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3 years ago

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Joakim is icing 30 cupcakes. He spreads mint icing on 1/5 of the cupcakes and chocolate on 1/2 of the remaining cupcakes. The re

st will get vanilla icing. How many cupcakes have vanilla icing?

2 answers:

yanalaym [24]3 years ago

7 0

Answer: 12 cupcakes have vanilla icing.

Step-by-step explanation:

We know that Joakim is icing a total number of 30 cupcakes.

According to the exercise, Joakim spreads mint icing on $\frac{1}{5}$ of the cupcakes.

Let be "m" the number of cupcakes that have mint icing. Then

$m=(30)(\frac{1}{5})=6$

He spreads chocolate on $\frac{1}{2}$ of the remaining. The remaining is:

$30\ cupcakes-6\ cupcakes=24\ cupcakes$

Let be "c" the number of cupcakes that have chocolate. Then

$c=(24)(\frac{1}{2})=12$

Let be "v" the number of cupcakes that have vanilla icing.

Since the rest will get vanilla icing, we know that:

$v=12$

Sauron [17]3 years ago

7 0

Answer:

9 cupcakes.

Step-by-step explanation:

We have been given that Joakim is icing 30 cupcakes. He spreads mint icing on 1/5 of the cupcakes.

Let us find the number of cupcakes with mint icing.

$\text{Cupcakes with mint icing}=30\times \frac{1}{5}$

$\text{Cupcakes with mint icing}=6$

We have been given that 1/2 of the remaining cupcakes have chocolate icing.

$\text{Cupcakes with chocolate icing}=\frac{30}{2}$

$\text{Cupcakes with chocolate icing}=15$

We have been given that remaining cakes have vanilla icing.

$\text{Cupcakes with vanilla icing}=30-(6+15)$

$\text{Cupcakes with vanilla icing}=30-21$

$\text{Cupcakes with vanilla icing}=9$

Therefore, 9 cupcakes will have vanilla icing.

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Answer:

reflection across BC
the image of a vertex will coincide with its corresponding vertex
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Step-by-step explanation:

We want to identify a rigid transformation that maps congruent triangles to one-another, to explain the coincidence of corresponding parts, and to identify the theorems that show congruence.

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<h3>1.</h3>

Triangles GBC and ABC share side BC. Whatever rigid transformation we use will leave segment BC invariant. Translation and rotation do not do that. The only possible transformation that will leave BC invariant is <em>reflection across line BC</em>.

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<h3>2.</h3>

In part 3, we show ∆GBC ≅ ∆ABC. That means vertices A and G are corresponding vertices. When we map the congruent figures onto each other, <em>corresponding parts are coincident</em>. That is, vertex G' (the image of vertex G) will coincide with vertex A.

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<h3>3.</h3>

The markings on the figure show the corresponding parts to be ...

side AB and side GB
side AC and side GC
angle ABC and angle GBC
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And the reflexive property of congruence tells us BC corresponds to itself:

side BC and side BC

There are four available congruence theorems applicable to triangles that are not right triangles

SSS -- three pairs of corresponding sides
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SSS

Corresponding sides are listed above. Here, we list them again:

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2 years ago

Read 2 more answers

26.48 divided by 2 and the method

e-lub [12.9K]

Answer:

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Step-by-step explanation:

If you put it into the calculator, you get the answer. Or you could do long division, but that's difficult for no reason. You could also divide each number individually by 2, as they are all even, so that works too.

Hope this helps!

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3 years ago

Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a

Ivan

Answer:

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At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

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$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

$P(X$

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}$

The cumulative distribution for this function is given by:

$F(X) = 1- e^{-\lambda x}, x\ geq 0$

We know the value for the mean on this case we have that :

$mean = \frac{1}{\lambda}$

$\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367$

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: $Var(X) =\frac{1}{\lambda^2}$

And the deviation would be:

$Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725$

And the mean is given by $Mean = 2.725$

Two deviations correspond to 5.540, so we want this probability:

$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

$P(X$

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4 years ago

The number of square feet per house is normally distributed with a population standard deviation of 154 square feet and an unkno

11111nata11111 [884]

Answer: (1500.66,1599.34)

Step-by-step explanation:

Confidence interval for population mean:

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As per given , we have

n= 16

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Required confidence interval: (1500.66,1599.34)

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3 years ago

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