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brilliants [131]
3 years ago
6

GOOD MORNING PEOPLE :) CAN SOMEONE PLEASE HELP ME WITH THIS MATH QUESTION PLEASE ASAP:)

Mathematics
2 answers:
Alik [6]3 years ago
7 0
It's actually NOT b, if you do the math
correctly you should choose A
Ann [662]3 years ago
5 0

Answer:

B

Step-by-step explanation:

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Help pleaseeeeeee!!!!!
Semmy [17]

Answer:

the function R is always 2 above on the y axis to the function G

Step-by-step explanation:

adding 2 to x makes the function go up on the y axis

8 0
3 years ago
Need help fast pls thank you
Mama L [17]

Answer:

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
) Determine the probability that a bit string of length 10 contains exactly 4 or 5 ones.
yanalaym [24]

Answer: 0.4512

Step-by-step explanation:

A bit string is sequence of bits (it only contains 0 and 1).

We assume that the  0 and 1 area equally likely to any place.

i.e. P(0)= P(1)= \dfrac{1}{2}

The length of bits : n = 10

Let X = Number of getting ones.

Then , X \sim Bin(n=10,\ p=\dfrac{1}{2})

Binomial distribution formula : P(X=x)=^nC_x p^x q^{n-x} , where p= probability of getting success in each event and q= probability of getting failure in each event.

Here , p=q=\dfrac{1}{2}

Then ,The probability that a bit string of length 10 contains exactly 4 or 5 ones.

P(X= 4\ or\ 5)=P(x=4)+P(x=5)\\\\=^{10}C_4(\dfrac{1}{2})^{10}+^{10}C_4(\dfrac{1}{2})^{10}

=\dfrac{10!}{4!6!}(\dfrac{1}{2})^{10}+\dfrac{10!}{5!5!}(\dfrac{1}{2})^{10}

=(\dfrac{1}{2})^{10}(\dfrac{10!}{4!6!}+\dfrac{10!}{5!5!})

=(\dfrac{1}{2})^{10}(210+252)

=(0.0009765625)(462)

=0.451171875\approx0.4512

Hence, the  probability that a bit string of length 10 contains exactly 4 or 5 ones is 0.4512.

3 0
4 years ago
What is the x-coordinate for the minimum point in the function f(x) = 4 cos(2x − π) from x = 0 to x = π?
melamori03 [73]
When you are looking at a graph, a minimum point would be where the curve is decreasing, then begins to increase. Right at the point where it switches, the slope is a horizontal line, or 0. We can take the derivative is f(x), then look for all the x values where the slope (which is equal to the first derivative) is equal to zero.

f'(x) = 2 * -4sin(2x - pi)

The 2 comes from the derivative of the inside, 2x-pi.

So now set the derivative equal to 0.

-8sin(2x-pi) = 0

We can drop the -8 by dividing both sides by -8.

sin(2x-pi) = 0

This can be rewritten as arcsin(0) = 2x-pi

So when theta equals 0, what is the value of sin(theta)? At an angle of 0, there is just a horizontal line pointing to the right on the unit circle with length of 1. Sine is y/h, but there is no y value so it is just 0. If arcsin(0) = 0, we can now set 2x-pi = 0

2x = pi

x = pi/2

This is a critical number. To find the minimum value between 0 and pi, we need to find the y values for the endpoints and the critical number.

f(0) = -4

f(pi/2) = 4

f(pi) = -4

So the minimum points are at x=0 and x=pi




5 0
4 years ago
What is the constant of proportionality in this proportional relationship?
lora16 [44]

Answer:

7/9

Step-by-step explanation:

1 1/5 / 2 = 7/9 = 0.777

6 0
3 years ago
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