When do you need to use two number lines to compare two fractions

Mathematics

1 answer:

OLEGan [10]3 years ago

4 0

Depends on the fraction to determine your answer

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A total of 150 new recruits join an army training course. After one month, 30 have dropped out. what is the percentage decrease

Pachacha [2.7K]

So what will it be a percentage of? it will be the percentage of the original recruits, of the original 150.

for this, we need to divide the number of recruits that left over all recruits, and multiply by 100%: $\frac{30}{150} *100%= \frac{1}{5} *100%=20%$

so the number of trainees decreased by 20 %

4 0

3 years ago

Read 2 more answers

Sergeu [11.5K]

Answer:

5 pints per hour.

Step-by-step explanation:

Morning:

Given:

Number of strawberries packed in every 4 minutes = 3

∴ Number of strawberries packed in every one minute = $\frac{3}{4}$

So, number of strawberries packed in every 60 minutes = $\frac{3}{4}\times 60=45$

1 hour = 60 minutes.

So, strawberries are packed at a rate of 45 pints per hour in the morning.

Afternoon:

Number of strawberries packed in every 3 minutes = 2

∴ Number of strawberries packed in every one minute = $\frac{2}{3}$

So, number of strawberries packed in every 60 minutes = $\frac{2}{3}\times 60=40$

1 hour = 60 minutes.

So, strawberries are packed at a rate of 40 pints per hour in the afternoon.

Difference between the packing rates is given as:

Morning rate - Afternoon rate = 45 - 40 = 5 pints per hour.

So, she packed 5 more pints in one hour in the morning than in the afternoon.

3 0

3 years ago

Find the following integral

ololo11 [35]

There's nothing preventing us from computing one integral at a time:

$\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2$

$\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2$

$\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx$

Expand the integrand completely:

$x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x$

Then

$\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}$