Plz help me!!!!!!!!! Whoever solves correct will mark brainlist

Mathematics

1 answer:

OleMash [197]2 years ago

7 0

Answer:

in picture

Step-by-step explanation:

Brainliest please~

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Which function below has the following domain and range?

Lelechka [254]

Answer:

{(2,0),(-5,1),(7,8),(6,0),(-7,1)

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2 years ago

Y varies directly with x, and y = 7 when x = 2. What is the value of y when x = 5?

____ [38]

Y varies directly with x" means y = constant * x or y = kx ----- (1)1. when x = -1/2, y = 2Put it in (1):2 = k * (-1/2)multiply by 2:4 = k * (-1)4 = -kk = -4y = -4x
"find the value of y when x= -0.3"y = -4x = (-4) * (-0.3) = 1.2

So for each problem, start with y = kxThey give you one x and one y value.Put it in y = kx and find k.Then put x = -0.3 and find y.

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3 years ago

The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds. the prob

Anni [7]

Given:
μ = 200 lb, the mean
σ = 25, the standard deviation

For the random variable x = 250 lb, the z-score is
z = (x-μ)/σ =(250 - 200)/25 = 2

From standard tables for the normal distribution, obtain
P(x < 250) = 0.977

Answer: 0.977

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3 years ago

A 5-card hand is dealt from a perfectly shuffled deck. Define the events: A: the hand is a four of a kind (all four cards of one

TiliK225 [7]

In a hand of 5 cards, you want 4 of them to be of the same rank, and the fifth can be any of the remaining 48 cards. So if the rank of the 4-of-a-kind is fixed, there are $\binom44\binom{48}1=48$ possible hands. To account for any choice of rank, we choose 1 of the 13 possible ranks and multiply this count by $\binom{13}1=13$ . So there are 624 possible hands containing a 4-of-a-kind. Hence A occurs with probability

$\dfrac{\binom{13}1\binom44\binom{48}1}{\binom{52}5}=\dfrac{624}{2,598,960}\approx0.00024$

There are 4 aces in the deck. If exactly 1 occurs in the hand, the remaining 4 cards can be any of the remaining 48 non-ace cards, contributing $\binom41\binom{48}4=778,320$ possible hands. Exactly 2 aces are drawn in $\binom42\binom{48}3=103,776$ hands. And so on. This gives a total of

$\displaystyle\sum_{a=1}^4\binom4a\binom{48}{5-a}=886,656$

possible hands containing at least 1 ace, and hence B occurs with probability

$\dfrac{\sum\limits_{a=1}^4\binom4a\binom{48}{5-a}}{\binom{52}5}=\dfrac{18,472}{54,145}\approx0.3412$

The product of these probability is approximately 0.000082.

A and B are independent if the probability of both events occurring simultaneously is the same as the above probability, i.e. $P(A\cap B)=P(A)P(B)$ . This happens if

the hand has 4 aces and 1 non-ace, or
the hand has a non-ace 4-of-a-kind and 1 ace

The above "sub-events" are mutually exclusive and share no overlap. There are 48 possible non-aces to choose from, so the first sub-event consists of 48 possible hands. There are 12 non-ace 4-of-a-kinds and 4 choices of ace for the fifth card, so the second sub-event has a total of 12*4 = 48 possible hands. So $A\cap B$ consists of 96 possible hands, which occurs with probability