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4 years ago

Is the answer b? If not then what’s the answer

Mathematics

2 answers:

solmaris [256]4 years ago

5 0

The correct answer is b. Hope i helped you!!!

m_a_m_a [10]4 years ago

3 0

I would say the answer is b, because it is a rhombus gape and all sides are congruent and there are 4 sides

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Which of the following has the same value

ioda

When solving (-2)5(-2)-8, you would first, following PEMDAS ofcourse, multiply -2 and 5, getting -10. Then you would multiply -2 and -10, which would be positive 20. Lastly, you would subtract 8 from 20, which equals 12.

I cannot find an answer within the options you gave me, but I am 100% sure that (-2)5(-2)-8 equals 12.

Hope this helped in some way! Please mark brainliest! Thank you v much! :)

4 0

4 years ago

a certain flight arrives on time percent of the time. suppose flights are randomly selected. use the normal approximation to the

iVinArrow [24]

The probability that exactly 128 flights are on time is 88.27% and the probability that atleast 1 flight is on time is 0.022.

Given that the chances of reaching the flight on time be 88%.

We are required to find the probability that exactly 128 flights are on time, atlleast 1 flight on time.

Probability is basically the calculation of likeliness of happening an event among all the events possible. It lies between 0 and 1.

Probability=Number of items/Total items.

It cannot be negative.

Probability that the flight will be on time if one flight is selected=0.88

Probability that exactly 128 flights are on time=

128=1145*P/100

P=12800/145

P=88.27%

b) Atleast 1 flight is on time

=( $128C_{0}(0.88)^{0} (0.12)^{1}$ + $128C_{1} (0.88)^{1} (0.12)^{11}$ )/128

=(0.12+128*0.88*0.245)/128

=(0.12+2.575)/128

=0.022

Hence the probability that exactly 128 flights are on time is 88.27% and the probability that atleast 1 flight is on time is 0.022.

Learn more about probability at brainly.com/question/24756209

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Other options are incomplete and not found anywhere else.

8 0

1 year ago

. The product of 3V5 and 6v5 is

rjkz [21]

The product of 3v5 and 16v5 is 18v^10

3 0

3 years ago

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 7.9 years, and standard deviation of

Darya [45]

If you randomly purchase one item, the probability it will last longer than 9 years is; 11.12%

<h3>How to find the probability from z-score?</h3>

The formula for Z-score is;

z = (x' - µ)/σ

where;

x' = sample mean

µ = population mean

σ = standard deviation

Thus;

z = (9 - 7.9)/0.9

z = 1.22

From online p-value from z-score calculator, we have;

probability = 0.1112 = 11.12%

Read more about Z-score probability at; brainly.com/question/25638875

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6 0

2 years ago

Use lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. f(x,y = xyz; x^

Snezhnost [94]

I'm assuming the constraint involves some plus signs that aren't appearing for some reason, so that you're finding the extrema subject to $x^2+2y^2+3z^2=96$ .

Set $f(x,y,z)=xyz$ and $g(x,y,z)=x^2+2y^2+3z^2-96$ , so that the Lagrangian is

$L(x,y,z,\lambda)=xyz+\lambda(x^2+2y^2+3z^2-96)$

Take the partial derivatives and set them equal to zero.

$\begin{cases}L_x=yz+2\lambda x=0\\L_y=xz+4\lambda y=0\\L_z=xy+6\lambda z=0\\L_\lambda=x^2+2y^2+3z^2-96=0\end{cases}$

One way to find the possible critical points is to multiply the first three equations by the variable that is missing in the first term and dividing by 2. This gives

$\begin{cases}\dfrac{xyz}2+\lambda x^2=0\\\\\dfrac{xyz}2+2\lambda y^2=0\\\\\dfrac{xyz}2+3\lambda z^2=0\\\\x^2+2y^2+3y^2=96\end{cases}$

So by adding the first three equations together, you end up with

$\dfrac32xyz+\lambda(x^2+2y^2+3z^2)=0$

and the fourth equation allows you to write

$\dfrac32xyz+96\lambda=0\implies \dfrac{xyz}2=-32\lambda$

Now, substituting this into the first three equations in the most recent system yields

$\begin{cases}-32\lambda+\lambda x^2=0\\-32\lambda+2\lambda y^2=0\\-32\lambda+3\lambda z^2=0\end{cases}\implies\begin{cases}x=\pm4\sqrt2\\y=\pm4\\z=\pm4\sqrt{\dfrac23}\end{cases}$

So we found a grand total of 8 possible critical points. Evaluating $f(x,y,z)=xyz$ at each of these points, you find that $f(x,y,z)$ attains a maximum value of $\dfrac{128}{\sqrt3}$ whenever exactly none or two of the critical points' coordinates are negative (four cases of this), and a minimum value of $-\dfrac{128}{\sqrt3}$ whenever exactly one or all of the critical points' coordinates are negative.

6 0

3 years ago