All categories

3 years ago

36 is _______% less than 60?

2 answers:

6 0

<span>36 is _______% less than 60?

36=60x
(divide both sides by 60)

x=.6

.6 as a percent is 60

60 percent subtracted from 100 is 40

40 is your answer

hope this helps

</span>

Arte-miy333 [17]3 years ago

4 0

60*(1-x)=36
60-60x=36
-60x=36-60
-60x=-24
x=24/60
x=4/10
x=2/5
x=0.4

36 is 40% less than 60

You might be interested in

I will give the BRAINIEST!!!<br><br>Which of these statements best describes the relation shown?

ludmilkaskok [199]

Answer:

i believe the answer is C

Step-by-step explanation:

i dont know its just what i would choose im not the smartests in math

3 0

2 years ago

Enter a range if values for x.

Blizzard [7]

Answer:

See the attachment photo!

3 0

2 years ago

2x-4y=6 and 3x+y=-5 what is the point of intersection

VMariaS [17]

Answer:(-1,-2)

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int

Wittaler [7]

Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

Step-by-step explanation:

Salt in the tank is modelled by the Principle of Mass Conservation, which states:

(Salt mass rate per unit time to the tank) - (Salt mass per unit time from the tank) = (Salt accumulation rate of the tank)

Flow is measured as the product of salt concentration and flow. A well stirred mixture means that salt concentrations within tank and in the output mass flow are the same. Inflow salt concentration remains constant. Hence:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = \frac{d(V_{tank}(t) \cdot c(t))}{dt}$

By expanding the previous equation:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)$

The tank capacity and capacity rate of change given in gallons and gallons per minute are, respectivelly:

$V_{tank} = 220\\\frac{dV_{tank}(t)}{dt} = 0$

Since there is no accumulation within the tank, expression is simplified to this:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt}$

By rearranging the expression, it is noticed the presence of a First-Order Non-Homogeneous Linear Ordinary Differential Equation:

$V_{tank} \cdot \frac{dc(t)}{dt} + f_{out} \cdot c(t) = c_0 \cdot f_{in}$ , where $c(0) = 0 \frac{pounds}{gallon}$ .

$\frac{dc(t)}{dt} + \frac{f_{out}}{V_{tank}} \cdot c(t) = \frac{c_0}{V_{tank}} \cdot f_{in}$

The solution of this equation is:

$c(t) = \frac{c_{0}}{f_{out}} \cdot ({1-e^{-\frac{f_{out}}{V_{tank}}\cdot t }})$

The salt concentration after 8 minutes is:

$c(8) = 0.166 \frac{pounds}{gallon}$

The instantaneous amount of salt in the tank is:

$m_{salt} = (0.166 \frac{pounds}{gallon}) \cdot (220 gallons)\\m_{salt} = 36.52 pounds$

3 0

3 years ago

Rhian sat a maths test and an English test In the Maths test she scored 14 out of 17 In the English test she scored 31 out of 39

BabaBlast [244]

Answer:

Rhian did better in the Math test.

Step-by-step explanation:

Given: Rhian scored $14$ out of $17$ in a Math test. In the English test she scored $31$ out of $39$ .

To find: In which test did she do better?

Solution:

We have,

Rhian scored $14$ out of $17$ in a Math test.

Rhian scored $31$ out of $39$ in an English test.

So, first we need to find the percentage of the marks scored.

Percentage in the Math test $=\frac{14}{17}\times100=82.35 \%$ .

Percentage in the English test $=\frac{31}{39}\times100= 79.49\%$ .

Clearly, the percentage in the Math test is greater than the percentage in the English test.

So, Rhian did better in the Math test.

8 0

2 years ago