Solve for r. k=3r-7s

Mathematics

1 answer:

const2013 [10]4 years ago

5 0

R=7s/3+k/3

Hope this helps!

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Simply -3 3/4 divided by 2 1/2

Dima020 [189]

Answer:

$=-1\frac{1}{2}$

Step-by-step explanation:

Given the following question:

$-3\frac{3}{4} \div2\frac{1}{2}$

In order to find the answer, we first convert the mixed numbers into improper fractions. Then we use KCF (Keep, Change, Flip) and solve.

$-3\frac{3}{4} \div2\frac{1}{2}$
$-3\frac{3}{4} =12+3=-\frac{15}{4}$
$2\frac{1}{2}=2\times2=4+1=\frac{5}{2}$
$-\frac{15}{4} \div\frac{5}{2}$
$-\frac{15}{4} \times\frac{2}{5}$
$15\times2=30$
$4\times5=20$
$=-\frac{30}{20}$
$-\frac{30}{20} \div10=-\frac{3}{2}$
$-\frac{3}{2}=3\div2=-1\frac{1}{2}$
$=-1\frac{1}{2}$

Hope this helps.

4 0

2 years ago

What is the answer needdd it

Naddik [55]

I think it would be b or c

3 0

3 years ago

Life tests on a helicopter rotor bearing give a population mean value of 2500 hours and a population standard deviation of 135 h

stira [4]

Answer:

The percent of the parts are expected to fail before the 2100 hours is 0.15.

Step-by-step explanation:

Given :Life tests on a helicopter rotor bearing give a population mean value of 2500 hours and a population standard deviation of 135 hours.

To Find : If the specification requires that the bearing lasts at least 2100 hours, what percent of the parts are expected to fail before the 2100 hours?.

Solution:

We will use z score formula

$z=\frac{x-\mu}{\sigma}$

Mean value = $\mu = 2500$

Standard deviation = $\sigma = 135$

We are supposed to find If the specification requires that the bearing lasts at least 2100 hours, what percent of the parts are expected to fail before the 2100 hours?

So we are supposed to find P(z<2100)

so, x = 2100

Substitute the values in the formula

$z=\frac{2100-2500}{135}$