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2 years ago

8

A bakery sold a total of 400 cupcakes in a day, and 144 of them were vanilla flavored. What percentage of cupcakes sold that day

were vanilla flavored?

1 answer:

Lina20 [59]2 years ago

4 0

Answer:

i think it should be 36%

Step-by-step explanation:

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Christine basked a pie.she ate one sixth of the pie. Her brother ate one third of the pie and gave the leftovers to his friends

Sergeeva-Olga [200]

Answer:

Her brother gave his friends one half the pie.

Step-by-step explanation:

The whole pie is equal to 6/6, because that's equal to one. 1/3 = 1/6. 6/6 - 2/6 = 5/6. 5/6 - 2/6 = 3/6. 3/6 = 1/2.

8 0

3 years ago

Question #19- Use the answer bank below to fill in missing parts of the proof.

serious [3.7K]

Answer:

∠MPQ ≅ ∠MPR: Reason; Corresponding parts of congruent triangles are congruent (CPCTC)

∠PQR ≅∠PRQ: Reason; CPCTC

Step-by-step explanation:

$\overline{PQ}\cong \overline{PR}$ : Reason; Given

Draw $\overline{PM}$ so that M is the midpoint of $\overline{QR}$ : Reason; Two points determine a line

$\overline{QM}\cong \overline{RM}$ : Reason; Definition of midpoint

$\overline{PM}\cong \overline{PM}$ : Reason; Reflexive property

ΔPQM ≅ ΔPRM: Reason; Side Side Side (SSS) rule for triangle congruency

∠MPQ ≅ ∠MPR: Corresponding parts of congruent triangles are congruent CPCTC

∠PQR ≅∠PRQ: CPCTC

5 0

3 years ago

What percent is of 80 is 15?

pashok25 [27]

Answer:

The answer is 18.75

4 0

3 years ago

While mining, Jason found a large metal bar that weighed 24 ounces. Jason was also able to determine that the bar had 6 ounces o

tino4ka555 [31]

25% 6 divided by 24 will give you the percentage of copper in the bar

5 0

3 years ago

Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie

ludmilkaskok [199]

Answer:

$\lambda \geq 6.63835$

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. 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 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:

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

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

$P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}$

$P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}$

And replacing we have this:

$P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]$

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)$

And we want this probability that at least of 99%, so we can set upt the following inequality:

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99$

And now we can solve for $\lambda$

$0.01 \geq e^{-\lambda}(1+\lambda)$

Applying natural log on both sides we have:

$ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)$

$ln(0.01) \geq -\lambda+ln(1+\lambda)$

$\lambda-ln(1+\lambda)+ln(0.01) \geq 0$

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

$x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}$

Where :

$f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)$

$f'(x_n)=1-\frac{1}{1+\lambda}$

Iterating as shown on the figure attached we find a final solution given by:

$\lambda \geq 6.63835$

4 0

2 years ago