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Alecsey [184]
3 years ago
7

PLEASE HELP ASAP THIS IS DUE SOON

Mathematics
2 answers:
Alla [95]3 years ago
5 0

Answer:

True

Step-by-step explanation:

mote1985 [20]3 years ago
5 0

Answer:

True

Step-by-step explanation:

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I don’t know if the first one is right but could someone guess true or false?
victus00 [196]

Answer:

B) false D) true E) true

Step-by-step explanation:

Plug in numbers to disprove the first one.

Area formula for triangles says D is true

It's the Pythagorean's Theorem for Heaven's sakes.

3 0
4 years ago
find the quadratic equation if its solutions are 1.5 and -.25, and the equation coefficients are not decimals.
denpristay [2]

If a quadratic equation has solutions x_1 and x_2, then we can write it as

(x-x_1)(x-x_2)

So, in your case, we have (let me write 1.5 as 3/2 and -0.25 as -1/4)

\left(x-\dfrac{3}{2}\right)\left(x+\dfrac{1}{4}\right) = x^2-\dfrac{5}{4}x-\dfrac{3}{8}

If we don't want decimal coefficient, we can multiply the whole expression by 8:

8x^2-10x-3

3 0
3 years ago
How do you write 4540 million in standard form?
TEA [102]
4,540,000,000 here you go this is in stdrd form
7 0
3 years ago
Read 2 more answers
The complement of 17% is
ikadub [295]
38% :)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))
6 0
3 years ago
Read 2 more answers
3.- In a certain desert region the average number of persons who become seriously ill each year from eating a certain poisonous
Naily [24]

Answer:

P(X\geq 5)=1-P(X

Using the pmf we can find the individual probabilities like this:

P(X=0)=\frac{e^{-6.4} 6.4^0}{0!}=0.001662

P(X=1)=\frac{e^{-6.4} 6.4^1}{1!}=0.010634

P(X=2)=\frac{e^{-6.4} 6.4^2}{2!}=0.034029

P(X=3)=\frac{e^{-6.4} 6.4^3}{3!}=0.072595

P(X=4)=\frac{e^{-6.4} 6.4^4}{4!}=0.116151

And replacing we got:

P(X \geq 5) =0.76493

Step-by-step explanation:

Previous concepts

Let X the random variable that represent the number of people that will become sereiosly ill in two years. We know that X \sim Poisson(\lambda)

The probability mass function for the random variable is given by:

f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...

And f(x)=0 for other case.

For this case the value for \lambda would be:

\lambda = 3.2 \frac{ills}{year} *2 years = 6.4

For this distribution the expected value is the same parameter \lambda

E(X)=\mu =\lambda

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

P(X\geq 5)=1-P(X

Using the pmf we can find the individual probabilities like this:

P(X=0)=\frac{e^{-6.4} 6.4^0}{0!}=0.001662

P(X=1)=\frac{e^{-6.4} 6.4^1}{1!}=0.010634

P(X=2)=\frac{e^{-6.4} 6.4^2}{2!}=0.034029

P(X=3)=\frac{e^{-6.4} 6.4^3}{3!}=0.072595

P(X=4)=\frac{e^{-6.4} 6.4^4}{4!}=0.116151

And replacing we got:

P(X \geq 5) =0.76493

7 0
3 years ago
Read 2 more answers
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