749), i.e., (0.4958, 0.7422).

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