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3 years ago

Help me pls ?? 1. Suppose there is a raffle with numbered tickets 1 - 150. One ticket is randomly

Mathematics

1 answer:

padilas [110]3 years ago

3 0

A = 50% / equally likely
B = less likely / unlikely
C =less likely
D = Impossible

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Kamila [148]

Answer:

I just need points

Step-by-step explanation:

I just need points

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3 years ago

Choosing an odd number. 1-12. Put it in a fraction

xxMikexx [17]

2/6 is a fraction. the denominator has to be an even number.

7 0

3 years ago

Lesson: 1.08Given this function: f(x) = 4 cos(TTX) + 1Find the following and be sure to show work for period, maximum, and minim

Ber [7]

The given function is

$f(x)=4\cos \text{(}\pi x)+1$

The general form of the cosine function is

$y=a\cos (bx+c)+d$

a is the amplitude

2pi/b is the period

c is the phase shift

d is the vertical shift

By comparing the two functions

a = 4

b = pi

c = 0

d = 1

Then its period is

$\begin{gathered} \text{Period}=\frac{2\pi}{\pi} \\ \text{Period}=2 \end{gathered}$

The equation of the midline is

$y_{ml}=\frac{y_{\max }+y_{\min }}{2}$

Since the maximum is at the greatest value of cos, which is 1, then

$\begin{gathered} y_{\max }=4(1)+1 \\ y_{\max }=5 \end{gathered}$

Since the minimum is at the smallest value of cos, which is -1, then

$\begin{gathered} y_{\min }=4(-1)+1 \\ y_{\min }=-4+1 \\ y_{\min }=-3 \end{gathered}$

Then substitute them in the equation of the midline

$\begin{gathered} y_{ml}=\frac{5+(-3)}{2} \\ y_{ml}=\frac{2}{2} \\ y_{ml}=1 \end{gathered}$

The answers are:

Period = 2

Equation of the midline is y = 1

Maximum = 5

Minimum = -3

3 0

1 year ago

In 2011, the Institute of Medicine (IOM), a non-profit group affiliated with the Select one US National Academy of Sciences, rev

Sergio [31]

Answer:

B. Yes, there are at least 10 people with weak bones and 10 people with strong bones in each group.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

The correct answer is:

B. Yes, there are at least 10 people with weak bones and 10 people with strong bones in each group.

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3 years ago

BRAINLIEST TO THE BEST ANSWER !!!!

TiliK225 [7]

Which equation am I looking at

4 0

3 years ago