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Sveta_85 [38]
3 years ago
14

I need to know this please

Mathematics
2 answers:
anyanavicka [17]3 years ago
4 0
The answer is C



hope this helps :))
jenyasd209 [6]3 years ago
3 0
The answer should be C

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PLEASE HELP!!! 30 POINTS!!! NEED ANSWERS. WILL MARK BRAINIEST. FLAGGING NON RELEVANT ANSWER Follow the directions to convert the
kumpel [21]

Answer:

y-values of sampled wave:

0, 5, 5, -3, -5, -3, 2, 0, 0, -1, -2, 1, 1, -1, -1, -2, 2, 3, -1

Step-by-step explanation:

My teacher told me to put a dot on very cross or intersection that the line is on.

4 0
3 years ago
How to I solve problem 1?
Verdich [7]
#1

The uniforms are numbered 0, 1, 2, ..., 99. That's 100 numbers. Half of them are odd and half of them are even. So the probability that any one of the uniforms is odd is 1/2 just like the probability that any one uniform is even is 1/2.

(a) The numbers on the uniforms are independent of one another. That is, the number of her cross-country uniform does not in any way determine the number on her basketball uniform and vice versa. This means that we can find the probability that each is odd and multiply these together using what is called the counting principle. The probability that all are odd is:
(1/2)(1/2)(1/2)=1/8

(b) This is done the same way we did part (a). Since the probability of any one uniform being odd is the same as it being even (1/2), the answer here is the same: (1/2)(1/2)(1/2)=1/8

(c) This problem differs from that in (a) and (b). There is only one way for all three uniforms to be odd numbers: (odd, odd, odd) or all even (even, even, even). However, there are multiple ways for the uniforms to be two odd and one even. If the uniforms are listed in order: cross-country, basketball, softball we can get exactly one even in any of three ways:
even, odd, odd
odd, even, odd
odd, odd, even
The probability for any one of these possibilities is (1/2)(1/2)(1/2)=1/8 but since there are three way the probability that we get even exactly once is equal to (3)(1/8) = 3/8
7 0
3 years ago
It is believed that 11% of all Americans are left-handed. In a random sample of 100 students from a particular college with 1012
tia_tia [17]

Answer:

Step-by-step explanation:

We can calculate this confidence interval using the population proportion calculation. To do this we must find p' and q'

Where p' = 14/100= 0.14 (no of left handed sample promotion)

q' = 1-p' = 1-0.14= 0.86

Since the requested confidence level is CL = 0.98, then α = 1 – CL = 1 – 0.98 = 0.02/2= 0.01, z (0.01) = 2.326

Using p' - z alpha √(p'q'/n) for the lower interval - 0.14-2.326√(0.14*0.86/100)

= -2.186√0.00325

= -2.186*0.057

= 12.46%

Using p' + z alpha √(p'q'/n)

0.14+2.326√(0.14*0.86/100)

= 0.466*0.057

= 26.5%

Thus we estimate with 98% confidence that between 12% and 27% of all Americans are left handed.

6 0
3 years ago
The movement of a projectile, thrown upwards vertically, is described by the equation y = - 40x<img src="https://tex.z-dn.net/?f
kow [346]
At X = 2.5 the projectile hits the ground
7 0
3 years ago
Need help please. I need help with how to do these problems.
Vikki [24]
Look it up I will find it
3 0
3 years ago
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