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3 years ago

I need help with this question

Mathematics

2 answers:

Amiraneli [1.4K]3 years ago

7 0

Answer:

what is the question be more specific

Vika [28.1K]3 years ago

7 0

Download lightshot or Press the Windows key + Shift + S or Press the Windows key + PrtScn.

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“HOT PIZZA” is offering a 20% discount on all its pizzas

Anit [1.1K]

Answer:

p = £7.00

Step-by-step explanation:

Represent the original price by p. Then 0.80p = £5.60, or p = £7.00.

If a price is discounted by 20%, the discounted price is (1 - 0.20), or 0.80, times the original price.

6 0

3 years ago

Rewrite the equation in exponential form: log4(q) = m

Vesnalui [34]

log (2x)=3 I don't remember how I got my awn see

3 0

3 years ago

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

3 years ago

This is a list of names randomly selected to form the jury pool.

meriva

The term for the list of names randomly selected to form the jury pool is venire.

Venire is a writ issued by a judge to a sheriff directing the summons of prospective jurors. Also called venire facias . The panel of prospective jurors from which a jury is selected.

7 0

3 years ago

HELP DUE IN 10 MINS! Attached is the Question.

Nikitich [7]

Answer:

∠N ≅ ∠J - given
LN ≅ LJ - given
∠L ≅ ∠L - reflexive property

Postulate: ASA

Congruency Statement: ΔJML ≅ ΔNKL

7 0

3 years ago