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andrey2020 [161]
3 years ago
12

I am so bad at this, help me...

Mathematics
1 answer:
kondor19780726 [428]3 years ago
3 0

Answer: its 9 units ok yea


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Round to the nearest tenth :14.94<br><br> need answer ASAP, please explain your answer!
Diano4ka-milaya [45]
14.9


Step-by-step explanation:


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2 years ago
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Jana ran the first 3 and a half miles of a 5 mile race in 1 3rd of an hour. what was her average rate, in mph, for the first par
Lisa [10]

Answer:

always try to draw a diagram first

Step-by-step explanation:

Jana ran The first 3.5 miles in 1/3 of an hour. remember the problem tells us that we want our answer in miles per hour or mph. so we have to set up our problem so that our units match in the end. The diagram above shows 3.5 mi over 1/3 hour. so we must divide 3.5 miles by 1/3 hour

note:

3 \frac{1}{2} mi =  \frac{7mi}{2}

therefore

\frac{7mi}{2}  \:  \div  \:  \frac{1hr}{3}  =

\frac{7mi}{2}  \times  \frac{3}{1hr}  =  \frac{21mi}{2hr}  = 10 \frac{1}{2}  \frac{mi}{hr}

7 0
2 years ago
The​ _____ _____, denoted modifyingabove p with caret​, is given by the formula modifyingabove p with caretequals​_____, where x
elena-14-01-66 [18.8K]

Answer:

Sample proportion ; phat

x / n

Step-by-step explanation:

The sample proportion may be explained as a proportional representation of part of the population or larger sample which have a certain character or exhibit a particular trait.

The sample proportion is expressed Mathematically as ;

phat = x / n

x = number of subjects with a specified characteristic ;

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For instance ;

The proportion of black balls in a sample of 100 balls is 15 ; the sample proportion of bla k balls will be :

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2 years ago
HELP 5TH GRADE QUESTION!!!
babymother [125]

Answer:

d

Step-by-step explanation:

d

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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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