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2 years ago

PLZZZZ NEED HELP!!!!!

Mathematics

1 answer:

LuckyWell [14K]2 years ago

8 0

It's "all values except 4". 4 is a vertical asymptote and the function does not exist there at all. But the left side comes in from negative infinity and goes down to negative infinity, and on the right it comes down from positive infinity, but since it's getting closer and closer to the horizontal asymptote of y = 0, it's decreasing there too.

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nventing is a difficult way to make money. Only 5% of new patents earn a substantial profit. A certain city has just had30 indep

Arlecino [84]

Answer:

$P(X \geq 2) = 1-P(X$

And we can find the individual probabilities using the probability mass function and we got:

$P(X=0) = (30C0) (0.05)^0 (1-0.05)^{30-0} =0.2146$

$P(X=1) = (30C1) (0.05)^1 (1-0.05)^{30-1} = 0.3389$

And replacing we got:

$P(X \geq 2) = 1-P(X$

Step-by-step explanation:

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".

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

$X \sim Binom(n=30, p=0.05)$

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:

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

Solution to the problem

For this case we want this probability:

$P(X \geq 2)$

And we can use the complement rule and we got:

$P(X \geq 2) = 1-P(X$

And we can find the individual probabilities using the probability mass function and we got:

$P(X=0) = (30C0) (0.05)^0 (1-0.05)^{30-0} =0.2146$

$P(X=1) = (30C1) (0.05)^1 (1-0.05)^{30-1} = 0.3389$

And replacing we got:

$P(X \geq 2) = 1-P(X$

3 0

3 years ago

The weight of the chocolate and Hershey Kisses are normally distributed with a mean of 4.5338 G and a standard deviation of 0.10

Salsk061 [2.6K]

For the bell-shaped graph of the normal distribution of weights of Hershey kisses, the area under the curve is 1, the value of the median and mode both is 4.5338 G and the value of variance is 0.0108.

In the given question,

The weight of the chocolate and Hershey Kisses are normally distributed with a mean of 4.5338 G and a standard deviation of 0.1039 G.

We have to find the answer of many question we solve the question one by one.

From the question;

Mean(μ) = 4.5338 G

Standard Deviation(σ) = 0.1039 G

(a) We have to find for the bell-shaped graph of the normal distribution of weights of Hershey kisses what is the area under the curve.

As we know that when the mean is 0 and a standard deviation is 1 then it is known as normal distribution.

So area under the bell shaped curve will be

$\int\limits^{\infty}_{-\infty} {f(x)} \, dx$ = 1

This shows that that the total area of under the curve.

(b) We have to find the median.

In the normal distribution mean, median both are same. So the value of median equal to the value of mean.

As we know that the value of mean is 4.5338 G.

So the value of median is also 4.5338 G.

In the normal distribution mean, mode both are same. So the value of mode equal to the value of mean.

As we know that the value of mean is 4.5338 G.

So the value of mode is also 4.5338 G.

(d) we have to find the value of variance.

The value of variance is equal to the square of standard deviation.

So Variance = $(0.1039)^2$

Variance = 0.0108

Hence, the value of variance is 0.0108.

To learn more about normally distribution link is here

brainly.com/question/15103234

#SPJ1

3 0

10 months ago

Express the following in standard form (0.000000045)^4

snow_tiger [21]

0.00000004 to the power of 4

Your answer would be <u>0</u>

5 0

2 years ago

Read 2 more answers

How do you solve this inequality; 1/3x>-2?

Kisachek [45]

1/3x>-2=1>3x.-2
=1>-6x
=-1/6>x

4 0

3 years ago

Hi boys can you help me pls!!!!!!!

goblinko [34]

Answer: OPTION A.

Step-by-step explanation:

Given the following function:

$h(x)=-\frac{1}{4}x^2+\frac{1}{2}x+\frac{1}{2}$

You know that it represents the the height of the ball (in meters) when it is a distance "x" meters away from Rowan.

Since it is a Quadratic function its graph is parabola.

So, the maximum point of the graph modeling the height of the ball is the Vertex of the parabola.

You can find the x-coordinate of the Vertex with this formula:

$x=\frac{-b}{2a}$

In this case:

$a=-\frac{1}{4}\\\\b=\frac{1}{2}$

Then, substituting values, you get:

$x=\frac{-\frac{1}{2}}{(2)(-\frac{1}{4}))}\\\\x=1$

Finally, substitute the value of "x" into the function in order to get the y-coordinate of the Vertex:

$h(1)=y=-\frac{1}{4}(1)^2+\frac{1}{2}(1)+\frac{1}{2}\\\\y=0.75$

Therefore, you can conclude that:

<em> The maximum height of the ball is 0.75 of a meter, which occurs when it is approximately 1 meter away from Rowan.</em>

7 0

3 years ago