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Y_Kistochka [10]
3 years ago
15

How do you solve 372 divided by 12?

Mathematics
1 answer:
Anna007 [38]3 years ago
8 0
12/372= 31 is the answer.
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Which rate is the lowest price? <br> $3.00/6 lb<br> $2.50/5 lb<br> $2.40/4 lb<br> $2.40/6 lb
Oksi-84 [34.3K]
2.40/6 I believe tell me if it is right
7 0
2 years ago
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Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 75.6 Mbp
olga55 [171]

Answer:

a) 59.98

b) 2.99

c) 2.99

d) Significantly High

Step-by-step explanation:

Part a)

Highest speed measured = x = 75.6 Mbps

Average/Mean speed = \overline{x} = 15.62 Mbps

Standard Deviation = s = 20.03 Mbps

We need to find the difference between carrier's highest data speed and the mean of all 50 data​ speeds i.e. x - \overline{x}

x - \overline{x} = 75.6 - 15.62 = 59.98 Mbps

Thus, the difference between​ carrier's highest data speed and the mean of all 50 data​ speeds is 59.98 Mbps

Part b)

In order to find how many standard deviations away is the difference found in previous part, we divide the difference by the value of standard deviation i.e.

\frac{59.98}{20.03}=2.99

This means, the difference is 2.99 standard deviations or in other words we can say, the Carrier's highest data speed is 2.99 standard deviations above the mean data speed.

Part c)

A z score tells us that how many standard deviations away is a value from the mean. We calculated the same in the previous part. Performing the same calculation in one step:

The formula for the z score is:

z=\frac{x-\overline{x}}{s}

Using the given values, we get:

z=\frac{75.6-15.62}{20.03}=2.99

Thus, the Carriers highest data is equivalent to a z score of 2.99

Part d)

The range of z scores which are neither significantly low nor significantly​ high is -2 to + 2. The z scores outside this range will be significant.

Since, the z score for carrier's highest data speed is 2.99 which is well outside the given range, i.e. greater than 2, we can conclude that the  carrier's highest data speed​ is significantly higher.

3 0
3 years ago
Which expression is equivalent to –2(5x – 0.75)?
zzz [600]
-10x + 1.5 because you get rid of the parenthesis so -2 times 5 is -10 ( -10x ) and then -2 times -0.75 equals positive 1.5 due to two negatives canceling out to positive hope this helped!
4 0
2 years ago
Read 2 more answers
Simplify the following expression: (2x2+3x−4)−(4x2−7x+1)
Olegator [25]
The answer for this problem is B
6 0
3 years ago
Consider the differential equation y'' − y' − 20y = 0. Verify that the functions e−4x and e5x form a fundamental set of solution
KIM [24]

Answer:

Therefore e^{-4x} and e^{5x} are fundamental solution of the given differential equation.

Therefore  e^{-4x} and e^{5x} are linearly independent, since W(e^{-4x},e^{5x})=9e^x\neq 0

The general solution of the differential equation is

y=c_1e^{-4x}+c_2e^{5x}

Step-by-step explanation:

Given differential equation is

y''-y'-20y =0

Here P(x)= -1, Q(x)= -20 and R(x)=0

Let trial solution be y=e^{mx}

Then, y'=me^{mx}   and   y''=m^2e^{mx}

\therefore m^2e^{mx}-m e^{mx}-20e^{mx}=0

\Rightarrow m^2-m-20=0

\Rightarrow m^2-5m+4m-20=0

\Rightarrow m(m-5)+4(m-5)=0

\Rightarrow (m-5)(m+4)=0

\Rightarrow m=-4,5

Therefore the complementary function is = c_1e^{-4x}+c_2e^{5x}

Therefore e^{-4x} and e^{5x} are fundamental solution of the given differential equation.

If y_1 and y_2 are the fundamental solution of differential equation, then

W(y_1,y_2)=\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|\neq 0

Then  y_1 and y_2 are linearly independent.

W(e^{-4x},e^{5x})=\left|\begin{array}{cc}e^{-4x}&e^{5x}\\-4e^{-4x}&5e^{5x}\end{array}\right|

                    =e^{-4x}.5e^{5x}-e^{5x}.(-4e^{-4x})

                    =5e^x+4e^x

                   =9e^x\neq 0

Therefore  e^{-4x} and e^{5x} are linearly independent, since W(e^{-4x},e^{5x})=9e^x\neq 0

Let the the particular solution of the differential equation is

y_p=v_1e^{-4x}+v_2e^{5x}

\therefore v_1=\int \frac{-y_2R(x)}{W(y_1,y_2)} dx

and

\therefore v_2=\int \frac{y_1R(x)}{W(y_1,y_2)} dx

Here y_1= e^{-4x}, y_2=e^{5x},W(e^{-4x},e^{5x})=9e^x ,and  R(x)=0

\therefore v_1=\int \frac{-e^{5x}.0}{9e^x}dx

       =0

and

\therefore v_2=\int \frac{e^{5x}.0}{9e^x}dx

       =0

The the P.I = 0

The general solution of the differential equation is

y=c_1e^{-4x}+c_2e^{5x}

7 0
3 years ago
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