What addition doubles fact can help you find 4+3? Explain how you know?

You and a friend play a game where you each toss a balanced coin. If the upper faces on the coins are both tails, you win $1; if

Zina [12379]

Answer: The mean and variance of Y are $0.25 and $6.19 respectively.

Step-by-step explanation:

The scenario is as follows: You and a friend participate in a game involving tossing a fair coin.

The sample space for tossing two coins is {TT, HT, TH, HH}

Let Y represent the earnings from one round of the game.

If both faces are heads, you win $1; therefore, P(Y=1)=P(TT)= $\dfrac{1}{4}=0.25$

You win $6 if both faces are heads, so P(Y=6)=P(HH)= $\dfrac{1}{4}=0.25$

If the faces do not match, you lose $3 which means P(Y=1)=P(TH, HT)= $\dfrac{2}{4}=0.50$

To find the expected value to win: E(Y)= $\sum_{i=1}^{i=3} y_ip(y_1)$

$=1\times0.25+6\times0.25+(-3)\times0.50=0.25$

Thus, the mean of Y: E(Y)= $0.25

$E(Y^2)=\sum_{i=1}^{i=3} y_i^2p(y_i)\\\\=1^2\times0.25+6^1\times0.25+(-3)^2\times0.5\\\\=0.25+1.5+4.5=6.25$

Variance = $E[Y^2]-E(Y)^2$

$=6.25-(0.25)^2=6.25-0.0625=6.1875\approx6.19$

Therefore, variance of Y = $ 6.19

6 0

1 month ago

In how many ways can you put seven marbles in different colors into two jars? Note that the jars may be empty.

tester [12383]

Answer:

128

Step-by-step explanation:

Imagine you have two jars and seven marbles.

The result is $2^{7}$ .

That totals 128

3 0

28 days ago

Evan deposits $500 in a savings account that earns interest. Let f(t)=500 and g(t)=1.05t, where t represents the time, in years,

zzz [12365]

Answer:d

Step-by-step explanation:

8 0

29 days ago

The time for a visitor to read health instructions on a Web site is approximately normally distributed with a mean of 10 minutes

Svet_ta [12734]

Response:

a) The average is 10 and the variance is 0.0625.

b) 0.6826 = 68.26% likelihood that the average time of visitors falls within 15 seconds of 10 minutes.

c) 10.58 minutes.

Step-by-step clarification:

To figure this out, we must comprehend the normal probability distribution alongside the central limit theorem.

Normal Probability Distribution

Issues involving normal distributions can be resolved using the z-score formula.

For a data set with mean $\mu$ and standard deviation $\sigma$ , the z-score related to a measure X is defined as:

$Z = \frac{X - \mu}{\sigma}$

The z-score illustrates how many standard deviations the measure is positioned from the mean. Once the z-score is calculated, we refer to the z-score table to find the p-value matching this z-score. This p-value reflects the likelihood that the measurement is less than X, which means it indicates the percentile of X. To determine the chance that the measure exceeds X, we subtract the p-value from 1.

Central Limit Theorem

The Central Limit Theorem states that for a normally distributed random variable X, defined by mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of sample means of size n can be approximated as a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For non-normal distributions, this theorem applies if n is at least 30.

In this case, the distribution has a mean of 10 minutes and a standard deviation of 2 minutes.

This indicates that $\mu = 10, \sigma = 2$

Assuming 64 visitors independently access the site.

This implies that $n = 64, = \frac{2}{\sqrt{64}} = 0.25$

a. The expected value and variance of the average time of the visitors.

Employing the Central Limit Theorem, the mean is 10 and the variance is (0.25)^2 = 0.0625.

b. The chance that the mean visit duration is within 15 seconds of 10 minutes.

15 seconds = 15/60 = 0.25 minutes, thus between 9.75 and 10.25 seconds, which corresponds to the p-value of z when X = 10.25 minus the p-value of z when X = 9.75.

X = 10.25

$Z = \frac{X - \mu}{\sigma}$

Through the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{10.25 - 10}{0.25}$

$Z = 1$

$Z = 1$ has a p-value of 0.8413.

X = 9.75

$Z = \frac{X - \mu}{s}$

$Z = \frac{9.75 - 10}{0.25}$

$Z = -1$

$Z = -1$ has a p-value of 0.1587.

0.8413 - 0.1587 = 0.6826.

0.6826 = 68.26% chance that the average visitor time is within 15 seconds of 10 minutes.

c. The value surpassed by the average visitor time with a probability of 0.01.

$Z = \frac{X - \mu}{s}$ This corresponds to the 99th percentile, denoting X when Z equals a p-value of 0.99, hence X when Z = 2.327.

$2.327 = \frac{X - 10}{0.25}$

$X - 10 = 2.327*0.25$

$X = 10.58$

Thus, the result is 10.58 minutes.

6 0

20 days ago

You’re working in a pharmacy, and are making a table to help with dosage amounts for a certain drug. The recommended dosage is 4

lawyer [12517]

Answer: Children who weigh 10 pounds require 12 teaspoons, those at 15 pounds need 18 teaspoons, and for 30 pounds, it is necessary to have 36 teaspoons.

Step-by-step explanation:

We have a recommended dose of 40 mg per 2.2 pounds of weight, given in three doses throughout the day every 8 hours.

Thus, for a single pound, the daily required dosage is calculated as $\dfrac{40}{2.2}=18.18\ ml$ .

Each individual dose corresponds to $\dfrac{18.18}{3}=6.06\ ml$ of the drug.

This means for one pound, the dosage needed per dose is 6.06 ml.

Thus, for 10 pounds, the calculation is 10 x 6.06 ml = 60.6 ml

To convert ml to teaspoons, knowing that 1 teaspoon equals 5 ml, we find: (60.6 ml ÷ 5) = 12.12 ≈ 12 teaspoons.

For a weight of 15 pounds, the drug requirement is 15 x 6.06 ml = 90.9 ml.

This translates to (90.9 ml ÷ 5) = 18.18 ≈ 18 teaspoons.

For the weight of 30 pounds, it is calculated as 30 x 6.06 ml = 181.8 ml.

Converting this gives us (181.8 ml ÷ 5) = 36.36 ≈ 36 teaspoons.

Therefore, children weighing 10 pounds need 12 teaspoons, those at 15 pounds require 18 teaspoons, and those at 30 pounds need 36 teaspoons.

7 0

2 months ago