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3 years ago

11

Please I need help!!! My grade depends on it!

1 answer:

balu736 [363]3 years ago

8 0

A binomial distribution has the following properties;
1. There is a fixed number of trials.
2. Each trial is independent of the others.
3.Each trial has a constant probability of success.
4. There are only two possible outcomes, 'success' and 'failure'.

Statements B and D meet the requirements for the variable X to have a binomial distribution.

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Please can I have an explanation also, I am terrible at these kinds of questions!

wlad13 [49]

Answer:

<em>The fraction of the beads that are red is</em>

Step-by-step explanation:

<u>Algebraic Expressions</u>

A bag contains red (r), yellow (y), and blue (b) beads. We are given the following ratios:

r:y = 2:3

y:b = 5:4

We are required to find r:s, where s is the total of beads in the bag, or

s = r + y + b

Thus, we need to calculate:

$\displaystyle \frac{r}{r+y+b} \qquad\qquad [1]$

Knowing that:

$\displaystyle \frac{r}{y}=\frac{2}{3} \qquad\qquad [2]$

$\displaystyle \frac{y}{b}=\frac{5}{4}$

Multiplying the equations above:

$\displaystyle \frac{r}{y}\frac{y}{b}=\frac{2}{3}\frac{5}{4}$

Simplifying:

$\displaystyle \frac{r}{b}=\frac{5}{6} \qquad\qquad [3]$

Dividing [1] by r:

$\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{1}{1+y/r+b/r}$

Substituting from [2] and [3]:

$\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{1}{1+3/2+6/5}$

Operating:

$\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{1}{\frac{10+3*5+6*2}{10}}$

$\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{10}{10+15+12}$

$\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{10}{37}$

The fraction of the beads that are red is $\mathbf{\frac{10}{37}}$

8 0

3 years ago

Shawna lives in an apartment 9 4/5 miles from the hospital where she works. Her brother rents a room in a house 7 2/5 miles from

tamaranim1 [39]

Answer:

2 2/5 miles

Step-by-step explanation:

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2 years ago

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Degger [83]

The answer to this is 10.

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Please do all of these :)

wariber [46]

A SCATTER PLOT IS JUST A DOT PLOT

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Natalija [7]

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