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4 years ago

Plssssdsm helpppppppppppppp

Mathematics

1 answer:

zhuklara [117]4 years ago

8 0

The y intercept is -2. Remember when x is substituted as 0, you can find the y intercept

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Put the following into vertex form: y=-2x^2+4x+6

Ymorist [56]

This is how your answer would look like in vertex form:
y= -2 (x - 1)^2 + 8

4 0

3 years ago

A cone has a volume of 90ft3. The diameter of the cone is 6ft. What is the height of the cone? Round your answer to the nearest

Anvisha [2.4K]

V=πr²(1/3)(h)
d=6, therefore r=3 and
90=(3)²(1/3)(h) ... I didn't include the pi because it isn't in the answer
90=3h
h=30

8 0

3 years ago

Someone please help asap, I will mark as brainliest. Thank you!!

defon

Answer:

the answer is c go ahead and click it

Step-by-step explanation:

8 0

2 years ago

A certain kind of sheet metal has, on average, 3 defects per 18 square feet. Assuming a Poisson distribution, find the probabili

Fofino [41]

Answer:

75.8% probability that a 31 square foot metal sheet has at least 4 defects.

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.

In this problem, we have that:

3 defects per 18 square feet.

So for 31 square feet, we have to solve a rule of three

3 defects - 18 square feet

x defects - 31 square feet

$18x = 3*31$

$x = \frac{31*3}{18}$

$x = 5.17$

So $\mu = 5.17$

Assuming a Poisson distribution, find the probability that a 31 square foot metal sheet has at least 4 defects.

Either it has three or less defects, or it has at least 4 defects. The sum of the probabilities of these events is decimal 1.

$P(X \leq 3) + P(X \geq 4) = 1$

We want $P(X \geq 4)$

$P(X \geq 4) = 1 - P(X \leq 3)$

In which

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = \frac{e^{-5.17}*(5.17)^{0}}{(0)!} = 0.0057$

$P(X = 1) = \frac{e^{-5.17}*(5.17)^{1}}{(1)!} = 0.0294$

$P(X = 2) = \frac{e^{-5.17}*(5.17)^{2}}{(2)!} = 0.0760$

$P(X = 3) = \frac{e^{-5.17}*(5.17)^{3}}{(3)!} = 0.1309$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0057 + 0.0294 + 0.0760 + 0.1309 = 0.242$

Finally

$P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.242 = 0.758$

75.8% probability that a 31 square foot metal sheet has at least 4 defects.

8 0

4 years ago

A student reports three trials to measure the density of an unknown metal. The percent error in these trials is given in the tab

Tatiana [17]

I think that the most accurate one is trial 3 since it is the least incorrect and closest percent-wise to the actual correct density of the metal.
Hope this helps!

3 0

3 years ago