Which is the correct label for the angle?<br> С<br> B

Haddie flipped a fair coin 12 times. The coin landed on heads 11 times and tails 1 time. She flips the coin a 13th time. How lik

Tpy6a [65]

Answer:

50/50. there is no outside force that decides otherwise

6 0

3 years ago

Darren has the option of investing in either Stock A or Stock B. The probability of the return of Stock A being 25% is 0.45, 14%

Lunna [17]

Answer:

C)A is 15.95% ,B is 11.85%

Step-by-step explanation:

We know that the expected value in probability distribution is given as

Lets X is the expected value then

$X= \sum_{i=1}^{i=n} X_i P_i$

For stock A

X=0.25 x 0.45+0.14 x 0.25+0.04 x 0.3

X=0.1595

So the expected return for A is 15.95%

For stock 9

X=0.3 x 0.3+0.09 x 0.25+0.02 x 0.3

X=0.1185

So the expected return for B is 11.85%

So our option C will be the answer of that problem.

8 0

3 years ago

Solve integral (2x-5)⁷ dx

trapecia [35]

Answer: 128x^7dx-2240x^6dx+16800x^5dx-70000x^4dx+175000x^3dx-262500x^2dx+218750xdx-7812

Step-by-step explanation:

1) Expand : (2x-5\right)^7dxquad 128x^7dx-2240x^6dx+16800x^5dx-70000x^4dx+175000x^3dx-262500x^2dx

2) Distribute Parentheses

3) Apply your minus/plus rules

+(-a)= -a

128x^7dx-2240x^6dx+16800x^5dx-70000x^4dx+175000x^3dx-262500x^2dx+218750xdx-78125dx

4 0

3 years ago

There are two major tests of readiness for college: the act and the sat. act scores are reported on a scale from 1 to 36. the di

Olin [163]

Solution: We are given:

ACT scores follow normal distribution with $\mu=20.9,\sigma =4.8$

SAT scores follow normal distribution with $\mu=1026,\sigma=209$

Now, let's find the z score corresponding to Joe's SAT score 1351.

$z=\frac{x-\mu}{\sigma}$

$=\frac{1351-1026}{209}$

$=\frac{325}{209}$

$=1.56$

Therefore, Joe's SAT score is 1.56 standard deviations above the mean.

Now, we have find the Joe's ACT score, which will be 1.56 standard deviations above the mean.

Therefore, we have:

$z=\frac{x-\mu}{\sigma}$

$1.56=\frac{x-20.9}{4.8}$

$1.56 \times 4.8 = x - 20.9$

$7.488=x-20.9$

$x=20.9+7.488$

$x=28.388 \approx 28.4$

Therefore, Joe's equivalent ACT score to SAT score 1351 is 28.4

7 0

3 years ago

In response to the increasing weight of airline passengers, the Federal Aviation Administration in 2003 told airlines to assume

Rufina [12.5K]

Answer:

The approximate probability that the total weight of the passengers exceeds 4432 pounds is 0.2709

Step-by-step explanation:

$\mu = 192$

$\sigma = 31$

No. of passengers = n = 21

We are supposed to find the approximate probability that the total weight of the passengers exceeds 4432 pounds

$x = \frac{4432}{21} \sim 211$

$Z=\frac{x-\mu}{\sigma}\\Z=\frac{211-192}{31}$

Z=0.612

refer z table

P(Z<211)=0.7291

P(Z>211)=P(Z>4432)=1-0.7291=0.2709

Hence the approximate probability that the total weight of the passengers exceeds 4432 pounds is 0.2709

7 0

3 years ago

Which is the correct label for the angle? С B