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3 years ago

Jenny ran 6km. one mile is approximately 1.6 km. about how many miles did jenny run

Mathematics

1 answer:

Diano4ka-milaya [45]3 years ago

5 0

To find this answer you must divide 6km by 1.6km
the answer should be 3.75 miles

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Ms. Wong ordered a box of 50 shirts to sell in her clothing store. She quickly sold 30 of the shirts. Which equations can be use

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Answer:

answer = 20

Step-by-step explanation:

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3 years ago

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John bought a computer scanner and supplies for $215.70, which will

lutik1710 [3]

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2 years ago

Work out an estimate for square root 1.13 + 2.84 x 9.33

Feliz [49]

hi there!

 $\large \rm = \color{brown} 27.6272 \: \\ \tt in \: \: alternate \: \: form \\ \\ \large \tt \color{cyan} \frac{17267}{625} , \color{red}27 \frac{392}{625}$ 

<h2 />

Step-by-step explanation:

 $\tt1.13 + 2.84 \times 9.33$ 

multiply the number

 $\tt \color{green}1.13 + 26.4972$ 

add the number

 $= \tt \color{green}27.6272 \\ \\ \tt \color{pink}or \: \: in \: \: alternate \: \: form \\ \\ \tt \color{purple} \frac{17267}{625} \tt \color{blue} \: , \: 27 \frac{392}{625}$ 

<h2>hope it helps</h2>

#Carryonlearning

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6 0

2 years ago

153 of the 258 high school students surveyed at a local school said they went outside more during school hours as elementary sch

Juli2301 [7.4K]

Answer-

It is reasonable that the state education department claims the percentage for the entire state is 63%.

Solution-

153 out of the 258 high school students surveyed at a local school said they went outside more during school hours as elementary school students than they do now as high school students.

Here,

x = 153,

n = 258,

Then,

$\text{sample proportion}=p=\dfrac{153}{258}=0.5930=59.30\%$

And,

$\text{Margin Error}=2\times \sqrt{\dfrac{p(1-p)}{n}}$

Putting the values,

$\text{ME}=2\times \sqrt{\dfrac{(0.5930)(1-0.5930)}{258}}\\\\=2\times \sqrt{\dfrac{(0.5930)(0.407)}{258}}\\\\=2\times \sqrt{\dfrac{0.2414}{208}}\\\\=2\times 0.0341\\\\=0.0682\\\\=6.82\%$

The interval would be,

$=(59.30-6.82,\ 59.30+6.82)\\\\=(52.48,\ 66.12)$

As population proportion p=63% lies in this interval, so it is reasonable.

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4 years ago

Given the function below, evaluate f f(x) = (96/-55

katrin2010 [14]

I think it’s suppose to remain as a fraction f(x)= -96/55

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3 years ago