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3 years ago

Please help thank you

Mathematics

1 answer:

Aleks04 [339]3 years ago

3 0

Answer:

The Angle-Side-Angle Postulate (ASA) states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent.

Step-by-step explanation:

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You have a coin that is not weighted evenly and therefore is not a fair coin. Assume the true probability of getting heads when

Alexandra [31]

Answer:

$X \sim Binom(n=157, p=0.52)$

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:

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

And we want this probability:

$P(X$

And we can use the following Excel code to find the exact answer:

"=BINOM.DIST(75,157,0.52,TRUE)"

And we got 0.1633

The other way to solve the problem is using the normal approximation

We need to check the conditions in order to use the normal approximation.

$np=157*0.52=81.64 \geq 10$

$n(1-p)=157*(1-0.52)=75.36 \geq 10$

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

$E(X)=np=157*0.52=81.64$

$\sigma=\sqrt{np(1-p)}=\sqrt{157*0.52(1-0.52)}=6.26$

We want this probability:

$P(X$

And using the continuity correction we have this:

$P(X$

We can use the z score given by this formula $Z=\frac{x-\mu}{\sigma}$ .

$P(X< 76.5)=P(\frac{X-\mu}{\sigma}< \frac{76.5-81.64}{6.26})=P(Z < -0.821)=0.206$

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".

Solution to the problem

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

$X \sim Binom(n=157, p=0.52)$

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:

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

And we want this probability:

$P(X$

And we can use the following Excel code to find the exact answer:

"=BINOM.DIST(75,157,0.52,TRUE)"

And we got 0.1633

The other way to solve the problem is using the normal approximation

We need to check the conditions in order to use the normal approximation.

$np=157*0.52=81.64 \geq 10$

$n(1-p)=157*(1-0.52)=75.36 \geq 10$

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

$E(X)=np=157*0.52=81.64$

$\sigma=\sqrt{np(1-p)}=\sqrt{157*0.52(1-0.52)}=6.26$

We want this probability:

$P(X$

And using the continuity correction we have this:

$P(X$

We can use the z score given by this formula $Z=\frac{x-\mu}{\sigma}$ .

$P(X< 76.5)=P(\frac{X-\mu}{\sigma}< \frac{76.5-81.64}{6.26})=P(Z < -0.821)=0.206$

4 0

4 years ago

Find the equation of the tangent line and normal line to the curve at a given point y=x^4+2e^x

Oliga [24]

The slope of the line is given by the derivative

m = dy/dx = 4x^3 + 2e^x

using the point slope form of a straight line
y - y1 = m(x - x1) where m is the above and (x1,y1) is the point of contact of the tangent we have

y - y1 = (4x^3 + 2e^x)(x - x1) which is equation of the tangent

4 0

3 years ago

What integer best represents that the temperature outside increased by 12°F?

fgiga [73]

Answer:

-11.11*C

Step-by-step explanation:

12°F − 32) × 5/9 = -11.11*C

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3 years ago

In the following pie chart, if the monthly savings of the family is $3000, what is the monthly expenditure on clothes?

sladkih [1.3K]

10% of 3000 is 300.
Theirs monthly expense on clothes is $300.

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3 years ago

Rewrite 1/4 book 2/5 hour as a unit rate.

4vir4ik [10]

Answer:

I think it is c thhgfghiss

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