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3 years ago

5miles=8km how many miles is in 80km

Mathematics

1 answer:

Aneli [31]3 years ago

8 0

Answer: 50 miles

Step-by-step explanation:

Figuring this out is fairly simple, all you need to do is find out how many sets of 8 km are in 80km and multiply that number by 5. So you divide 80 by 8 and you get 10, then multiply 10 by 5 and you'll get 50. There are 50 miles in 80km.

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Simplify 12x + 8x - 3x

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when two numbers have the same variable, we can add and subtract just like normal numbers.

12x + 8x = 20x - 3x = 17x

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2 years ago

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A chemist is conducting an experiment which involves a certain chemical reaction. He begins the experiment with 666 milliliters

Ket [755]

Answer:

8 Minutes

Step-by-step explanation:

since we are trying to find how many minutes have passed we first need to find how much substance was used during the time. we can do that by subtracting the leftover substance from the amount we started with.

666-562 = 104

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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adoni [48]

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3 years ago

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Rent and other associated housing costs, such as utilities, are an important part of the estimated costs of attendance at colleg

zhenek [66]

Answer:

96% confidence interval estimate for the mean monthly rent of all unmarried BYU students in winter 2018 is [335.89 , 356.10].

Step-by-step explanation:

We are given that a group of researchers at the BYU Off-Campus Housing department want to estimate the mean monthly rent that unmarried BYU students paid during winter 2018.

During March 2018, they randomly sampled 314 BYU students and found that on average, students paid $346 for rent with a standard deviation of $86.

So, the pivotal quantity for 95% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = average rent paid by 314 BYU students = $346

s = sample standard deviation = $86

n = sample of students = 314

$\mu$ = population mean monthly rent of all unmarried BYU students

So, 96% confidence interval for the population mean monthly rent, $\mu$ is ;

P(-2.082 < $t_3_1_3$ < 2.082) = 0.96

P(-2.082 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.082) = 0.96

P( $-2.082 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.082 \times {\frac{s}{\sqrt{n} } }$ ) = 0.96

P( $\bar X -2.082 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X +2.082 \times {\frac{s}{\sqrt{n} } }$ ) = 0.96

96% confidence interval for $\mu$ = [ $\bar X -2.082 \times {\frac{s}{\sqrt{n} } }$ , $\bar X +2.082 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $346 -2.082 \times {\frac{86}{\sqrt{314} } }$ , $346 +2.082 \times {\frac{86}{\sqrt{314} } }$ ]

= [335.89 , 356.10]

Therefore, 96% confidence interval estimate for the mean monthly rent of all unmarried BYU students in winter 2018 is [335.89 , 356.10].

5 0

4 years ago