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3 years ago

Please help i wanna move to the next grade,10 points

Mathematics

1 answer:

AfilCa [17]3 years ago

4 0

Answer:

C -1

Step-by-step explanation:

$- 3x + 2 < 5$

$5 - 2 = 3$

$- 3x \div - 3$

$3 \div - 3 = - 1$

$x = - 1$

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If x is an even number, the function f(x) = 2x − 1 gives an odd number. Identify the set of odd numbers corresponding to this se

Alex787 [66]

X = {0, 2, 4, 6, 8}.

=> f(x) = 2x - 1

x                  f(x) = 2x - 1

0                   2*0 - 1 = - 1

2                   2*2 - 1 = 3

4                   2*4 - 1 = 7

6                   2*6 - 1 = 11

8                   2*8 - 1 = 15

Answer: {-1, 3, 7, 11, 15}

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3 years ago

Can someone help me with this question please?

Simora [160]

Answer:

Each sequence has a difference of eight numbers so we add 8 numbers in each sequence

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3 years ago

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There are 10 golf balls in a rectangular prism. Each golf ball has a diameter of 1.25 inches. Find the volume of the golf balls

lara31 [8.8K]

The answer is D.

The formula for the volume of a sphere is $\frac{4}{3} \pi r^{3}$

You would divide the diameter in half which would give you your radius (r).
Then you plug in the radius value to find the volume of ONE ball.
You would then multiply by 10 to find the volume of TEN golf balls.

D is the closest answer to the volume of ten golf balls

8 0

3 years ago

The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

If the mean number of arrivals is 10

This means that $\mu = 10$

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347$

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is $P(X \leq 10)$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045$

$P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045$

$P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023$

$P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076$

$P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189$

$P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378$

$P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631$

$P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901$

$P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126$

$P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251$

$P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831$

58.31% probability that there are no more than 10 arrivals.

8 0

3 years ago

HELP PLEASE PLEASE PLEASE

CaHeK987 [17]

Answer:

If you beg for help like that you won't get any.

Step-by-step explanation:

6 0

2 years ago