Which equations could be used to solve for the unknown lengths of ABC? Check all that apply.

with the salary cap in the NFL, it is said that on "any given Sunday" any team could beat any other team. If we assume every wee

prohojiy [21]

Answer:

$P(X \geq 1) = 1-P(X$

And we can find the individual probability like this:

$P(X=0) = 16C0 (0.5)^0 (1-0.5)^{16-0} = 0.0000153$

And replacing we got:

$P(X \geq 1) = 1-P(X$

Step-by-step explanation:

Assuming the following question: With the salary cap in the NFL, it is said that on any given Sunday any team could beat any other team. If we assume every week of the 16 week season a team has a 50% chance of winning, what is the probability that a team will have at least 1 win?

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=16, p=0.5)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

And we want this probability:

$P(X \geq 1)$

And using the complement rule we got:

$P(X \geq 1) = 1-P(X$

And we can find the individual probability like this:

$P(X=0) = 16C0 (0.5)^0 (1-0.5)^{16-0} = 0.0000153$

And replacing we got:

$P(X \geq 1) = 1-P(X$

7 0

3 years ago

Please help with algebra class I’m struggling

adell [148]

Answer:

Step-by-step explanation:

For what mileages will company A (which has a flat rate of $111) be less than the cost for company B

Company A < Company B

111 < 75 + 0.80m

solving

111 - 75 < 75 - 75 + 0.80m

36 < 0.80m

36/0.80 < 0.80m/0.80

as we are dividing by a positive number, we don't have to change the direction of the < sign.

45 < m

or m > 45

so if Ali intends to drive more than 45 miles, Company A's vehicle will be less expensive to rent

5 0

3 years ago

Suppose that the probability of getting a job offer (from a specific company) is 0.17. Suppose that we observe empirically that

yarga [219]

Answer:

P (X | CI) = 0.7313

Step-by-step explanation:

X = event of getting a job

CI = Campus interview

P (X | CI) = p ( CI ∩ X) / P ( CI ) = P( CI | X)P(X) / P( CI | X)P(X) + P( CI | X')P(X')

P (X | CI) = 0.93 * 0.17 / 0.93 * 0.17 + 0.07 * (1 - 0.17)

P (X | CI) = 0.93 * 0.17 / 0.93 * 0.17 + 0.07 * 0.83

P (X | CI) = 0.1581 / 0.1581 + 0.0581

P (X | CI) = 0.1581 / 0.2162

P (X | CI) = 0.73126735

P (X | CI) = 0.7313

8 0

4 years ago

• Hari bought 25 kg of pure ghee at Rs 180 per kg and 25 kg vanaspati ghee of Rs.80 per kg. He mixed both types of ghee and sold

Basile [38]

Step-by-step explanation:

as given,

hari bought,

amount of pure ghee he bought = 25 kg

cost of pure ghee per kg = Rs. 180

so, cost of 25 kg of pure ghee = 180 × 25 = Rs 4,500

amount of vanaspati ghee = 25 kg

cost of per kg of vanaspati ghee = Rs. 80

so, cost of 25 kg of vanaspati ghee = 80 × 25 = Rs. 2,000

he mixed both the ghee and got ( 25 + 25 ) = 50 kg of mixed ghee.

he sold them in cost = Rs. 150 per kg

so, total SP of mixed ghee = 150 × 50 = Rs. 7,500

so,

total CP = ( 4500 + 2000 ) = Rs. 6,500

total SP = Rs. 7,500

so,

Profit = SP - CP

→ 7,500 - 6,500 = Rs. 1000

Profit / gain = ( gain × 100 / CP)

→ (1000 × 100 / 6,500 )

→ 1000/65 = 15.38 ( approx.)

therefore, it's a profit of 15.38%.

hope this answer helps you dear! and may u have a great day ahead! take care!

8 0

3 years ago

Please help with this, i tried to solve for x, and it didnt give me any of those answers

BartSMP [9]

The answer is A. x= -7

7 0

3 years ago