Find three consecutive integers whose sum is 51.

6x7-8/4 using the order of operations will reward 10 pts best answer will be marked brainliest.

Phantasy [73]

Answer:

40

Step-by-step explanation:

6x7-8/4 =42 - 2 = 40

8 0

3 years ago

The average number of mosquitoes in a stagnant pond is 80 per square meter with a standard deviation of 12. If 36 square meters

svp [43]

Answer:

a) 18.4%

Step-by-step explanation:

Assuming a normal distribution, the z-score (z) for the probability of the average of the mosquitoes count being 'X' is given by:

$z(X)=\frac{X-\mu }{\frac{\sigma}{\sqrt{n}}}$

Where 'μ' is the distribution mean, 'σ' is the standard deviation and 'n' is the number of square meters chosen.

For X = 81.8 and n=36:

$z(X)=\frac{81.8 - 80}{\frac{12}{\sqrt{36}}}\\z(X)= 0.9$

A z-score of 0.9 is equivalent to the 81.59 th percentile in a normal distribution.

Therefore, the probability (P) that the average of those counts is more than 81.8 mosquitoes per square meter is:

$P= 100\% - 81.59\%\\P=18.41\%$

8 0

3 years ago

Please help explain good luck

Tcecarenko [31]

Answer:

27

Step-by-step explanation:

Remember to follow PEMDAS. Note that, when multiplying or dividing:

1) one positive and one negative = negative number

2) two positive = positive number

3) two negative = positive number

(-7)(-5) = -7 * -5 = 35

+(-8) = +1 * -8 = -8

Subtract:

35 - 8 = 27

27 is your answer.

~

8 0

3 years ago

Read 2 more answers

Solve the equation <img src="https://tex.z-dn.net/?f=%2835%20x%5E%7B4%7D%20y%2B14%20x%5E%7B5%7D%20y-2%20y%5E%7B3%7D-4x%20y%5E%7B

jeka94

$\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0$

$M_y=35x^4+14x^5-6y^2-12xy^2$
$N_x=35x^4-6y^2$

$\dfrac{N_x-M_y}N=\dfrac{-14x^5+12xy^2}{7x^5-6xy^2}=-2$

This suggests an integrating factor depending on $x$ only is possible, and given by

$\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^{2x}$

Distributing across the ODE, we end up with

$\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^{2x}}_{M^*}\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^{2x}}_{N^*}\,\mathrm dy=0$

The equation is now exact, with

${M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^{2x}$

Now we find the solution:

$F_x=M^*$
$F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^{2x}\,\mathrm dx$
$F=(7x^5y-2xy^3)e^{2x}+f(y)$

$F_y=N^*$
$(7x^5-6xy^2)e^{2x}+f'(y)=(7x^5-6xy^2)e^{2x}$
$f'(y)=0$
$\implies f(y)=C$

The general solution is then

$F(x,y)=(7x^5y-2xy^3)e^{2x}=C$

7 0

3 years ago

Determine the quadratic inequality that produces the graph below. HELP ASAP!

svetlana [45]

Answer:

<em> believe it's option C but not 100 percent sure.</em>

Step-by-step explanation:

I'm not good at explaining.

4 0

3 years ago