Ask question

Ask question

All categories

BlackZzzverrR [31]

3 years ago

15

A couple plans to have four children. Using a tree diagram , obtain the sample space. List the elements that make up the sample

space. (Use "B" for "boy" and "G" for "girl.")
A. BBBG, BBGB, BBGG, BGBB, BGBG, BGGB, BGGG, GBBB, GBBG, GBGB, GBGG, GGBB, GGBG, GGGB

B. BBBB, BBBG, BBGB, BBGG, BGBB, BGBG, BGGB, BGGG, GBBB, GBBG, GBGB, GBGG, GGBB, GGBG, GGGB, GGGG

C. BBBB, BBB, BB, B

D. BBBB, BBBG, BBGG, BGGG, GGGG

1 answer:

blagie [28]3 years ago

4 0

<span>Correct answer is B.

There are $2^4 = 16$ points in the sample space, so choices C and D are automatically out.

Choice A is out because there is a possibility to have all 4 boys, but BBBB is not in choice A.

So the answer is choice B</span>

You might be interested in

I need help solving this problem

Olin [163]

3/5 is the correct answer for sin s

8 0

2 years ago

X4-3x3+3x2-3x+6÷x-2 synthetic division

Dima020 [189]

2 | 1 3 3 -3 6
| 0 2 10 26 46
___________

1 5 13 23 52

x^3+5x^2+13x+23+ (52/(x-2))

5 0

3 years ago

Read 2 more answers

What is the simplest form of 46/8

PilotLPTM [1.2K]

23/4

If you divide the numerator and denominator by 2, you get simplest form.

7 0

3 years ago

Read 2 more answers

How to find (c,d,e)<br>Please do a step by step working.Thank you.

Natasha2012 [34]

(a)
The inverse is when you swap the variables and solve for y.
g(t) = 2t - 1 (Note: g(t) represents y)
rewrite as: y = 2t - 1
swap the variables: t = 2y - 1
solve for y: t + 1 = 2y
   $\frac{t + 1}{2}$ = y
Answer for (a): $g^{-1}(t)$ = $\frac{t + 1}{2}$

(b)
Same steps as part (a) above:
h(t) = 4t + 3
rewrite as: y = 4t + 3
swap the variables: t = 4y + 3
solve for y: $y =\frac{t - 3}{4}$

Answer for (b): $h^{-1}(t)$ = $\frac{t - 3}{4}$

(c)
$g^{-1} ( h^{-1}(t)) = g^{-1} (\frac{t - 3}{4})$
replace all t's in the $g^{-1}(t)$ equation with $\frac{t - 3}{4}$
$g^{-1} (\frac{t - 3}{4})$ = $\frac{ \frac{t-3}{4} + 1}{2}$
= $\frac{ \frac{t-3}{4} + \frac{4}{4}}{2}$ = $\frac{ \frac{t - 3 + 4}{4}}{2}$ = $\frac{ \frac{t + 1}{4}}{2} = \frac{t + 1}{8}$
Answer for (c): $g^{-1} ( h^{-1}(t))$ = $\frac{t + 1}{8}$

(d)
h(g(t)) = h(2t - 1) = 4(2t - 1) + 3 = 8t - 4 + 3 = 8t - 1
Answer for (d): h(g(t)) = 8t - 1

(e)
h(g(t)) = 8t - 1
   y = 8 t - 1
   t = 8y - 1
t + 1 = 8y
$\frac{t + 1}{8}$ = y
Answer for (e): inverse of h(g(t)) = $\frac{t + 1}{8}$

8 0

3 years ago

The doubling period of a bacterial population is 20 20 minutes. At time t = 100 t=100 minutes, the bacterial population was 9000

Nuetrik [128]

Answer:

The initial population was 2810

The bacterial population after 5 hours will be 92335548

Step-by-step explanation:

The bacterial population growth formula is:

$P = P_0 \times e^{rt}$

where P is the population after time t, $P_0$ is the starting population, i.e. when t = 0, r is the rate of growth in % and t is time in hours

Data: The doubling period of a bacterial population is 20 minutes (1/3 hour). Replacing this information in the formula we get:

$2 P_0 = P_0 \times e^{r 1/3}$

$2 = e^{r \; 1/3}$

$ln 2 = r \; 1/3$

$ln 2 \times 3 = r$

$2.08 \% = r$

Data: At time t = 100 minutes (5/3 hours), the bacterial population was 90000. Replacing this information in the formula we get:

$90000 = P_0 \times e^{2.08 \; 5/3}$

$\frac{9000}{e^{2.08 \; 5/3}} = P_0$

$2810 = P_0$

Data: the initial population got above and t = 5 hours. Replacing this information in the formula we get:

$P = 2810 \times e^{2.08 \; 5}$

$P = 92335548$

3 0

3 years ago