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3 years ago

Pllllllllzzzzzzz answer this!!!♥️♥️

Mathematics

1 answer:

Umnica [9.8K]3 years ago

6 0

Answer:

Use the regression calculator to compare the teams’ number of runs with their number of wins.

A 2-column table with 9 rows. Column 1 is labeled R with entries 808, 768, 655, 684, 637, 619, 613, 609, 563. Column 2 is labeled W with entries 93, 94, 66, 81, 86, 75, 61, 69, 55.

What is the y-intercept of the trend line, to the nearest hundredth?

Step-by-step explanation:

Pinky

2 years ago

Dude , you are suppose to answer it not question it!

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Choose yes or no to tell whether the expression represent a 20% discount off the price of an item that originally cost d dollars

Vaselesa [24]

Answer: B. d - 0.2

Step-by-step explanation: 20%= 20/100= 0.2

Which means = d-0.2

Hope this helped.

8 0

3 years ago

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Lee drove 420 miles and used 15 gallons of gasoline. How many miles did Lee’s car travel per gallon of gasoline?

olasank [31]

Answer:

28 miles per gallon

Step-by-step explanation:

3 0

3 years ago

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Kirsten has 9 syrup containers from a local cafe. There are 6 milliliters of syrup per container.

ololo11 [35]

Answer: 54 mL

Step-by-step explanation:

Simply do 9(number of containers)*6(Syrup per container) to get 54 mL of syrup.

Hope it helps <3

3 0

3 years ago

In a list of households, own homes and do not own homes. Five households are randomly selected from these households. Find the p

hichkok12 [17]

Answer:

The probability of success for this case would be:

$p =\frac{9}{15}= 0.6$ representing the proportion of homes that are own homes

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

$X \sim Binom(n=15, p=0.6)$

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:

$P(X=3)$

And uing the probability mass function we got:

$P(X=3)= 15C3 (0.6)^3 (1-0.6)^{15-3}= 0.00165$

Step-by-step explanation:

Adduming the following info: In a list of 15 households, 9 own homes and 6 do not own homes. Five households are randomly selected from these 15 households. Find the probability that the number of households in these 5 own homes is exactly 3.

Round your answer to four decimal places

P (exactly 3)=

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

The probability of success for this case would be:

$p =\frac{9}{15}= 0.6$ representing the proportion of homes that are own homes

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

$X \sim Binom(n=15, p=0.6)$

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:

$P(X=3)$

And uing the probability mass function we got:

$P(X=3)= 15C3 (0.6)^3 (1-0.6)^{15-3}= 0.00165$

4 0

2 years ago

The probability that a male will be colorblind is .042. Find the probabilities that in a group of 53 men, the following are true

Dvinal [7]

Answer:

See below for answers and explanations

Step-by-step explanation:

Part A

$\displaystyle P(X=x)=\binom{n}{x}p^xq^{n-x}\\\\P(X=5)=\binom{53}{5}(0.042)^5(1-0.042)^{53-5}\\\\P(X=5)=\frac{53!}{(53-5)!*5!}(0.042)^5(0.958)^{48}\\\\P(X=5)\approx0.0478$

Part B

$P(X\leq5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\P(X\leq5)=\binom{53}{1}(0.042)^1(1-0.042)^{53-1}+\binom{53}{2}(0.042)^2(1-0.042)^{53-2}+\binom{53}{3}(0.042)^3(1-0.042)^{53-3}+\binom{53}{4}(0.042)^4(1-0.042)^{53-4}+\binom{53}{1}(0.042)^5(1-0.042)^{53-5}\\\\P(X\leq5)\approx0.9767$

Part C

$\displaystyle P(X\geq 1)=1-P(X=0)\\\\P(X\geq1)=1-(1-0.042)^{53}\\\\P(X\geq1)\approx1-0.1029\\\\P(X\geq1)\approx0.8971$

6 0

2 years ago